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Bab 4

Pengumpulan dan pengolahan data

4.1. Pengumpulan data4.1.1. Data anthropometri 

4.1.1.1. Data anthropometri  posisi berdiri

Untuk membuat suatu produk yang ergonomis kita memerlukan dimensi tubuh yang sesuai dengandimensi produk yang akan dibuat. Data anthropometri  posisi berdiri sangat berguna untuk 

membuat produk yang sesuai saat kita berdiri.a) Data tinggi badan berdiri

Tabel 3.4.1. Data anthropometri TBB

  1 2 3 4 5 6 7 8 9 10

1 154,0 1!,4 151, 15," 1"#, 1"1," 14!,1 14,! 1",0 15#,1

21"$,# 1#,1 1"!,$ 1"5, 15#,$ 1"!,4 1"!,! 10,0 155,1 154,3

3 1",# 1$,3 1!0," 1!1,1 1"3,$ 1"0," 1$,0 1!0,3 1"3,0 154,!

4 1$, 1#,5 1!1,3 1!3,$ 1"4,4 1!3,3 1$,4 1!0,1 1"0,! 1"0,3

5 10,5 1"5,0 153,0 13,3 151,4 14!,# 14$,$ 1"$,5 14,0 1"5,!

6 14",1 15,0 15$,# 155,1 1"#,$ 1!3,5 154, 154," 15!,! 1#,!

7 1"5,5 151," 10,$ 1"4, 15!,1 1#," 153, 15#,5 155, 15,1

8 1!#,! 1!,4 1"3,5 1!0," 1"3,3 1"4,0 15$,4 1"$,4 11,# 14",$

9 1"4,4 1#,4 150, 1"3,0 15,# 10," 1!3,5 1!3,# 1!3,4 1!0,3

10 10,# 1$,$ 1!0,$ 1!1,5 151,0 151,0 1"0,! 1"1,$ 1"!,0 15,"

 b) Data tinggi mata berdiriTabel 3.4.#. Data anthropometri T%B

  1 2 3 4 5 6 7 8 9 10

1 11,4 1#,4 13!,5 1#4,! 150,! 11,# 143,$ 13!,$ 1$,! 14$,

2 134,# 1##,0 1$,4 14$," 13,0 1#",$ 1#4,! 14$,! 1#!,4 1#,"

3 1#5,1 14!," 14", 15!,5 15," 133,$ 1#3,0 1#",4 15,# 1"1,3

4 1#"," 10," 15#,3 1#,4 153,# 10, 13,0 10,4 15!," 13,0

5 15, 1#!,$ 1#",1 15,# 150,! 131,5 143, 14,$ 13",# 153,3

6 131,1 14!,$ 1#$,$ 1!," 141,3 134,3 130," 10," 13,5 154,1

7 134,! 134,$ 1#,4 14#, 130,4 131,# 14,$ 133,3 150,! 14#,4

8 134,4 1#$,$ 133,5 14#,$ 13$,1 1#3,3 1"1,3 15,1 143,1 15$,$

9 10,# 150, 15#,! 131,0 130,5 153,0 1#,0 1#4,5 1, 13!,$

10 11,4 14!,1 14,0 143," 1#",$ 13,0 13!,4 150,3 13#, 134,!

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&) Data tinggi bahu berdiri

Tabel 3.4.3. Data anthropometri T'B

1 2 3 4 5 6 7 8 9 10

1 15#." 14#.3 14!.# 140.5 13!." 14$ 14#.3 1#$. 13.5 13".#

2 143.3 133." 135.! 14.3 1#5 14$." 150.5 13!.3 13.4 151.4

3 153.5 150.5 153.# 13#.! 130 130.4 151.1 1#5.# 143." 145.

4 143.4 13.4 14#. 145.3 150.1 1#4 1#!.1 135 13.$ 134.#

5 141.$ 1#5.$ 1##.! 1#!.4 144." 150." 135 1#5.5 143.# 14.#

6 1#5. 135 154.# 130.4 143.4 1#4.$ 145.5 150." 1#.1 1#4."

7 140." 13.4 143.$ 14! 1##. 15#.5 154.$ 155 133.# 13!

8133.! 14. 13$.# 133 153." 1##.1 154.3 131." 13$.3 1##.#

9 130.1 13!.5 14".1 1#4 145 14.# 151.$ 1#$.$ 133.4 143.

10 130.4 154.$ 145 13$.3 1#!.3 153.4 14.! 13".5 14!." 13.4

d) Data tinggi (angkauan tangan

Tabel 3.4.4. Data anthropometri TT

 1 2 3 4 5 6 7 8 9 10

1 #04.3 #0!." ##$.! #1.5 #13. ##5.$ #04. #3!.1 1$!.4 ##4.!

2 1$. ##5.$ ###.4 #31.! #04.# 1$1.1 ###." #10. 1$4 1$."

3 1$4.5 #10.5 ##3.4 #0$. #13 ##.1 ##0.# 1$1.$ #1$.$ 1$#.1

4 #01." ##5." 1!!.3 ##".5 #01.1 #30.1 #04.# ##4.! #40.# 1!$.$

5 #01." #35. 1$!.5 1$3. #3$.4 ##!.# ##. #35.1 #03." #0.!

6 ##1.5 #3!." #41.5 ##! #11." ##1.! 1$. #11. #30.# 1!.!

7 #3." #33.4 #0". ###.# 1!".1 1$3. 1$0.! ##0.$ #05.# 1$4.

8 #1!." 1$#. #40. #1!.4 #34 #0$.4 #04." #11." ##4.5 ##$.#

9 1$!.! 1$!.# #11.# #0 1$".5 ##!.$ #03.5 #3.$ 1!.5 #1

10 1!".! #1!.3 ##".3 ##5.5 #3!.# 1!".4 ##0.3 1$!.1 #04.1 1!.$

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e) Data (angkauan tangan

Tabel 3.4.5. Data anthropometri T

 1 2 3 4 5 6 7 8 9 10

1 "" ,$ $,3 "4,0 4,3 5," ",4 !,$ $,3 "4,$ !0,!

2 5, 5,0 3,0 "#,! 3," 5,4 "5,3 !0,# ", !0,0

3 "0,0 3," "1,0 "4,4 "!,3 "!,4 "!," ""," "$,0 $,$

4 "4,4 "1,$ "5, "3," ",# ,0 3,4 3,# "!,# "5,!

5 "3," 1, "0," !0,1 ",5 $,0 "5,0 !1,0 ,3 3,3

6 "!,1 1," "5,$ "",! 1,$ #,3 "1,! "1,3 !,! ",5

7 $,1 $," "0," ",5 "1,0 $,3 #,3 "",# "4,0 !,!

8 !, "4,! "4,4 "1, #,3 "#,! ",5 ",5 "3,0 "$,!

9 "",5 ",$ ,# $," "3, "0,1 "!," 1,4 "0,$ 5,$

10 #,$ ",1 "#,5 !0," $,0 ,1 "1,5 !0,0 5,$ #,#

*) Data rentang tanganTabel 3.4.. Data anthropometri +T

  1 2 3 4 5 6 7 8 9 10

1 1!5," 153, 1!," 11," 14$,# 1$#,4 1", 15!,1 150,$ 14,"

2 1!",1 15$," 1"0,4 10,4 10,0 153,# 154," 1!1,$ 11,0 1$0,$

3 1"!,0 1$3,! 1!",! 1!4,$ 1!5,0 1"#,4 1!4,$ 1"#,5 15",3 1$,"

4 1!#, 154,1 1$#,$ 1$4,0 1"3,3 1",1 1, 1"4,0 1"$,# 15","

5 1"4,3 10,# 1$0,4 10,$ 1$#,5 15$," 1"4, 1!,! 15," 1"1,$

6 1"3," 153,$ 14",$ 1"3,$ 1!3,0 15,4 1"", 11,1 1",$ 1$1,

7 14",# 14!,$ 1!3,4 1"5,# 155,1 14!, 1!, 1!",! 154,0 1",5

8 1"$,0 15,0 1," 15$,$ 1!$,5 13,5 1!1,0 1"3,0 1$#,$ 1!3,4

9 13,4 1$3,5 1"," 1!$,3 1!0,4 1",0 1!0,# 1!3,0 1!,3 15$,0

10 1"!,1 1!4,3 1!0," 10,3 1!1,1 1",! 1$3,4 15$,! 1$0,1 1$#,0

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g) Data tinggi siku berdiri

Tabel 3.4.". Data anthropometri TB

 1 2 3 4 5 6 7 8 9 10

1 10$,# 110,5 11#,# 10, 111 105,4 $,4 $" 10",4 11#

2 100,4 101,3 103,$ 110,4 11#,4 10#, 10#," 10",! 10,4 $!,4

3 10$,3 10!,5 10," 11#,4 $,4 $",# 101,! 103,! 110,4 105

4 10$,# 10,! $$," 103," $,$ 111 10#,# $$,1 10#,$ 10",3

5 103,! 10!,1 10!,! 10#," 10!, 10! 105,$ 110 10!,3 $",4

6 10,4 110,$ 101,4 101,5 110,# 10!,4 10," 101," 105,$ 101

7 $",4 10",$ 104,! 100,! $" 10!," 10$," 110,! 111,1 105,5

8 101,$ 104, 10$,3 11#,5 $",# 10," $$," 10#, 10! 10#,#

9 $, 104 10,# 11#,# 110," 105,5 $",! 101,# $!,1 10$,"

10 104,4 10"," 104," 11#,4 $",! 104,5 10",3 10",1 $$,$ $,4

h) Data tebal badanTebal 3.4.!. Data anthropometri TB

 1 2 3 4 5 6 7 8 9 10

1 44,3 43," 51,1 54,1 5#, 40,1 50,! 54,0 4$,0 41,5

24#,3 50,! 43, 40,1 41,$ 4,3 4,# 54,1 4!,4 4$,5

3 3$,4 4",3 4",! 53,1 3$,5 5#,3 41,0 3$, 51,$ 53,!

4 4!,3 40,3 5#,1 43,5 50,! 45,$ 53,4 4#,4 4",3 43,1

5 50," 51,3 40, 3$,$ 53,0 41, 4", 4!,1 40,$ 41,4

6 4!,1 4#,3 40, 53,3 3$, 44," 50," 41,0 44,4 40,#

7 44," 43,$ 43,# 4,4 44,$ 41,3 4!," 40,0 4",0 51,3

8 44,0 4!," 5#,0 44,1 45,3 54,0 4!,! 51,1 44,1 4,4

9 4,# 51,5 4", 4#,3 43,3 4#,! 53," 51,! 4#,3 40,5

10 4$,1 4",0 43,5 50,4 54,4 53,1 51,# 45,1 4",$ 51,1

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i) Data pan(ang lengan ba-ah

Tebal 3.4.$. Data anthropometri /B

 1 2 3 4 5 6 7 8 9 10

1 #5,3 #"," #5,5 #4 #",1 #!,$ #", #!," #,1 #4,

2 #,$ #$,5 #4,1 #4,3 #$, #,! #4,1 #",! #",1 #",1

3 #$,1 #4,5 #,1 #4,4 #$,3 #4 #",! #5,# #!,! #5

4 #,! #",5 #!,# #4,4 #$ #5,1 #",! #5, #," #,$

5 #4,$ #$,# #5," #,# #!,$ #5,# #!,5 #!,$ #$,4 #,4

6 #5 #! #,4 #5 #5,! #5,# #5,# #4,4 #$,1 #5,#

7 #,$ #5,! #,$ #!,$ #",1 #$," #$,1 #, #!,5 #4,4

8 #",1 #!,$ #4,5 #!,! # #4,5 #",$ #4 #",4 #5,5

9 #!,! #, #",# #" #4,5 #4, #$ #!,# #!,# #!,3

10 #", #4,4 #!,3 #!,! #,3 #", #! #4,3 #4,3 #4,!

4.1.1.2. Data anthropometri  posisi dudu 

Data anthropometri posisi duduk berhubungan dengan perhitungan dimensi tubuh manusia dalam posisi duduk. Dengan data ini kita dapat membuat produk yang sesuai pada saat tubuh manusia

sedang duduk, dan bisa membuat produk lebih nyaman.a) Data tinggi duduk tegak 

Tabel 3.4.10. Data anthropometri TDT

1 2 3 4 5 6 7 8 9 10

1 !0,0 "!,3 $0,1 !3,1 !3,# !, !1,# "!,# !#,0 !",#

2 !#,$ !4,1 "$,1 $0,# $1," !"," "!,4 $1,$ "!,# !#,!

3 !0,4 "$,0 !4,$ !5,! !1,3 !3,1 !!,# $0,4 "!,5 !#,3

4 "!,# !!,$ "$,! !!,4 !, "$," $1,# !3,1 $#,0 !0,#

5 "!,! !,3 !,4 "$,$ !",# $0,3 !$,5 !1, !1," $0,3

6 !$,# !0, !3, !",1 "!,0 "$,! !0,3 !3,# !1,# !!,0

7 $1,1 $1,$ !5,5 !",0 "!,0 !3,# !3,5 $0, !#,# $1,

8 !0,1 !",0 !!,# "$,4 $1,# !5,$ $1, !4,1 !1, !,0

9 !", !!,$ $1,3 !3,$ !4,5 !4,3 !#,! !5,0 "!,$ $1,#

10 $0, !#,# !1,# !0,1 !,3 !,5 "$,$ !3,3 !,5 !#,!

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 b) Data tinggi mata duduk 

Tabel 3.4.11. Data anthropometri T%D

1 2 3 4 5 6 7 8 9 10

1 "#,5 !#,3 $,0 !#,0 "4,5 !,0 "$, ",! "",1 !#,3

2 "!,1 !, !#,! !,4 $,$ "5,5 !1,! "3,$ !3," ",1

3 !0," !, "1,$ !,4 !1,0 "3,4 "5,0 ," ,$ !3,0

4 "5,3 !3,0 "4, ",3 "3,5 !1,# "!,1 "4,! "5,0 !#,4

5 "5,$ ",3 !#,# !#,0 "4,1 ,3 "4,4 $,4 "1, $,4

6 !, "#, "5,0 "3,4 !#,5 $,3 !0,# ",$ "0,0 "$,0

7 $, ",# "4,! ",5 !0, !#,4 "#,3 $,# !3,1 !#,1

8"$,1 "5," !1,! !0,! "#,5 "3,5 "0,# !#,4 $,1 "1,3

9 "," "1,$ !1,# "!,3 !3,1 "5,! ",1 !0,3 "1,1 "#,"

10 !#," "4,3 !0,! "0,1 "0,3 !0,1 $,5 "$,! "#,# "!,#

&) Data tinggi bahu duduk 

Tabel 3.4.1#. Data anthropometri TBD

1 2 3 4 5 6 7 8 9 10

1 0,$ 55,3 1,$ #,1 55,1 , ",4 "," 5$,! 5$,5

2

4, " 1,! 4 1 ,! 3,# #,# #,# 4,#

3 "," 1,1 1,4 53,3 5",5 3," 3,4 5,! 53," 3

4 #, ! ,# #,# 53 5",! 5,5 5$,! 4,! 0,"

5 , ,# 3,# 4,$ #," 4,3 53, ",3 53,5 1

6 5,$ 5!," 5$,5 54," 5,4 4,5 ,5 1,# 4 !,3

7 5!," !,3 5,5 !,1 5", 53,3 0,3 53 55,! 0,

8 !,5 !, 5$,$ !,5 5",5 !,! 1,1 5",1 # 53,$

9 3,5 5,1 4,5 !,4 55,$ 0,# 3 54,4 !,!

10 ",1 0," 5,# 53, 5!,5 ,! ,1 0, 0,5 55

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d) Data lebar bahu

Tabel 3.4.13. Data anthropometri /B

1 2 3 4 5 6 7 8 9 10

1 40, 41,4 3$, 40, 4,# 3!,3 35,4 44," 3",! 3,5

2 45,1 3",# 3,3 3,# 40,5 3,0 3!,! 44,1 41,! 45,0

3 44,4 3$,# 41," 3!,1 4#,1 3!, 35,! 34, 41,! 34,$

4 3!,3 41,1 3,1 3!,5 3$,1 3,4 3",# 34,0 41,3 3!,"

5 34,$ 3,1 41,4 4#, 45,3 3$,$ 3,1 40,! 35,3 45,0

6 3,4 3!,! 3", 35, 40,# 44,3 44,5 3!," 3",! 3$,3

7 45,0 44," 3$,4 3,! 40,0 40,# 44,! 34,! 40,! 35,0

8 3!,! 34,1 43,1 3$,5 40," 3,# 34," 4#,$ 3!,0 35,3

9 4,# 34,! 4,0 3,$ 40," 3,# 4#," 43,5 41," 3",5

10 35,# 35,1 3",1 4#,5 4,0 4#,# 3$,4 45,! 44,5 3",5

e) Data lebar sandaranTabel 3.4.14. Data anthropometri /

1 2 3 4 5 6 7 8 9 10

1 #5,$ ##,# #4,4 #4," #1,0 1!," #0,4 #,0 #0," 1$,#

2 1!,5 #0,! #1,$ #5,$ ##,3 ##," #5," 1$, #1,! #4,3

3 ##,5 #3,5 #4, #0,0 ##,5 #1,5 #3," #3, ##,4 #4,#

4 #5,5 #3, ##,3 #0,4 #3,5 #3,0 #5, #5," #5, #1,0

5 1!,5 ##,! #4,4 #5,5 #5,5 ##,4 ##,0 #0,5 ##,0 #3,0

6 #5,5 #4,! ##,3 #3,5 #3," #4,0 #3,3 #0,0 #4," #4,1

7 1!,4 ##,0 ##, #0,5 ##,0 ##,0 #1,# 1$,$ #1,3 #3,

8 1$,5 #0,$ #1,! #0,# #3,# 1!,! #3,5 #3, 1!," #5,3

9 1$,1 1!,! 1$,! 1$,4 #4,$ #1,4 #0,5 1!, #1,$ #3,$

10 #0," #1, #3,1 ##,! #3,# #3,# #4," #5,! #0,5 #4,1

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*) Data siku ke siku

Tabel 3.4.15. Data anthropometri 

1 2 3 4 5 6 7 8 9 10

1 33,$ 3,! 45,3 43,3 34,3 4",1 44,1 3$,1 3#," 3#,!

2 3,$ 35,3 41," 3#,1 41, 43,! 4#,$ 41,5 3#,$ 34,#

3 3$,5 33," 45,3 43,4 3$,$ 34,$ 45, 34,$ 44," 4#,4

4 34,$ 4#," 3$,4 40,0 4", 45,5 45,# 4,# 45," 4,0

5 3#,3 3!,5 3$," 44,1 4,5 34,! 4,0 3!, 34,4 33,!

6 45, 44,# 3$,4 40, 45,3 35," 3#,$ 33,$ 40,0 3",!

7 35,5 43,3 34, 44,3 43,4 40,0 44,0 4,1 3#,# 4",!

8 31,5 45,3 3!,0 34,1 3,0 3!,$ 4#,1 44,1 40, 31,5

9 41,$ 45,3 4",1 31, 45,4 4#,3 4#,4 33, 45,4 33,"

10 41,5 3$,0 45,0 4,# 3#, 35,! 34,$ 4#, 3#,! 31,#

g) Data lebar pinggulTabel 3.4.1. Data anthropometri /

1 2 3 4 5 6 7 8 9 10

1 40, 41,$ 3#, #, 34," 43,0 #!, 34,5 40," #!,

2 3, 3#,$ #,4 3!,5 #$,3 43,1 #",$ 3!,4 #!,0 30,"

3 4#,1 #5,! 3", 33, 35, 30,4 #!,$ 30," 3#,3 35,1

4 3#," 33,# #",0 40,4 30,! 31,5 41,3 #",# 30, #5,$

5 #5,3 41,5 3$,1 30,# #,5 34,3 30,5 30,4 34,4 #!,4

6 34,$ 4#,# 31,4 #!,1 #,0 #!, 35,# 43,5 33,! 34,5

7 3$,1 3,1 #$," 3"," 3#," 3!," 34,1 #!,1 #$,$ 41,#

8 40,5 4#,! 43,! 3!,5 35,5 3!,4 34," 3!,0 3#,0 3!,1

9 3",! 41,1 31, #,4 35,4 #$,$ 33,# 3",5 3$, 35,1

10 41,$ 33,4 43," 41,! #,! #",1 4#,5 #4," 3$,0 3",0

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h) Data tinggi sandaran

Tabel 3.4.1". Data anthropometri T

1 2 3 4 5 6 7 8 9 10

1 44,# 44,5 4,5 50,0 4,0 4",1 43,3 4!,0 3$,0 3!,0

2 4,4 3!,0 5#,5 4$,5 43,0 44,3 40,5 41,0 4,5 4",0

3 4,0 4",1 43,3 4!,0 3$,0 3!,0 50,5 4!,5 4#,! 41,0

4 43,0 44,3 40,5 41,0 4,5 4",0 4#,0 40,5 4!,0 4,5

5 3$,0 3!,0 50,5 4!,5 4#,! 41,0 41,0 4,0 4#, 53,0

6 4,5 4",0 4#,0 40,5 4!,0 4,5 41,5 41,0 44,0 4,0

7 4#,! 41,0 41,0 4,0 4#, 53,0 4,0 44,# 44,5 4,5

8 4!,0 4,5 41,5 41,0 44,0 4,0 44,5 4,4 3!,0 5#,5

9 4#, 53,0 4,0 44,# 44,5 4,5 50,0 4,0 4",1 43,3

10 44,0 4,0 44,5 4,4 3!,0 5#,5 4$,5 43,0 44,3 40,5

i) Data tinggi pinggang

Tabel 3.4.1!. Data anthropometri T

1 2 3 4 5 6 7 8 9 10

1 1!,4 #0,# #0,0 #0,4 1!,4 #0,# 1",0 ##,! 1!,4 ##,

2 1$,# 1!,$ #1,! 1$,! #0,! #3,3 #5,0 #4,# #4,5 1!,1

3 1$,1 1!, 1!,0 #0,3 #1,0 #1, 1$," 1",5 #0,# 1!,3

4 #1,1 #0,# ##,0 #0,$ #1,$ #1, #4,0 ##,$ #0,! #1,4

5 1",1 #3,1 #4,4 1",5 #0,3 #0,# #3,5 1"," #1,! #1,3

6 1$,$ #0,4 1"," 1!,3 ##,4 #3," #0,$ #1,4 #0,# 1$,!

7 ##,$ 1$,3 1$, #0,$ 1",5 1!,4 1",$ #0,3 1",4 #4

8 #3," #1," 1$,3 1!,4 #3,$ #0,3 1$,# 1!,$ ##,$ #4,

9 1$,0 ##,$ #4,$ #4,1 1$,4 #0, #3,5 #4,5 #3,$ 1!,"

10 ##,1 ##,# #4,$ #0,5 1!,5 #1,4 #0,4 #0,3 #0, ##,!

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 () Data pan(ang sandaran

Tabel 3.4.1$. Data anthropometri 

1 2 3 4 5 6 7 8 9 10

1 #"," #4,0 #4,# #5,$ ##,0 #," #1," #$,! #1," #1,#

2 #, ##,$ ##,0 3#,# #4,3 #!,3 31, 31," 30,! #",#

3 3#,5 30,0 #",# ##,4 #$, #!,1 #,! 3#,5 #,! 31,0

4 #3,4 #",4 31,$ #1,! #5,0 #,5 #3,# #!, 3#,4 #",

5 #",4 #!,0 #,3 #4,5 #,5 31,# 3#,4 #$,0 #$,3 #3,#

6 #4,4 #$,$ 30,$ #3," #,3 #!,5 31," #4,1 #5, #5,"

7 #$,5 #,0 #4, #!,# #4,5 #!, #$,3 #1,! #3," #4,3

8 #3,5 #5,1 31,0 #1,5 31,4 31,0 #,4 #",# #",3 #,#

9 #4,0 #,4 #3,! #5,0 #3,1 #!,$ 3#,4 #1,0 #4,3 #$,1

10 30,3 31,0 3#,1 #", #,4 #3,# ##," ##,3 30, #5,3

k) Data tinggi siku dudukTabel 3.4.#0. Data anthropometri TD

1 2 3 4 5 6 7 8 9 10

1 1!,! 1",# #," #3,! #4, 1$,# ##,$ #0,3 ##,3 #1,"

2 #,1 ##,5 #5," ##," 1",4 ##,0 ##,$ ##,3 1$, ##,1

3 #",0 #,0 #3,1 1$,4 #,1 #1," #1,! #3,$ 1",$ #0,5

4 1!,# #4,3 #4,0 1,5 #,0 #5, #1,$ #1,# #1,$ #3,4

5 #5,4 #5,0 ##,0 1",# 1$,! #3,! #,# #,3 1!,# #4,$

6 #0,3 #5,$ 1$," ##,0 1$,# 1, ##,4 #0,3 #4,# #1,5

7 #4,# #1,3 #5,5 #4,! ##,5 1",1 #1," #5," ##," #3,

8 #4, #1, #0,4 #4,# #4," 1$,1 #, #0,$ 1!,# #4,4

9 #,# 1"," 1!,$ #0,$ 1",# 1!,1 1$,4 ##," #",0 #0,$

10 1," 1!, #0,3 1$, #0,# #0,$ #4,$ #3,5 #0,1 ##,

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l) Data tebal paha

Tabel 3.4.#1. Data anthropometri T'

1 2 3 4 5 6 7 8 9 10

1 #1,! #$,5 #5,$ #",! #1,1 #3,# #$,1 #$,$ #",5 #,!

2 #,# #,1 #!,3 ##,$ #3,5 #!,0 #$," #1,$ 1$,! #3,0

3 1$, #1,4 #3,$ ##,1 #0,0 #1,5 #$,1 #,3 1$,1 #",1

4 #4,4 #!,4 #3,0 #3,3 #5,1 ##,5 #3,# #0,5 #0,0 #",3

5 #0,$ #4, #5,5 #0,1 #3,# ##,0 #0,0 #1,$ #,5 #,!

6 #," #0, #5," #!,0 #3,$ #3,3 #!,! #$,$ #$,! #4,4

7 #5,$ #, #",1 #1," #3,1 #1,5 #1,4 ##," #1," #1,3

8 #!, 1$,4 #$, #0,# #4,4 #$,3 #!,$ #3, #",! #!,

9 #!, #3," #!,# #0,4 #0,3 #4, #$," #,5 ##,0 #1,#

10 #",3 #4,4 #3,5 #!,# #5," #3,4 #",5 #,! #",0 #0,5

m) Data tinggi poplitealTebal 3.4.##. Data anthropometri T/

1 2 3 4 5 6 7 8 9 10

1 44,$ 4#,! 4!,3 41,1 45, 4"," 41,! 3$,4 43,# 4",!

2 3$, 4#,5 44,5 44, 41,! 3$,4 3$,! 4,# 4,! 3$,4

3 3!,3 3!,# 43,! 40,1 3$,# 4!,5 43,0 4!,# 4",1 4",#

4 43, 4!,# 4#,3 43,$ 40,1 4#,0 3$,$ 43,1 41,0 41,0

5 4",3 40," 41,! 4,4 44,5 44,# 41,! 4,0 4!,# 40,$

6 41,! 4#,! 3!,# 4"," 4,4 41,5 4", 45,0 45,$ 4!,0

7 4",5 45,0 3$,4 4#,4 4",1 40,1 40,3 3!, 3!, 4",3

8 4!,$ 45,0 4",5 41, 3$,4 43,5 4#,$ 41,5 3!," 40,1

9 41,1 3$," 45," 45," 3!,3 3$,3 40, 4",0 43,! 3!,!

10 43,1 44,# 4#,$ 43, 45,# 4", 43, 43,5 44,0 4!,"

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n) Data pantat popliteal

Tebal 3.4.#3. Data anthropometri /

1 2 3 4 5 6 7 8 9 10

1 4#, 53,1 43, 41, 43,0 50, 53, 5,5 5, 43,1

2 4!,$ 4",1 41,3 50,5 3$,5 44,1 5,# 45,0 4#,1 44,0

3 5,3 5,$ 4!,4 4",! 4$,3 45," 4!,1 5, 53,3 3$,!

4 41,1 55,1 50,# 50,$ 4$,5 4!,# 40, 51,! 44,1 4,5

5 53,0 5#,5 55,4 53,0 51, 43,5 5#,0 40,! 4$,4 51,$

6 5, 4!,4 5#,$ 51,! 53," 41,3 41,$ 4#,1 4#, 45,!

7 44," 50,! 54,! 51,5 4",# 5#, 51," 4,# 4"," 45,#

8 4!,# 4,1 45,4 45, 51,! 41,1 50,1 55,! 50,# 54,

9 44,1 41," 43,4 4$,1 43,! 44,0 41, 4#,$ 54,! 5,0

10 40,5 53,3 54,3 40," 5, 41, 55,3 40,! 51,$ 4!,"

o) Data tebal lututTebal 3.4.#4. T/

1 2 3 4 5 6 7 8 9 10

1 1",0 1,3 1," 1,4 1",5 ##,1 1",$ 1!," #0, 1",#

2 ##,# #1,3 #1," #1,$ #0,3 1,0 1,# ##,$ ##,0 #0,4

3 1!,! #0,3 1,5 1$,0 ##,0 1!,1 #1,5 1, 1!,3 1!,#

4 ##,$ #0,5 1$,5 1", #0,4 #1,! ##,# 1", #0,0 ##,"

5 1!,0 1!,# 1$, #0,3 1$,5 #0,0 1$,$ ##, 1$,3 #0,!

6 1",3 #0,$ 1",0 #0, #1,1 #0,3 1," #1,! 1"," #1,4

7 1$,5 1$," #0,$ 1",0 ##,$ #0,0 #1,# 1$,$ #1,# #0,1

8 #1,0 1$,1 1",4 1,4 #1,1 #1,$ #0,! ##,# 1!,3 1!,#

9 ##,0 1",! 1,0 #0," #1,3 #1,! #0," 1!,$ 1,5 #0,0

10 1$," 1$,1 1",# 1!,5 1",3 #1,3 ##,$ 1,3 ##,0 1,0

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 p) Data pantat ke lutut

Tebal 3.4.#5. Data anthropometri /

1 2 3 4 5 6 7 8 9 10

1 4,5 "3,5 "#,0 "1,# 4,5 "3,5 "0,0 4,0 "1,0 3,0

2 ,5 ",5 4,5 "5,0 ,5 ",5 3,0 ,0 "3,5 5,5

3 "1,0 3,0 "0,0 4,0 "1,0 3,0 ,4 ",5 4,5 4,5

4 "3,5 5,5 3,0 ,0 "3,5 5,5 #,0 !,5 "0,5 "#,0

5 4,5 4,5 ,4 ",5 4,5 4,5 "4,0 !,0 0,5 5,5

6 "0,5 "#,0 #,0 !,5 "0,5 "#,0 "$,0 ,5 ,0 1,5

7 0,5 5,5 "4,0 !,0 0,5 5,5 55,! "3,5 "#,0 ",5

8 ,0 1,5 "$,0 ,5 ,0 1,5 "3,5 ",5 5,0 5!,0

9 "#,0 ",5 55,! "3,5 "#,0 ",5 "1,# 4,5 "3,5 "0,0

10 5,0 5!,0 "3,5 ",5 5,0 5!,0 "5,0 ,5 ",5 3,0

 4.1.1.3. Data anthropometri  !ari tangan

Data anthropometri (ari tangan digunakan untuk membuat produk yang berhubungan dengan (aritangan sehingga produk bisa lebih nyaman dan tepat saat digunakan.

a) Data pan(ang (ari 1Tabel 3.4.#. Data anthropometri 1

1 2 3 4 5 6 7 8 9 10

1 ,4 4,$ ",# ,3 ,3 5, 4,$ ,# ",5 5,"

2 ,0 5,3 5," , ,1 ,1 ,0 5,0 5,$ ,3

3 5,1 ,0 5,5 ",1 5,1 ,$ ,5 4,! ",5 5,!

4 ",# 5,! ," ,5 ",# ,0 ", ,! ",# 5,$

5 ,! 5,1 ,3 ,0 ",3 4,! ",# 5,1 5,$ ,5

6 5,! 5,! ",4 ," 5,0 5,1 ,0 5,0 ",1 ",#

7 5,! ,3 ,1 ", ",0 ",3 ",5 5," 5,4 ",0

8 5,! 5,3 ,3 ",3 ,4 ",5 5,0 ,# ,# 5,

9 5,! , 5, ",1 5,3 ,5 5," 5,3 5,# 5,1

10 5,! ",3 ", ,! ," ,3 5,4 ,1 ",4 5,0

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 b) Data pan(ang (ari #

Tabel 3.4.#". Data anthropometri #

1 2 3 4 5 6 7 8 9 10

1 ,! ,4 ,$ ,! ,4 ",3 ," ",1 ",4 ,4

2 ," ,4 ",0 ",$ ,# ",$ ",5 ",3 ",# ",0

3 ",$ ",$ ",# !,0 ,4 ",4 ",4 !,0 ",4 ",5

4 ",4 ",5 ",$ ",1 ,! ,! ,4 ",! ",0 ,4

5 , ,# ",4 ,5 "," ",5 ,# ",$ !,0 ","

6 ,$ , ,3 ,$ ,# ",4 ",! ",1 "," !,0

7 "," ",4 ",3 ",$ ",0 ",5 ",5 ",# ", ,4

8 "," ",4 ",$ ,$ ",# ",1 !,0 ", ",3 ,$

9 ,3 ,3 ,$ ,4 , ",# ",3 ,3 ",# !,0

10 ,! ",5 ",5 , ,5 "," ",0 ", ",# ,5

&) Data pan(ang (ari 3

Tabel 3.4.#!. Data anthropometri 3

1 2 3 4 5 6 7 8 9 10

1 !,# !,# !,$ ",1 !, !,1 ",3 $,0 !,3 ",

2 ",1 ",! !,$ ",1 ",3 ",4 !,0 $,0 ",$ !,!

3 ",# ",1 !,3 !,5 !,4 ",5 !,4 ",1 ",# ",#

4 !,3 !,! !,3 !,! !, $,0 ",3 ",1 ",0 ",#

5 ",$ "," !,# ",$ $,0 ",! ", !,4 ",1 ",#

6 !,0 !, ",5 ",! ",! !, !,3 !,! !,1 !,4

7 ",3 ",0 !," ",! !, "," ",$ ",# ",1 ",3

8 ",4 ",! ", !, !,5 ",# !,3 ",$ !,5 ",4

9 !,3 ", ", ",5 !,4 !,# ",# ",! ",1 !,$

10 !,1 !,$ ",$ "," !,1 !,0 ",0 "," !,1 ",3

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d) Data pan(ang (ari 4

Tabel 3.4.#$. Data anthropometri 4

1 2 3 4 5 6 7 8 9 10

1 !,4 !,4 ",4 ! ,$ !,4 ,3 ,5 !,# !,1

2 ,5 ,3 ",$ ,# "," ,# ",0 ",4 !,4 !,0

3 ,! ,1 ",$ !,# ,0 ,4 ",0 ,3 ", ,5

4 !,1 ",1 ",# !,3 ,$ !,5 ,$ ",0 ,$ ,5

5 ,0 ",1 ,3 ,$ ,# ",1 !,4 ",3 ,5 !,3

6 "," ",4 ",4 , ",0 , ",4 ,4 ,1 !,0

7 ",1 ",4 ",$ !,# !,4 !,3 ",5 ,# ,# !,0

8 ",4 ",$ , ",0 ",$ ",# !,5 !,4 ",4 ",5

9 !,3 ",3 ", ",$ ,! !,4 ," ",0 !,# !,3

10 ",5 ",5 ,$ !,# ",! ,1 !,# ,1 ",! ,#

e) Data pan(ang (ari 5

Tabel 3.4.30. Data anthropometri 5

1 2 3 4 5 6 7 8 9 10

1 , ,3 5," , 5,$ ,3 ,3 5, 4,$ ,5

2 , , ,4 ,# 5,0 ,! 5,3 5,5 ,4 ,5

3 , ,5 5," ,5 5,$ 5,1 ,0 ," ,# 5,$

4 5," 5,# 5,1 4,$ ,0 ," ,$ 4,$ , ,"

5 ,5 5,1 5, 5,5 5,$ ,4 5,3 5,0 ,4 ,

6 5,3 5,# 5,$ ,0 ," 5,1 ,3 , 5,5 5,4

7 5,1 4,$ 5,0 ,5 4,$ ,$ 5,! ,! 5,1 ,!

8 5," 5,! ,1 ,# 5," ," 5,1 , 5,0 5,1

9 ,3 ,5 , 5,! 5,! 4,$ 5, ,5 , ,

10 ,$ ,# 5, , 5,$ ,3 5,1 5, ,0 5,0

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*) Data pangkal ke tangan

Tabel 3.4.31. Data anthropometri pangkal ke tangan

1 2 3 4 5 6 7 8 9 10

1 10,4 10,$ $,0 $,4 $,0 $,5 10," $, 11,4 $,$

2 11,$ $,! 11," 10,1 10,$ 10,0 10,4 10,3 11,3 $,3

3 10," $,! $,$ 11,3 10,5 $,4 10,4 11,! 11,5 11,"

4 $, 10,0 10,# 10,# 11,# 11,# 11,# 11,! 11,! $,3

5 $,$ 11,0 11,3 $,0 11,! $,1 10," 11,0 $,# 10,4

6 10,4 $,4 $,$ $,! 10,# 11,5 $, 10,0 10," 11,!

7 $,3 $," 10,4 11,3 $,0 10,$ 1#,0 $,! $,4 $,5

8 $,3 $,! $,4 $,3 11," 11,1 10,# 11,! $,! 11,0

9 10,5 10,5 10,4 $,4 $,4 10," 10, $,# 11, 10,4

10 10,! 11,3 $,! 1#,0 10,0 11,$ $,3 $," 10," $,1

g) Data lebar #,3,4,5Tabel 3.4.3#. Data anthropometri lebar #,3,4,5

1 2 3 4 5 6 7 8 9 10

1 ,4 , ,! ",1 ",! ",$ ,4 ,3 ,3 ,3

2 ",5 ,5 ,# ",3 ",1 ,0 ", ",$ ", ,$

3 ",5 , ",4 ",1 ",# ,# ",4 "," ",3 ,5

4 ", ,0 ",3 ",! ",0 ",3 , "," ,# ",1

5 "," ",# ",0 ," , ",$ ,$ ,1 ,! ",

6 ",3 "," ",1 ",# ",$ ",3 ",0 ",4 ,$ ,0

7 ," ,# ",# ,4 ,0 ",! ,$ ",# ,! ,"

8 ",$ ,! ,5 ,0 ," , ",1 ," , ",!

9 ", ",! ,5 ,4 ,5 ",3 ,5 ",! ,3 ,3

10 ",! , ," ,# ",4 ,1 , ," "," ,"

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h) Data lebar tangan

Tabel 3.4.33. Data anthropometri lebar tangan

1 2 3 4 5 6 7 8 9 10

1 $,3 10,3 11,0 $,5 $,4 10,4 $,! $,# $,4 10,4

2 $,$ $,0 10,4 11,0 10, $,$ $,$ 10,5 $,# 11,0

3 $,$ 10, 10,! $,1 10,3 10," 10,$ $,! 10, $,#

4 10,0 10,0 $,# 10,! 10,0 $,0 $, $,4 $,# 10,#

5 10,$ 10,3 10,# $,! $,5 $, $,! $,! 10,0 10,"

6 $,! $,5 $,3 $,0 10,# 10,0 $," $," 10,$ $,!

7 $,3 10,1 10,1 $,0 $,0 $," $,0 10,0 10," 10,4

8 $,! 10,4 10," 10,4 $,1 $,4 10,5 $, $,4 10,4

9 $,# $,4 10,4 10,1 10,! $,5 10,$ $,3 10,4 10,$

10 $,0 $,3 10,! 10," 10,4 $,5 $,5 $,3 $," 10,

i) Data pan(ang telapak tangan

Tabel 3.4.34. Data anthropometri pan(ang telapak tangan

1 2 3 4 5 6 7 8 9 10

1 1$,# 1!,# 1!, 1!, 1",# 1$,$ 1",1 1",3 1,# 1,5

2 1$,4 1$,# 1!,5 #0,0 1$,3 1",4 1!," 1,3 1$,3 1,!

3 1," 1",3 1," 1$,4 1",# 1," 1,$ 1"," 1$,0 1",3

4 1,4 1,4 1$," 1$,$ 1!,0 1$,# 1$,# 1",3 1", 1,4

5 1",$ 1!,5 1",4 1!,$ 1$,4 #0,0 1!,3 1",5 1$,! 1$,$

6 1",# 1"," 1!,# 1,3 1",# 1!,1 1!," 1",0 1$,1 1,

7 1!,5 1$,! 1",$ 1$,5 1,# 1$,3 1!,5 1$,# 1$, 1,!

8 1!," 1!,1 1",$ 1!,# 1$, 1,4 1!,! 1,! 1",3 1",

9 1, 1$,3 1$,5 1!,1 1",5 1!,$ 1,5 1!, 1$ 1!,5

10 1$, 1$,1 1!,$ 1,4 1", #0,0 1!,5 1!,3 1$,# 1",4

4.1.2. Data display "arna

Tes display -arna ini dilakukan dengan &ara praktikan menuliskan (elas apa tidak (elasnya suatu-arna yang dikombinasikan dengan -arna lain. Tu(uan tes display -arna ini untuk mengetahuidisplay -arna yang (elas menurut praktikan yang selan(utnya akan diproses untuk membuat peran&angan display poster).

Tabel 3.4.35. Data display -arna

 2o. /atar arna elas Tidak elas

1 bubu 6oklat  

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# %erah   3 uning   4 7range   5 'i(au  

Biru   "

6oklat

buabu   ! %erah   $ uning   10 7range   11 'i(au   1# Biru

13

uning

buabu   14 6oklat   15 7range   1 %erah   1" 'i(au  

1! Biru   1$

7range

6oklat   #0 buabu   #1 uning   ## %erah   #3 'i(au   #4 Biru   #5

%erah

buabu   # 6oklat   #" 7range   #! 'i(au   #$ uning   30 Biru   31

'i(au

buabu   3# 6oklat   33 uning   34 7range   35 %erah   3 Biru   3"

Biru

buabu   3! 6oklat   3$ 'i(au   40 %erah  

41 7range   4# uning  

4.1.3. Data tes buta "arna

Tes buta -arna ini dilakukan dengan &ara pratikan menuliskan angka atau gambar yang terdapat pada buku tes buta -arna yang dipegang oleh asisten dan diganti setiap 5 detik. Tes buta -arna ini ber*ungsi mengetahui apakah seseorang mengalami kelainan buta -ana atau tidak dan kemudianakan dibuat peran&angan display poster.

Tabel 3.4.3. Data hasil tes buta -arna

 2o a-aban

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ldi D-i 8ang Benar  

1 1# 1#

# ! !

3 5 5

4 #$ #$

5 "4

" "

" 45 45

! # #

$ #

10 1! 1

11

1# 35 35

4.1.4. Data atual ruangan

a) Tingkat kebisingan dan pen&ahayaantingkat kebisingan dan intensitas &ahaya sangat berpengaruh dalam akti9itas disuatu ruangan.

Untuk mengetahui tingkat kebisingan dan tingkat intensitas ruangan diperlukan pengamatan

menggunakan lux meter dan sound level , diba-ah ini merupakan hasil pengamatan tes tingkat

kebisingan dan pen&ahayaan di ruang sekretariat (urusan sek(ur).

Tabel 3.4.3". Tes kebisingan dan pen&ahayaan ruang sekretariat (urusan

atuan kun(ungan : 5 menit

'ari : ertama enin, " Desember #015)

Tempat penelitian : +uang sekretariat (urusan 2o Bilangan random am   Lux meter Sound level  

1 1 0!:05 0.#$ ".00

# 3 0!:15 0.#! 3.3

3 " 0!:35 0.30 0.#

4 10 0!:50 0.## 1.

5 11 0!:55 0.30 1.$

14 0$:00 0.#3 "."

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" 15 0$:15 0.#$ 4.1

! 1" 0$:#5 0.#3 3.

$ 1! 0$:30 0.30 3.5

10 #0 0$:40 0.30 0.

11 #3 0$: 55 0.30 4.5

1# #4 10:00 0.# 3.!

13 # 10:10 0.30 .5

14 #! 10:#0 0.30 5.!

15 30 10: 30 0.30 0.3

1 3# 10:40 0.#! 1.!

1"34 10:50

0.30 3.1

1! 3 11:00 0.30 0.4

1$ 3" 11:05 0.#3 "."

#0 40 11:#0 0.#$ 1.$

#1 41 11:#5 0.30 0.#

## 4# 11:30 0.30 3.5

#3 44 11:40 0.30 .5

#4 4 11:50 0.## 5.!

#5 4$ 1# :00 0.# 4.5

# 50 1# :05 0.#! 3.1

#" 5# 1# :15 0.30 3.4

#! 53 1# :#0 0.30 "."

#$ 54 1# :#5 0.## 3.

30 5 1#:35 0.3# 1.!

31 5" 1#:45 0.30 0.4

3# 5$ 1#:55 0.#" 5.!

33 3 13:00 0.#3 .5

34 3 13:00 0.30 0.#

35 " 13:#0 0.31 1.!

3 ! 13:30 0.3# 3.

3" $ 13:35 0.## 4.5

3! "1 13:45 0.30 5.!

3$ "4 14:00 0.30 .5

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40 "4 14:00 0.#4 1.!

41 "! 14:#0 0.# 5.!

4# !0 14:30 0.#! 0.3

43 !1 14:35 0.30 3.5

44 !# 14:40 0.#$ 1.!

45 !3 14:45 0.30 .5

4 !4 14:50 0.30 0.3

4" !5 14:55 0.3# 5.!

4! ! 15:00 0.#5 4.5

4$ !$ 15:15 0.30 5.!

50$0 15:#0

0.## 4."

Tabel 3.4.3!. Tes kebisingan dan pen&ahayaan ruang sekretariat (urusan

Tes kebisingan dan pen&ahayaan

atuan kun(ungan : 5 menit

'ari : edua elasa, ! Desember #015):

Tempat penelitian : +uang sekretariat (urusan

 2o Bilangan random am   Lux meter Sound level  

1 1 0!:05 0.34 "."

# 3 0!:15 0.34 $.#

3 " 0!:35 0.34 5.1

4 10 0!:50 0.3# "3.#

5 11 0!:55 0.35 !.3

14 0$:00 0.34 4.

" 15 0$:15 0.35 #.5

! 1" 0$:#5 0.35 4.5

$ 1! 0$:30 0.35 5.4

10 #0 0$:40 0.35 4.!

11 #3 0$: 55 0.35 .1

1# #4 10:00 0.35 .!

13 # 10:10 0.34 4.#

14 #! 10:#0 0.35 #.3

15 30 10: 30 0.35 3.0

1 3# 10:40 0.35 ".3

1" 34 10:50 0.3# 4.0

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1! 3 11:00 0.35 4.4

1$ 3" 11:05 0.35 .5

#0 40 11:#0 0.35 !.$

#1 41 11:#5 0.35 !.$

## 4# 11:30 0.3 3.3

#3 44 11:40 0.35 ".5

#4 4 11:50 0.34 5.3

#5 4$ 1# :00 0.34 .5

# 50 1# :05 0.35 4.5

#" 5# 1# :15 0.35 !."

#!53 1# :#0

0.35 ."

#$ 54 1# :#5 0.3# 5.4

30 5 1# :35 0.3 3.5

31 5" 1# :45 0.3 .!

3# 5$ 1# :55 0.35 !."

33 3 13:00 0.35 "#.1

34 3 13:00 0.35 4.

35 " 13:#0 0.3" $.3

3 ! 13:30 0.3 4.5

3" $ 13:35 0.3 #.3

3! "1 13:45 0.3" 4.1

3$ "4 14:00 0.35 1.!

40 "4 14:00 0.3# ".!

41 "! 14:#0 0.35 3."

4# !0 14:30 0.35 !.4

43 !1 14:35 0.35 5.5

44 !# 14:40 0.35 4.3

45 !3 14:45 0.31 .#

4 !4 14:50 0.35 .$

4" !5 14:55 0.3" $.!

4! ! 15:00 0.34 !.3

4$ !$ 15:15 0.35 5.1

50 $0 15:#0 0.3$ #.$0

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 b) /uas ruang sekretariat (urusan/ebar ; 3,$ meter an(ang ; 3," meter Tinggi ; 3,# meter 

<ambar 3.4.1. +uang sekretariat (urusan

&) apasitas ruanganapasitas ruangan sekretariat (urusan kurang lebih dapat menampung 3 sampai 5 orang.

d) =ungsi ruangan

=ungsi /aboratorium dalam proses pendidikan, yaitu:1) ebagai tempat untuk menyelesaikan administrasi perkuliahan, pengambilan kartu u(ian

dan pemin(aman buku.#) ekretariat (urusan sebagai -adah logistik ringan yang menun(ang proses perkuliahan.3) ekretariat (urusan sebagai tempat bagi mahasis-a yang butuh in*ormasi mengenai

 perkuliahan ataupun beasis-a.

e) kti9itas dalam ruangan

kti9itas yang dilakukan di lab sekretariat (urusan adalah sebagai berikut:1) %elakukan pemin(aman buku.#) %enyelesaikan administrasi perkuliahan.3) Bertanya in*ormasi mengenai e9ent yang ada di kampus ataupun diluar kampus.4) %entranskip nilai mahasis-a.

4.2. Pengolahan data4.2.1. Peran#angan ruang seretariat !urusan

4.2.1.1. Peran#angan produ 

4.2.1.1.1. $e!a omputer

ada praktikum kali ini produk yang akan diran&ang yaitu produk me(a komputer. gar dapatmenghasilkan produk me(a komputer yang ergonomis diperlukan pengukuran antrhopometri

dimensi tubuh berdiri dan dimensi tubuh duduk. engukuran yang dilakukan antara lain sebagai

 berikut:a) angkauan tangan (t)

Tabel 3.4.3$ angkauan tangan

% 

n1 2 3 4 5 6 7 8 9 10   ´ x

1 "",$ $,3 "4 4,3 5," ",4 !,$ $,3 "4,$ !0,! "1,#5

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2 5, 5 5 "#,! 3," 5,4 "5,3 !0,# ", !0 "0,$

3 "0 3," 3," "4,4 "!,3 "!,4 "!," ""," "$ $,$ "3,3!

4 "4,4 "1,$ "1,$ "3," ",# 3,4 3,# "!,# "5,! "0,5"

5 "3," 1, 1, !0,1 ",5 $ "5 !1 ,3 5,3 "1,01

6 "!,1 1," 1," "",! 1,$ #,3 "1,! "1,3 !,! ",5 $,1$

7 $,1 $," $," ",5 "1 $,3 #,3 "",# "4 !,! $,!

8 !, "4,! "4,! "1, #,3 "#,! ",5 ",5 "3 "$,! "1,#"

9 "",5 ",$ ",$ $," "3, "0,1 "!," 1,4 "0,$ 5,$ "#,1

10 #,$ ",1 ",1 !0," $ ,1 "1,5 !0 5,$ #,# $,#5

  ∑ ´ x "0!,$

1) %enghitung rararata subgroup

x = ∑ xk

x = ∑ 708,9

10

x =70,89  

#) %enghitung standar de9iasi

σ =√∑ ( x1−´ x )2

 N 

σ = √ (77,9-70,89 )2  + .... + (62,2 – 70,89 )2

100

σ = 5,79

3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup

σx =σ

√ k

σx =5,79

√ 10σx = 1,83

4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!

Z = 1 – [1 - β

2   ]Z = 1 – [1 – 0, 98

2   ]

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Z = 1 – 0, 01

Z = 0,99

Berdasarkan tabel distribusi normal, nilai yang paling mendekati untuk harga > ; 0,$$

adalah #,0 maka >t ; #,0

5) U(i keseragaman datai) Batas kontrol atas B)

B= x+(Z t  x σx )

B =70,89+(2,06 x 1,!3)

B ; "4,ii) Batas kontrol ba-ah BB)

BB= x-(Zt  x σ x )

BB = x-(2,06 x 1,83)

BB ; ",1#

1 # 3 4 5 " ! $ 10".1#

!.1#

$.1#

"0.1#

"1.1#

"#.1#

"3.1#

"4.1#

"1.#5"0.$

"3.3!

"0.5""1.01

$.1$

$.!

"1.#"

"#.1

$.#5

U(i keseragaman data

A bar B BB # bar

<ambar 3.4.#. U(i keseragaman data

) U(i ke&ukupan data

N'=[

Zt

α √N∑ x i

2

- (∑ x i )2

∑ xi ]2

N'=[

2,06

0,04 √ 505115,83 - 502539,21

7089   ]2

 2C ; 13,0eterangan: 2C 2, data &ukup

") Distribusi *rekuensi

i) angkauan +)

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+ ; DmaA Dmin

+ ; !1,00 E 1,40+ ; 1$,0

ii) Banyak kelas

; 1 F 3,3 log n ; 1 F 3,3 log 100

; ",Dibulatkan ke atas, sehingga ; $

iii) Gnter9al G)

I=R

I=19,60

9

I=2,18

Tabel 3.4.40. Distribusi *rekuensi

!) %ean

x =!"i.#$

!"i

x =7079,080

100

x =70,79

$) osisi median

osisi median ;1

2 x N 

osisi median ;1

2 x100

osisi median ; 50

10) %edian

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x =%&$+ [1

2N- (*$-1 )

"$]  x I

~x =67,935+[12

 x 100- (35 )

51   ]  x 2,18

x =67,935+[50-35

51   ]x 2,18

x = 70,05

11) %odus

x =%&$) +

[   ( "$) - "$)-1 )( "$)- "$)+1 )+ (" $) - "$)-1) ] x I

x = 67,935+[17-9

(17-10) + (17-9) ]  x 2,18

x =67,935+[87+8 ]  x 2,18

x =69,30

1#) ersentili) /etak 5

%tk 5  =I

100  N

%(tk 5  =5

100  100

%tk 5  = 5

ersentil 5 terletak pada data ke5 pada interval  ke1

ii) /etak 50

%tk 50  =I

100  N

%(tk 50  =50

100  100

%tk 50  = 50

ersentil 50 terletak pada data ke50 pada interval  ke4

iii) /etak $5

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%tk 95  =I

100  N

%tk 95  =95

100

  100

%tk 95  = 95

ersentil $5 terletak pada data ke$5 pada interval  ke$

i9) 2ilai 5

5  = %&5 + I [I

100N - *k5-1

"5]

5  = 61,395, + 2,18

[5

100100 - 0

13   ]5  = 61,89

9) 2ilai 50

50 = %&50 + I [I

100N - *k50-1

" 50]

50 = 67,935 + 2,18 [50100

100 - 35

16   ]50 = 70,05

9i) 2ilai $5

95  = %&95 + I [I

100N - *k95-1

" 95]

95  = 78,835 + 2,18 [95

100100 - 91

9   ]50  = 80,01

 b) +entang tangan rt)

Tabel 3.4.41. +entang tangan

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% 

n1 2 3 4 5 6 7 8 9 10   ´ x  

1 1!5," 153, 1!, 11," 14$, 1$#, 1", 15!, 150, 14," 1,1

2 1!",1 15$, 10, 10,4 10 153, 154," 1!1, 11 1$0,$ 1,$3

3 1"! 1$3, 1!", 1!4,$ 1!5 1"#, 1!4,$ 1"#, 15", 1$," 1"!,3

4 1!#, 154, 1$#, 1$4 1"3, 1", 1, 1"4 1"$, 15"," 1"5,05

5 1"4,3 10, 1$0, 10,$ 1$#, 15$, 1"4, 1!, 15, 1"1,$ 1"#,!

6 1"3," 153, 14", 1"3,$ 1!3 15, 1"", 11, 1", 1$1, 1"0,5

7 14",# 14!, 1!3, 1"5,# 155, 14!, 1!, 1!", 154 1",5 1,33

8 1"$ 15 1, 15$,$ 1!$, 13, 1!1 1"3 1$#, 1!3,4 1"4,4$

9 153,4 1$3, 1", 1!$,3 1!0, 1" 1!0,# 1!3 1!, 15$ 1"4,1!

10 1"!,1 1!4, 1!0, 10,3 1!1, 1", 1$3,4 15$, 1$0, 1$# 1"$,

  ∑ ´ x 1."#4,"3

1) %enghitung ratarata subgroup

=∑ i

k

  =1726,73

10

  = 172,67

#) %enghitung standar de9iasi

σ =√∑(i- )

2

N

= √(185,7 – 172,67 )

2 + (153,6 – 172,67 )

2 + / + (192,0 – 172,67 )

2

100

; 13,4"

3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup

σ  =σ

√ k

σ  =13,47

√ 10

σ  = 4,26

4) 'itung nilai > tabel >t), dengan ? ; 0,$

Zt = Z[1-[1- β

2   ]]

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Zt = 1 [1- [1-0,96

2   ]]Zt = 0,98

>t ; #,0

5) U(i keseragaman datai) Batas kontrol atas B)

= + (Zt   σ )

= 172,67+ (2,06 4,26 )

= 1!1,45

ii) Batas kontrol ba-ah BB)

= - (Zt   σ) = 172,67- (2,06 4,26 )

= 163,90

1 # 3 4 5 " ! $ 10155.00

10.00

15.00

1"0.00

1"5.00

1!0.00

1!5.00

1.11".$3

1"!.31"5.05

1"5.051"#.!0

1"0.50

1.33

1"4.4$1"5.1!

U(i keseragaman data

B BB bar  

+ata sbgrup

<ambar 3.4.3. eta kontrol u(i keseragaman data +T

) U(i ke&ukupan data dengan H ; 0,04

N'=[

Zt

α √N∑ i

2- (∑ i )

2

∑ i]

2

N'=

[

2,06

0,04 √ 100(2999751,95) – (1726,73 )2

1726,73

  ]

2

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N'=[

2,06

0,04 √ 299975195 – 298159649,29

1726,73   ]2

N'=[

51,5 1347,401726,73   ]

2

N'= 40,20

eterangan: 2C 2, data &ukup

") Distribusi *rekuensii) angkauan +)

R = $- $I

R = 194 - 147,20

R = 46,80  

ii) Banyak kelas )

= 1 + 3,3

= 1 + 3,3 90

= 7,4 7

iii) Interval G)

I =

R

I =46,80

7

I = 6,69

Tabel 3.4.4#. Tabel distribusi *rekuensi rentang tangan +T)

 Interval  &i ' ()( *)( ()B *)B )$ &i.)$

14",#0 153,!! ! ! 14",#0 153,!! 14",1$5 153,!!5 150,54 1#04,3#

153,!$ 10,5" 1$ #" 153,!$ 10,5" 153,!!5 10,5"5 15",#3 #$!",3"

10,5! 1",# 11 3! 10,5! 1",# 10,5"5 1",#5 13,$# 1!03,1#

1",#" 1"3,$5 1# 50 1",#" 1"3,$5 1",#5 1"3,$55 1"0,1 #04",3#

1"3,$ 1!0,4 1 1"3,$ 1!0,4 1"3,$55 1!0,45 1"",30 #!3,!0

1!0,5 1!",33 1" !3 1!0,5 1!",33 1!0,45 1!",335 1!3,$$ 31#",!3

1!",34 1$4,0# 1" 100 1!",34 1$4,0# 1!",335 1$4,0#5 1$0,! 3#41,5

+&i 100 +&i.)$ 1"#4!,3#

!)   Mean

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 Mean  =∑ " i &

∑ " i

 Mean =

17248,32

100

 Mean  = 172,48

$) %ediani) /etak median

/etak median  =1

2  N

/etak median  =1

2  100

/etak median

 = 50

%edian terletak pada nilai ke50 pada Interval  ke4

ii) 2ilai median

%edian  = %&$+ [1

2N - *k$-1

"$]×I

%edian  = 167,26 + [50 - 38

12   ]× 6,69

%edian  = 167,26 +

[1212 ]

×6,69

%edian  = 167,26 + 6,69

%edian  =  1"3,$5

10) %odus

%odus = %&$ + [

"$ - "$-1

( "$ - "$-1 )+( "$ - "$+1 ) ]× I

%odus = 167,26 + [

19 - 8(19 - 8 )  + (19 - 11) ]   ×  ,$

%odus  = 167,26 + [11

 11 + 8 ]   ×  ,$

%odus  = 167,26 + [11

19 ]× ,$

%odus  = 167,26 + 3,87

%odus  = 171,14

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11) ersentili) /etak 5

%tk 5  =I

100  N

%(tk 5  =5

100  100

%tk 5  = 5

ersentil 5 terletak pada data ke5 pada Interval  ke1

ii) /etak 50

%tk 50  =I

100  N

%(tk 50  =50100

  100

%tk 50  = 50

ersentil 50 terletak pada data ke50 pada Interval  ke4

iii) /etak $5

%tk 95  =I

100  N

%tk 95  =95

100  100

%tk 95  = 95

ersentil $5 terletak pada data ke$5 pada Interval  ke"

i9) 2ilai 5

5  = %&5 + [I

100N - *k5-1

"5]× I

5  = 147,195 +

[

5 - 0

8

  ]  ×  ,$

5  = 147,195 + [58 ]×  ,$

5  = 147,195 + 4,18 

5  = 151,38 

9) 2ilai 50

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50 = 167,26 + [I

100N - *k50-1

"50]×I

50  = 167,26 + [50 - 3812   ]   ×  ,$

50 = 167,26 + [12

12 ]×6,69

50 = 167,26 + 6,69

50 = 173,96

9i) 2ilai $5

95  = %&95 +[I100

N - *k95-1

" 95 ]×I

95  = 187,33 + [95 - 100

17   ]×6,69

95  = 187,33 + [-517 ]×6,69

95  = 187,33 + (-1,97)

95  = 185,37

&) Tinggi siku duduk tsd)Tabel 3.4.43. Tabel data tinggi siku duduk tsd)

  n

 1 2 3 4 5 6 7 8 9 10   ´ x

11!,!

0

1",#

0

#,"

0

#3,!

0

#4,

0

1$,#

0

##,$

0

#0,3

0

##,3

0

#1,"

0#1,"5

2#,1

0

##,5

0

#5,"

0

##,"

0

1",4

0

##,0

0

##,$

0

##,3

0

1$,

0

##,1

0##,33

3

#",0

0

#,0

0

#3,1

0

1$,4

0

#,1

0

#1,"

0

#1,!

0

#3,$

0

1",$

0

#0,5

0 ##,"4

41!,#

0

#4,3

0

#4,0

0

1,5

0

#,0

0

#5,

0

#1,$

0

#1,#

0

#1,$

0

#3,4

0##,30

5#5,4

0

#5,0

0

##,0

0

1",#

0

1$,!

0

#3,!

0

#,#

0

#,3

0

1!,#

0

#4,$

0##,!!

6#0,3

0

#5,$

0

1$,"

0

##,0

0

1$,#

0

1,

0

##,4

0

#0,3

0

#4,#

0

#1,5

0#1,#1

7#4,#

0

#1,3

0

#5,5

0

#4,!

0

##,5

0

1",1

0

#1,"

0

#5,"

0

##,"

0

#3,

0##,$1

8#4,

0

#1,

0

#0,4

0

#4,#

0

#4,"

0

1$,1

0

#,

0

#0,$

0

1!,#

0

#4,4

0##,4"

9#,#

0

1","

0

1!,$

0

#0,$

0

1",#

0

1!,1

0

1$,4

0

##,"

0

#",0

0

#0,$

0#0,$0

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101,"

0

1!,

0

#0,3

0

1$,

0

#0,#

0

#0,$

0

#4,$

0

#3,5

0

#0,1

0

##,

0#0,"4

∑  ##0,#

3

1) %enghitung ratarata subgroup

=∑ i

k

  =220,23

10

  = 22,02

#) %enghitung standar de9iasi

σ =

√∑

(i-  )2

N

= √(18,8 – 22,02 )2  + (17,2 – 22,02 )2  + / + ( 48,7 – 22,02)2

100

; #,!5

3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup

σ  =σ

√ k

σ  = 2,85√ 10

σ  = 0,90

4) 'itung nilai > tabel >t), dengan ? ; 0,$

Zt = Z[1-[1- β

2   ]]Zt = 1 [1- [1-0,96

2   ]]Zt = 0,98

>t ; #,0

5) U(i keseragaman datai) Batas kontrol atas B)

= + (Zt   σ )

= 22,02+ (2,06 0,90 )

=#3,!"

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ii) Batas kontrol ba-ah BB)

= - (Zt   σ)

= 22,02 - (2,06 0,90 )

= 20,16

) U(i ke&ukupan data dengan H ; 0,04

N'=[

Zt

α √N∑ i

2- (∑ i )

2

∑ i]

2

N

'

=

[

2,06

0,04 √ 100(49315,83) – (2202,30 )2

2202,30

  ]

2

N'=[

2,06

0,04 √ 493158300 – 4850125,29

2202,30   ]2

N'=[51,5 285,40

2202,30   ]2

  N'= 44,5

eterangan: 2C 2, data &ukup

<ambar 3.4.4. eta kontrol u(i keseragaman data

") Distribusi *rekuensi

i) angkauan +)

R = $- $ I

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R = 26,60 - 16,50

R = 10,10  

ii) Banyak kelas )

= 1 + 3,3

= 1 + 3,3 100

= 7,6

Dibulatkan men(adi ;!

iii)   Interval G)

I =R

I =10,10

8

I = 1,3 0

Tabel 3.4.44. Tabel distribusi *rekuensi tinggi siku duduk TD)

 Interval  &i & ()( *)( ()B *)B )$ &i.)$

1,50 1",!# $ $ 1,50 1",!# 1,4$5 1",!#4 1",1 154,44

1",!3 1$,15 $ 1! 1",!3 1$,15 1",!#4 1$,153 1!,4$ 1,40

1$,1 #0,4! 15 33 1$,1 #0,4! 1$,153 #0,4!# 1$,!# #$",#

#0,4$ #1,!1 13 4 #0,4$ #1,!1 #0,4!# #1,!11 #1,15 #"4,$0

#1,!# #3,13 1! 4 #1,!# #3,13 #1,!11 #3,140 ##,4! 404,55

#3,14 #4,4 1# " #3,14 #4,4 #3,140 #4,4$ #3,!0 #!5,5

#4,4" #5,"$ 1# !! #4,4" #5,"$ #4,4$ #5,"$! #5,13 301,0

#5,!0 #",1# 1# 100 #5,!0 #",1# #5,"$! #",1#" #,4 31",55

+&i 100 +&i.)$ ##0#,34

!)   Mean

 Mean  =∑ " i &

∑ " i

 Mean  =2202,34

100

 Mean  = 22,02

$) %edian

i) /etak median

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/etak median  =1

2  N

/etak median  =1

2

  100

/etak median  = 50

%edian terletak pada nilai ke50 pada Interval  ke5

ii) 2ilai median

%edian  = %&$()+ [1

2N - * k$()-1

"$()] I

%edian

 = 21,81 +

[50 - 46

18

  ]×  1,30

%edian  = 21,81 + [418 ]×1,3 0

%edian  = 21,81 + 0,29

%edian  =  ##,11

10) %odus

%odus = %&$) + [

"$) - "$)-1

( "$)- "$)-1)+ ( "$) - "$)+1 ) ]× I

%odus = 21,81 + [18 - 13

(18 - 13 )  + (18 - 12 ) ]   ×  1,30

%odus  = 21,81 + [5 5 + 6 ]   ×  1,30

%odus  = 21,81 +

[

5

11

 ]× 1,30

%odus  = 21,81 + 0,59

%odus  = 22,41

11) ersentili) /etak 5

%tk 5  =I

100  N

%tk 5  =5

100

  100

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%tk 5  = 5

ersentil 5 terletak pada data ke5 pada Interval  ke1

ii) /etak 50

%tk 50  =I100

  N

%tk 50  =50

100  100

%tk 50  = 50

ersentil 50 terletak pada data ke50 pada Interval  ke5

iii) /etak $5

%tk 95 =I

100

  N

%tk 95 =95

100  100

%tk 95 = 95

ersentil $5 terletak pada data ke$5 pada Interval  ke!

i9) 2ilai 5

5  = %&5 +

[

I

100N - *k5-1

"5

]× I

5  = 16,49 + [5 - 0

9   ]   ×  1,3 0

5  = 16,49 + [59 ]×  1,30

  5  = 16,49 + 0,72 

5  = 17,23 

9) 2ilai 50

50  = %& 50  + [I

100N - * k50-1

"50] I

50  = 21,81 + [50 - 46

18   ]   ×  1,30

50  = 21,81 + [418 ]×1,3 0

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50  = 21,81+ 0,29

50  = 22,11

9i) 2ilai $5

95  = %&95 +[I

100N - * k95-1

" 95] I

95  = 25,79 + [95 - 88

12   ]×1,30

95  = 25,79+[712 ]×1,3 0

95  = 25,79 + 0,76

95  = 26,55

4.2.1.1.2. (emari abinet

%eran&ang lemari kabinet yang sesuai data anthropometri  praktikan mata kuliah IJ. DataanthropometrI yang digunakan untuk meran&ang lemari kabinet adalah tinggi bahu berdiri, lebar  bahu, (angkauan tangan, an(ang ari 3 dan lebar (ari #, 3, 4 dan 5. erhitungannya sebagai berikut:a) Tinggi bahu berdiri

Tabel 3.4.45. Data pengolahan anthropometri T'B

n

1 2 3 4 5 6 7 8 9 10   ´ x

115#,"

0

14#,3

0

14!,#

0

140,5

0

13!,"

0

14$,0

0

14#,3

0

1#$,

0

13,5

0

13",#

0 141,"0

2143,3

0

133,"

0

135,!

0

14,3

0

1#5,0

0

14$,"

0

150,5

0

13!,3

0

13,4

0

151,4

0 141,04

3153,5

0

150,5

0

153,#

0

13#,!

0

130,0

0

130,4

0

151,1

0

1#5,#

0

143,"

0

145,

0 141,0

4143,4

0

13,4

0

14#,

0

145,3

0

150,1

0

1#4,0

0

1#!,1

0

135,0

0

13,$

0

134,#

0 13",0

5141,$

0

1#5,$

0

1##,!

0

1#!,4

0

144,"

0

150,"

0

135,0

0

1#5,5

0

143,#

0

14,#

0 13,43

61#5,

0

135,0

0

154,#

0

130,4

0

143,4

0

1#4,$

0

145,5

0

150,"

0

1#,1

0

1#4,"

0 13,05

7140,"

0

13,4

0

143,$

0

14!,0

0

1##,

0

15#,5

0

154,$

0

155,0

0

133,#

0

13!,0

0 14#,5#

8133,!

0

14,

0

13$,#

0

133,0

0

153,"

0

1##,1

0

154,3

0

131,"

0

13$,3

0

1##,#

0 13",5$

9130,1

0

13!,5

0

14",1

0

1#4,0

0

145,0

0

14,#

0

151,$

0

1#$,$

0

133,4

0

143,

0 13!,$"

1

0130,4

0

154,$

0

145,0

0

13$,3

0

1#!,3

0

153,4

0

14,!

0

13",5

0

14!,"

0

13,4

0 14#,0"

∑ 13$5,5

"

1) %enghitung ratarata subgroup

=∑ i

k

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  =1395,57

10

  = 139,56

#) %enghitung standar de9iasi

σ =√∑(i- )

2

N

  = √(152,70 – 139,56 )2 + (142,30 – 139,56 )2 + / + (136,40 – 139,56 )2

100

  ; $,5#

3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup

σ  =σ√ k

σ  =9,52

√ 10

σ  = 3,01

4) 'itung nilai > tabel >t), dengan ? ; 0,$

Zt = Z

[1-

[

1- β

2

  ]]Zt = 1 [1- [1-0,96

2   ]]Zt = 0,98

>t ; #,05) U(i keseragaman data

i) Batas kontrol atas B)

= + (Zt   σ)

= 139,56 + (2,06 3,01 ) = 145,"

ii) Batas kontrol ba-ah BB)

= - (Zt   σ)

= 139,56 + (2,06 3,01 )

= 133,35

) U(i ke&ukupan data dengan H ; 0,04

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N'=[

Zt

α √N∑ i

2- (∑ i )

2

∑ i]

2

N'=[

2,06

0,04 √ 100(1956693) – (13955,7 )2

13955,7   ]2

N'=[

2,06

0,04 √ 195669261 – 194761562,5

13955,7   ]2

N'=[51,5 952,73

13955,7   ]2

N'= 12,36

eterangan: 2C 2, data &ukup

1#5

130

135

140

145

150

141." 141.04 141.

13".13.43 13.05

14#.5#

13".5$13!.$"

14#.0"

U(i keseragaman data

B BB bar  

+ata subgroup

<ambar 3.4.5. eta kontrol u(i keseragaman data T'B

") Distribusi *rekuensii) angkauan +)

R = $- $I

R = 155,0 - 122,1

R = 32,9  

ii) Banyak kelas )

= 1 + 3,3

= 1 + 3,3 100

= 7,6  K !

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iii)   Interval G)

I =R

I = 32,98

I = 4,12

Tabel 3.4.4. Tabel distribusi *rekuensi T'B)

 Interval  &i ' ()( *)( ()B *)B )$ &i.)$

1##,10 1#,#1 14 14 1##,10 1#,#1 1##,0$5 1#,#15 1#4,1 1"3!,1"

1#,## 130,33 " #1 1#,## 130,33 1#,#15 130,335 1#!,#! !$",$3

130,34 134,45 11 3# 130,34 134,45 130,335 134,455 13#,40 145,35

134,4 13!,5" 15 4" 134,4 13!,5" 134,455 13!,5"5 13,5# #04","3

13!,5! 14#,$ 10 5" 13!,5! 14#,$ 13!,5"5 14#,$5 140,4 140,35

14#,"0 14,!1 1! "5 14#,"0 14,!1 14#,$5 14,!15 144," #05,5$

14,!# 150,$3 11 ! 14,!# 150,$3 14,!15 150,$35 14!,!! 13",3

150,$4 155,05 14 100 150,$4 155,05 150,$35 155,055 153,00 #141,$3

+&i 100 +&i.)$ 13$31,

!)   Mean

 Mean  =∑ " i &

∑ " i

 Mean  = 13931,66100

 Mean  = 13$,3#

$) %ediani) /etak median

/etak median  =1

2  N

/etak median  =1

2  100

/etak median  = 50

%edian terletak pada nilai ke50 pada Interval  ke5

ii) 2ilai median

%edian  = %&$+

1

2N - *k$-1

"$

× G

%edian  = 138,53+ [50 - 47

10   ]× 4,12

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%edian  = 138,53 + [310 ]×4,12

%edian  = 138,53 + 1,24

%edian  =  13$,""

10) %odus

%odus = %&$ + [

"$ - "$-1

( "$ - "$-1 )+( "$ - "$+1 ) ]× G

%odus = 142,65 + [18 - 10

(18 - 10 )  + (18 - 11 ) ]   ×  4,1#

%odus  = 142,65 +

[8 8 + 7 ]   ×  4,1#

%odus  = 142,65 + [815 ]× 4,1#

%odus  = 142,65 + 2,20

%odus  = 144,85

11) ersentili) /etak 5

%tk 5  =I100

  N

%tk 5  =5

100  100

%tk 5  = 5

ersentil 5 terletak pada data ke5 pada Interval  ke1

ii) /etak 50

%tk 50  =I

100  N

%tk 50  =50

100  100

%tk 50  = 50

ersentil 50 terletak pada data ke50 pada Interval  ke5

iii) /etak $5

%tk 95 =I

100  N

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%tk 95  =95

100  100

%tk 95 = 95

ersentil $5 terletak pada data ke$5 pada Interval  ke!

i9) 2ilai 5

5  = %&5 + [I

100N - *k5-1

"5]×  G

5  = 122,055 + [5 - 0

14   ]   ×  4,1#

5 = 122,055 +

[5

14

 ]×

 4,1#

5  = 122,055 + 1,47 

5  = 123,57 

9) 2ilai 50

50  = %& 50  + [I

100N - *k50-1

"50]×I

50  = 138,535 + [50 - 47

10   ]   ×  4,1#

50  = 138,535 + [310 ]×4,12

50  = 138,535 + 1,24

50 = 139,81

9i) 2ilai $5

95  = %&95 +[I

100N - *k95-1

" 95]×I

95  = 150,895 + [95 - 86

14   ]×4,12

95  = 150,895 +[914 ]×4,12

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95  = 150,895 + 2,65

95  = 153,58

 b) /ebar bahuTabel 3.4.4. Data pengolahan lebar bahu

n 

1 2 3 4 5 6 7 8 9 10   ´ x

1 40,0 41,40 3$,0 40,0 4,#0 3!,30 35,40 44,"0 3",!0 3,50 40,11

2 45,10 3",#0 3,30 3,#0 40,50 3,00 3!,!0 44,10 41,!0 45,00 40,10

3 44,40 3$,#0 41,"0 3!,10 4#,10 3!,0 35,!0 34,0 41,!0 34,$0 3$,1#

4 3!,30 41,10 3,10 3!,50 3$,10 3,40 3",#0 34,00 41,30 3!,"0 3!,0"

5 34,$0 3,10 41,40 4#,0 45,30 3$,$0 3,10 40,!0 35,30 45,00 3$,"4

6 3,40 3!,!0 3",0 35,0 40,#0 44,30 44,50 3!,"0 3",!0 3$,30 3$,3#

7 45,00 44,"0 3$,40 3,!0 40,00 40,#0 44,!0 34,!0 40,!0 35,00 40,15

8 3!,!0 34,10 43,10 3$,50 40,"0 3,#0 34,"0 4#,$0 3!,00 35,30 3!,33

9 4,#0 34,!0 4,00 3,$0 40,"0 3,#0 4#,"0 43,50 41,"0 3",50 40,#

10 35,#0 35,10 3",10 4#,50 4,00 4#,#0 3$,40 45,!0 44,50 3",50 40,53

L   ´ x 3$,0$

1) %enghitung ratarata subgroup

=∑ i

k

  =396,09

10

  = 39,61

#) %enghitung standar de9iasi

σ =√∑(i- )

2

N

  =

√(40,60 – 39,61

)

2 +

(41,40 – 39,61

)

2 + / +

(37,50 – 39,61

)

2

100

  ; 3,4"

3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup

σ  =σ

√ k

 σ  =3,47

√ 10

 σ  = 1,10

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4) 'itung nilai > tabel >t), dengan ? ; 0,$

Zt = Z[1-[1- β

2   ]]Zt = 1 [1- [1-0,96

2   ]]Zt = 0,98

>t ; #,0

5) U(i keseragaman datai) Batas kontrol atas B)

= + (Zt   σ )

= 39,60 + (2,06 1,10 ) = 41,!"

ii) Batas kontrol ba-ah BB)

= - (Zt   σ)

= 39,60 + (2,06 1,10 )

= 37,35

) U(i ke&ukupan data dengan H ; 0,04

N'=[

Zt

α √N∑ i

2- (∑ i )

2

∑ i]

2

N'=[

2,06

0,04 √ 100(158093,8) – (3960,9 )2

3960,9   ]2

N'

=

[2,06

0,04 √ 15809383 – 15688729

3960,9

  ]

2

N'=[51,5 347,35

3960,9   ]2

N'= 20,42

eterangan: 2C 2, data &ukup

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1 # 3 4 5 " ! $ 103.00

3".00

3!.00

3$.00

40.00

41.00

4#.00

40.11 40.10

3$.1#

3!.0" 3!.0"

3$."43$.3#

40.15

3!.33

40.#

U(i keseragaman data

B BB bar  

+ata subgroup

<ambar 3.4.. eta kontrol u(i keseragaman data /B

") Distribusi *rekuensii) angkauan +)

R = $- $I

R = 46,2 - 34

R = 12,2  

ii) Banyak kelas )

= 1 + 3,3

= 1 + 3,3 100

= 7,6  K !

iii)   Interval G)

I =R

I =12,2

8

I = 1,53

Tabel 3.4.4". Tabel distribusi *rekuensi lebar bahu

 Interval  &i ' ()( *)( ()B *)B )$ &i.)$

34,00 35,5# 14 14 34,00 35,5# 33,$$5 35,5#5 34," 4!,4

35,53 3",05 15 #$ 35,53 3",05 35,5#5 3",055 3,#$ 544,35

3",0 3!,5! 13 4# 3",0 3!,5! 3",055 3!,5!5 3",!# 4$1,

3!,5$ 40,11 15 5" 3!,5$ 40,11 3!,5!5 40,115 3$,35 5$0,#5

40,1# 41,4 13 "0 40,1# 41,4 40,115 41,45 40,!! 531,44

41,5 43,1" 11 !1 41,5 43,1" 41,45 43,1"5 4#,41 4,51

43,1! 44,"0 ! !$ 43,1! 44,"0 43,1"5 44,"05 43,$4 351,5#

44,"1 4,#3 11 100 44,"1 4,#3 44,"05 4,#35 45,4" 500,1"

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+&i 100 +&i.)$ 3$#,54

!)   Mean

 Mean  =∑ " i &

∑ " i

 Mean  =3962,54

100

 Mean  = 39,63

$) %ediani) /etak median

/etak median  =1

2  N

/etak median  =1

2  100

/etak median  = 50

%edian terletak pada nilai ke50 pada Interval  ke4

ii) 2ilai median

%edian  = %&$()+ [1

2N - *k$()-1

"$()]×I

%edian  = 38,545 + [50 - 42

15   ]× 1,53

%edian  = 38,545 + [815 ]×1,53

%edian  = 38,545 + 0,82

%edian  =  3$,3

10) %odus

%odus = %&$) + [

"$) - "$)-1

( "$)- "$)-1)+ ( "$) - "$)+1 ) ]× I

%odus = 38,545 + [15 - 13

(15 - 13)  + (15 - 13 ) ]   ×  1,53

%odus  = 38,545 + [2 2 + 2 ]   ×  1,53

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%odus  = 38,545 + [24 ]× 1,53

%odus  = 38,545 + 0,77

%odus  = 39,31

11) ersentili) /etak 5

%tk 5  =I

100  N

%tk 5  =5

100  100

%tk 5  = 5

ersentil 5 terletak pada data ke5 pada Interval  ke1

ii) /etak 50

%tk 50  =I

100  N

%tk 50  =50

100  100

%tk 50  = 50

ersentil 50 terletak pada data ke50 pada Interval  ke4

iii) /etak $5

%tk 95  =I

100  N

%tk 95  =95

100  100

%tk 95  = 95

ersentil $5 terletak pada data ke$5 pada Interval  ke!

i9) 2ilai 5

5  = %&5 + [I

100N - *k5-1

"5]× I

5  = 33,955 + [5 - 0

14   ]   ×  1,53

5  = 33,955 + [514 ]×  1,53

5  = 33,955 + 0,55 

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5  = 34,50 

9) 2ilai 50

50 = %& 50  + [I

100 N - *k50-1

"50 ]×I

50 = 38,545 + [50 - 42

15   ]   ×  1,53

50 = 38,545 + [815 ]×1,53

50 = 38,545 + 0,82

50 = 39,36

9i) 2ilai $5

95  = %&95 +[I

100N - *k95-1

" 95]×I

95  = 44,665 + [95 - 89

11   ]×1,53

95  = 44,665 + [611 ]×1,53

95  = 44,665 + 0,83

95  = 45,49

&) an(ang (ari 3Tabel 3.4.4". Data pengolahan anthropometrI  3

  n

 1 2 3 4 5 6 7 8 9 10   ´ x

1 !,#0 !,#0 !,$0 ",10 !,0 !,10 ",30 $,00 !,30 ",0 !,13

2 ",10 ",!0 !,$0 ",10 ",30 ",40 !,00 $,00 ",$0 !,!0 ",$3

3 ",#0 ",10 !,30 !,50 !,40 ",50 !,40 ",10 ",#0 ",#0 ",$

4 !,30 !,!0 !,30 !,!0 !,0 $,00 ",30 ",10 ",00 ",#0 !,04

5 ",$0 ","0 !,#0 ",$0 $,00 ",!0 ",0 !,40 ",10 ",#0 ",!!

6 !,00 !,0 ",50 ",!0 ",!0 !,0 !,30 !,!0 !,10 !,40 !,1$

7 ",30 ",00 !,"0 ",!0 !,0 ","0 ",$0 ",#0 ",10 ",30 ",

8",40 ",!0 ",0 !,0 !,50 ",#0 !,30 ",$0 !,50 ",40 ",$#

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9 !,30 ",0 ",0 ",50 !,40 !,#0 ",#0 ",!0 ",10 !,$0 ",!

10 !,10 !,$0 ",$0 ","0 !,10 !,00 ",00 ","0 !,10 ",30 ",!!

∑ 

"$,1!

1) %enghitung ratarata subgroup

=∑ i

k

  =79,18

10

  = 7,92

#) %enghitung standar de9iasi

σ =√∑ (i- )

2

N

  = √(8,20 – 7,92 )2 + (8,20 – 7,92 )2 + / + (7,30 – 7,92 )2

100

  ; 0,5$

3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup

σ  =σ

√ k

σ  =0,59

√ 10

σ  = 0,19

4) 'itung nilai > tabel >t), dengan ? ; 0,$

Zt = Z[1-[1- β

2   ]]

Zt = 1

[1-

[1-0,96

2   ]]Zt = 0,98

>t ; #,0

5) U(i keseragaman datai) Batas kontrol atas B)

= + (Zt   σ )

= 7,92 + (2,06 0,19 )

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= !,31

ii) Batas kontrol ba-ah BB)

= - (Zt   σ)

= 7,92 - (2,06 0,19 )

= 7,53

) U(i ke&ukupan data dengan H ; 0,04

N'=[

Zt

α √N∑ i

2- (∑ i )

2

∑ i]

2

N'=[

2,060,04 √ 100(6305,32) – (791,80 )2

791,80   ]2

N'=[

2,06

0,04 √ 630532 – 626947,24

791,80   ]2

N'=[51,5 59,87

791,80   ]2

N'

= 15,16

eterangan: 2C 2, data &ukup

1 # 3 4 5 " ! $ 10".00".#0

".40

".0

".!0

!.00

!.#0

!.40

!.13

".$3

".$

!.04 !.04

".!!

!.1$

".

".$# ".!

U(i keseragaman data

B BB bar  

+ata ,bgrup

<ambar 3.4.". eta kontrol u(i keseragaman data 3

") Distribusi *rekuensii) angkauan +)

R = $- $I

R = 9 - 7

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R = 2  

ii) Banyak kelas )

= 1 + 3,3

= 1 + 3,3 100

= 7,6≈7

iii)   Interval G)

I =R

I =2

7

I = 0,29

Tabel 3.4.4!. Tabel distribusi pan(ang (ari 3

 Interval  &i ' ()( *)( ()B *)B )$ &i.)$

",00 ",#! #0 #0 ",00 ",#! ,$$5 ",#!5 ",14 14#,!0

",#$ ",5" 1# 3# ",#$ ",5" ",#!5 ",5"5 ",43 !$,1

",5! ",! 1 4! ",5! ",! ",5"5 ",!5 ","# 1#3,5#

",!" !,15 14 # ",!" !,15 ",!5 !,155 !,01 11#,14

!,1 !,44 1 "! !,1 !,44 !,155 !,445 !,30 13#,!0

!,45 !,"3 10 !! !,45 !,"3 !,445 !,"35 !,5$ !5,$0

!,"4 $,0# 1# 100 !,"4 $,0# !,"35 $,0#5 !,!! 10,5

+&i 100 +&i.)$ "$#,!!

8) Mean

 Mean  =∑ " i &

∑ " i

 Mean  =792,88

100

 Mean  = 7,93

$) %ediani) /etak median

/etak median  =1

2  N

/etak median  =1

2  100

/etak median  = 50

%edian terletak pada nilai ke50 pada inter9al ke4ii) 2ilai median

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%edian  = %&$()+ [1

2N - *k$()-1

"$()]×I

%edian  = 7,87 + [50 - 4814   ]× 0,29

%edian  = 7,87 + [214 ]×0,29

%edian  = 7,87 + 0,04

%edian  =  ",$1

10) %odus

%odus = %&$ +

["$ - "$-1

( "$- "$-1 )+ ( "$ - "$+1 ) ]× I

%odus = 6,99 + [20 - 0

(20 - 0 ) + (20 - 12 ) ]   ×  0,#$

%odus  = 6,99 + [20

 20 + 8 ]   ×  0,#$

%odus  = 6,99 + [20

28 ]× 0,#$

%odus  = 6,99 + 0,20

%odus  = 7,20

11) ersentili) /etak 5

%tk 5  =I

100  N

%tk 5  =5

100  100

%tk 5  = 5

ersentil 5 terletak pada data ke5 pada inter9al ke1

ii) /etak 50

%tk 50  =I

100  N

%tk 50  =50

100  100

%tk 50  = 50

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ersentil 50 terletak pada data ke50 pada inter9al ke4

iii) /etak $5

%tk 95 =I

100  N

%tk 95 =95

100  100

%tk 95 = 95

ersentil $5 terletak pada data ke$5 pada inter9al ke"

i9) 2ilai 5

5  = %&5 +

[

I

100N - *k5-1

"5

]× I

5  = 6,99 + [5 - 0

20   ]   ×  0,#$

5  = 6,99 + [520 ]×  0,#$

5  = 6,99 + 0,07 

5  = 7,06 

9) 2ilai 50

50  = %& 50  + [I

100N - *k50-1

"50]×I

50  = 7,87 + [50 - 48

14   ]   ×  0,#$

50  = 7,87+ [214 ]×0,29

50  = 7,87 + (0.04)

50  = 7,91

9i) 2ilai $5

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95  = %&95 +[I

100N - *k95-1

" 95]×I

95  = 8,74 + [95 - 8812   ]×0,29

95  = 8,74 +[712 ]×0,29

95  = 8,74 + 0,16

95  = 8,90

d) /ebar (ari #, 3, 4 dan 5

Tabel 3.4.4$. Data pengolahan anthropometri lebar (ari #,3,4,5  n

 1 2 3 4 5 6 7 8 9 10   ´ x

1 ,4 , ,! ",1 ",! ",$ ,4 ,3 ,3 ,3 ,"$

2 ",5 ,5 ,# ",3 ",1 ,0 ", ",$ ", ,$ ",0

3 ",5 , ",4 ",1 ",# ,# ",4 "," ",3 ,5 ",0$

4 ", ,0 ",3 ",! ",0 ",3 , "," ,# ",1 ",0

5 "," ",# ",0 ," , ",$ ,$ ,1 ,! ", ",05

6 ",3 "," ",1 ",# ",$ ",3 ",0 ",4 ,$ ,0 ",1!

7 ," ,# ",# ,4 ,0 ",! ,$ ",# ,! ," ,"$

8 ",$ ,! ,5 ,0 ," , ",1 ," , ",! ,!"

9 ", ",! ,5 ,4 ,5 ",3 ,5 ",! ,3 ,3 ,$0

10 ",! , ," ,# ",4 ,1 , ," "," ," ,!5

∑ $.4

1) %enghitung ratarata subgroup

= ∑ ik

  =69,64

10

  = 6,97

#) %enghitung standar de9iasi

σ =√∑(i-  )

2

N

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= √(6,40 – 6,97 )

2 + (6,60 – 6,97 )

2 + / + (6,70 – 6,97 )

2

100

; 0,5"

3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup

σ  =σ

√ k

σ  =0,57

√ 10

σ  = 0,18

4) 'itung nilai > tabel >t), dengan ? ; 0,$

Zt = Z

[1-

[1- β2   ]]

Zt = 1 [1- [1-0,96

2   ]]Zt = 0,98

>t ; #,0

5) %enentukan B dan BB untuk u(i keseragaman data

i) Batas kontrol atas B)

= + (Zt   σ) = 6,97 + (2,06 0,18 )

= ",33

ii) Batas kontrol ba-ah BB)

= - (Zt   σ)

= 6,97 + (2,06 0,18 )

= 6,56

) U(i ke&ukupan data dengan H ; 0,04

N'=[

Zt

α √N∑ i

2- (∑ i )

2

∑ i]

2

N'=[

2,06

0,04 √ 100(4881,82) – (696,4 )2

696,4   ]2

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N'=[

2,06

0,04 √ 488182 – 484973

696,4   ]2

N'=[

51,5 56,65696,4   ]

2

N'= 17,55

1 # 3 4 5 " ! $ 10.#0

.40.0

.!0

".00

".#0

".40

."$

".0 ".0$ ".0 ".05".1!

."$.!" .$0 .!5

U(i keseragaman data

B BB bar  

+ata subgroup

eterangan: 2C 2, data &ukup

<ambar 3.4.!. eta kontrol u(i keseragaman data lebar (ari #, 3, 4 dan 5

") Distribusi *rekuensii) angkauan +)

R = $- $I

R = 7,90 - 6,00

R = 1,$0

ii) Banyak kelas )

= 1 + 3,3

= 1 + 3,3 100

  ; ", 8

iii)   Interval G)

I =R

I =1,9

8

I = 0,24

Tabel 3.4.50. Tabel distribusi *rekuensi lebar (ari #, 3, 4 dan 5

 Interval  &i ' ()( *)( ()B *)B )$ 'i.)$

,00 ,#3 1# 1# ,00 ,#3 5,$$5 ,#35 ,1# "3,3!

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,#4 ,4" $ #1 ,#4 ,4" ,#35 ,4"5 ,3 5",#0

,4! ,"1 ## 43 ,4! ,"1 ,4"5 ,"15 ,0 145,0$

,"# ,$5 ! 51 ,"# ,$5 ,"15 ,$55 ,!4 54,!

,$ ",1$ $ 0 ,$ ",1$ ,$55 ",1$5 ",0! 3,!

",#0 ",43 1 " ",#0 ",43 ",1$5 ",435 ",3# 11",04

",44 "," " !3 ",44 "," ",435 ","5 ",5 5#,!$

",! ",$1 1" 100 ",! ",$1 ","5 ",$15 ",!0 13#,5#

+&i 100 +&i.)$ $,4

!)   Mean

 Mean  =∑ " i &

∑ " i

 Mean  =696,46

100

 Mean  = 6,96

$) %ediani) /etak median

/etak median  =1

2  N

/etak median  =1

2  100

/etak median  = 50

%edian terletak pada nilai ke50 pada Interval  ke4

ii) 2ilai median

%edian  = %&$()+ [1

2N - *k$()-1

"$()]×I

%edian  = 6,715 + [50 - 43

8   ]× 0,24

%edian  = 6,715 + [78 ]×0,24

%edian  = 6,715 + 0,21

%edian  =  ,$3

10) %odus

%odus = %&$ + [

"$ - "$-1

( "$- "$-1 )+ ( "$ - "$+1 ) ]× I

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%odus = 6,475 + [22 - 9

(22 - 9 )  + (22 - 8 ) ]   ×  0,#4

%odus = 6,475 +

[13

 13 + 14

 ] ×

 0,#4

%odus  = 6,475 + [13

27 ]× 0,#4

%odus  = 6,475 + 0,11

%odus  = 6,59

11) ersentili) /etak 5

%tk 5  =

I

100   N

%tk 5  =5

100  100

%tk 5  = 5

ersentil 5 terletak pada data ke5 pada Interval  ke1

ii) /etak 50

%tk 50  =I

100  N

%tk 50  =50

100  100

%tk 50  = 50

ersentil 50 terletak pada data ke50 pada Interval  ke4

iii) /etak $5

%tk 95 =I

100  N

%tk 95 =

95

100   100

%tk 95 = 95

ersentil $5 terletak pada data ke$5 pada Interval  ke!

i9) 2ilai 5

5  = %&5 + [I

100N - *k5-1

"5]× I

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5  = 5,995 + [5 - 0

12   ]   ×  0,#4

5  = 5,995 +

[5

12

 ]×

 0,#4

5  = 5,995 + 0,10 

5  = 6,10 

9) 2ilai 50

50  = %& 50  + [I

100N - *k50-1

"50]×I

50  = 6,715 + [50 - 438   ]   ×  0,#4

50  = 6,715 + [78 ]×0,24

50  = 6,715 + 0,21

50  = 6,93

9i) 2ilai $5

95  = %&95 +[I

100N - *k95-1

" 95]×I

95  = 7,675 + [95 - 83

17   ]×0,24

95  = 7,675 + [12

17 ]×0,24

95  = 7,675 + 0,17

95  = 7,84

e) angkauan tanganTabel 3.4.51. angkauan tangan

% 

n1 2 3 4 5 6 7 8 9 10   ´ x  

1 "1,#5 $,3 "4,0 4,3 5," ",4 !,$ $,3 "4,$ !0,! "1,#5

2 "0," 5,0 5,0 "#,! 3," 5,4 "5,3 !0,# ", !0,0 "0,$

3 "4,11 3," 3," "4,4 "!,3 "!,4 "!," ""," "$,0 $,$ "3,3!

4 "0,$4 "1,$ "1,$ "3," ",# ,0 3,4 3,# "!,# "5,! "0,5"

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5 "1,"# 1, 1, !0,1 ",5 $,0 "5,0 !1,0 ,3 5,3 "1,01

6 "0,1 1," 1," "",! 1,$ #,3 "1,! "1,3 !,! ",5 $,1$

7 $,$ $," $," ",5 "1,0 $,3 #,3 "",# "4,0 !,! $,!

8 "1,#3 "4,! "4,! "1, #,3 "#,! ",5 ",5 "3,0 "$,! "1,#"

9 "1,0$ ",$ ",$ $," "3, "0,1 "!," 1,4 "0,$ 5,$ "#,1

10 $,"$ ",1 ",1 !0," $,0 ,1 "1,5 !0,0 5,$ #,# $,#5

+  ´ X  "11,4

1) %enghitung rararata subgroup

= ∑ k

= ∑ 711,46

10

=71,15  

#) %enghitung standar de9iasi

σ =√∑ ( x1−´ x )2

 N 

σ = √ (77,9-70,89 )2 + .... + (62,2 – 70,89 )2

100

σ = 5,63

3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup

σ =σ

√ k

σ =5,63

√ 10

σ = 1,78

4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!

Zt = Z[1-[1- β

2   ]]Zt = 1 [1- [1-0,96

2   ]]Zt = 0,98

>t ; #,0

5) U(i keseragaman data

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i) Batas kontrol atas B)

B= x+(Z t x σx )

B = 71,15+(2,06 x 1,"!)

B ; "4,!1ii) Batas kontrol ba-ah BB)

BB= x-(Zt  x σ x )

BB = 71,15-(2,06 x 1,78)

BB ; ",4!

) U(i ke&ukupan data

N'=[

Zt

α √N∑ i

2- (∑ i )

2

∑i]

2

N'=[

2,06

0,04 √ 50(5115,83) - (502539,21)

7089   ]2

 2C ; 1,0eterangan: 2C 2, data &ukup

1 # 3 4 5 " ! $ 10#.00

4.00

.00

!.00

"0.00

"#.00

"4.00

".00

"1.#5"0."

"4.11

"0.$4"1."#

"0.1$.$

"1.#3 "1.0$

$."$

U(i eseragaman data

B BB bar  

+ata subgroup

<ambar 3.4.$. eta kontrol u(i keseragaman data (angkauan tangan

") Distribusi *rekuensii) angkauan +)

+ ; DmaM Dmin

+ ; !1,00 E 1,40+ ; 1$,0

ii) Banyak kelas ; 1 F 3,3 log n ; 1 F 3,3 log 100

; ", K $

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iii) Interval  G)

I= R

 K 

I=19,60

9

I = 2,18

Tabel 3.4.5#. Tabel distribusi *rekuensi T

 Interval  &i ' ()( *)( ()B *)B )$ &i.)$

1,40 3,5" 13 13 1,40 3,5" 1,3$5 3,5"5 #,4$ !1#,31

3,5! 5,"5 " #0 3,5! 5,"5 3,5"5 5,"55 4," 45#,

5," ",$3 1# 3# 5," ",$3 5,"55 ",$35 ,!5 !0#,14

",$4 "0,11 15 4" ",$4 "0,11 ",$35 "0,115 $,03 1035,3!

"0,1# "#,#$ 10 5" "0,1# "#,#$ "0,115 "#,#$5 "1,#1 "1#,05

"#,30 "4,4" 1# $ "#,30 "4,4" "#,#$5 "4,4"5 "3,3$ !!0,#

"4,4! ",5 10 "$ "4,4! ",5 "4,4"5 ",55 "5,5" "55,5

", "!,!3 1# $1 ", "!,!3 ",55 "!,!35 "","5 $3#,$4

"!,!4 !1,01 $ 100 "!,!4 !1,01 "!,!35 !1,015 "$,$3 "1$,33

+&i 100 +&i.)$ 4$5,15

!)   Meanx =

!"i.#$

!"i

x =4695,15

100

x =46,95

$) %ediani) /etak median

/etak median ;

1

2 x N

/etak median ;1

2 x 100

/etak median ; 50

ii) 2ilai median

i=%&$ + [1

2N- (*$-1)

"$]   I

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i=70,115+ [1

2  100- 47

10   ]   2,18

i=70,115+ [50-4710   ] 2,18

i=70,115+ [ 0,26]   2,18

i=70,115+ 0,654

i=70,77

10) %odus

)=%& $) + [   ("$) -" $)-1)("$) - "$)+1 ) + ("$) - "$)-1 ) ]   I

=67,935+ [15-12

(15-10 )+ (15-12) ]   2,18

=67,935+ [ 3

5+3 ]   2,18

=67,935+0,82

=¿  !,"5

11) ersentili)/etak 5

%tk 5  =I

100  N

%tk 5  =5

100  100

%tk 5  = 5

ersentil 5 terletak pada data ke5 pada Interval  ke1

ii) /etak 50

%tk 50  =I100

  N

%tk 50  =50

100  100

%tk 50  = 50

ersentil 50 terletak pada data ke50 pada Interval  ke4

iii) /etak $5

%tk 95 =I

100  N

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%tk 95  =95

100  100

%tk 95 = 95

ersentil $5 terletak pada data ke$5 pada Interval  ke!

i9) 2ilai 5

5  = %&5 + [I

100N - *k5-1

"5] I

5  = 61,395 + [5

100100 - 0

10   ] 2,45

5  = #,#3

9) 2ilai 50

50  = %& 50+

I

100N - *k50-1

"50

I

50  = 70,115 + 2,18 [50

100100 - 36

17   ]   2,18

50 = 70,77

9i) 2ilai $5

95  = %&95 + [I

100N - *k95-1

" 95]   I

95

 = 78,335+

[

95

100100 - 89

11

  ] 2,18

50 = 79,80

4.2.1.1.3. Pintu

%eran&ang pintu yang sesuai data anthropometri praktikan mata kuliah IJ. DataanthropometrI yang digunakan untuk meran&ang pintu adalah tinggi badan berdiri tbb), tinggi bahu berdiri thb), lebar bahu, lebar (ari #,3,4,5 dan pan(ang (ari 3. /alu perhitungannya adalahsebagai berikut:a, Tinggi badan berdiri tbb)

Tabel 3.4.53. Data antrhopometri tinggi badan berdiri tbb)

-o 1 2 3 4 5 6 7 8 9 10   ×

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1154,00 1!,40 151,0 15,"0 1"#,0 1"1,"0 14!,10 14,!0 1",00 15#,10 11,0

21"$,#0 1#,10 1"!,$0 1"5,0 15#,$0 1"!,40 1"!,!0 10,00 155,10 154,30 1",53

31",#0 1$,30 1!0,"0 1!1,10 1"3,$0 1"0,"0 1$,00 1!0,30 1"3,00 154,!0 1"#,00

4 10,50 1"5,00 153,00 13,30 151,40 14!,#0 14$,$0 1"$,50 14,00 1"5,!0 1#,0

514",10 15,00 15$,#0 155,10 1"#,$0 1!3,50 154,0 154,"0 15!,!0 1#,!0 11,3"

61"5,50 151,"0 10,$0 1"4,0 15!,10 1#,"0 153,0 15#,50 155,0 15,10 10,13

71!#,!0 1!,40 1"3,50 1!0,"0 1"3,30 1"4,00 15$,40 1"$,40 11,#0 14",$0 1"0,0

81"4,40 1#,40 150,0 1"3,00 15,#0 10,"0 1!3,50 1!3,#0 1!3,40 1!0,30 1"1,"

910,#0 1$,$0 1!0,$0 1!1,50 151,00 151,00 1"0,!0 1"1,$0 1"!,00 15,"0 1!,0$

+× 14$4,51

1) %enghitung rararata subgroup

= ∑ k

= ∑ 1494,51

9

=  1,0

#) %enghitung standar de9iasi

σ=

√∑ (x1 - x )

2

N

σ = √ (154,00- 14 9 ,45 )2 + .... + (165,70 – 149,45 )2

90

σ = 10,!5

3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup

σ =σ

√ k

σ = 10,85

√ 9

σ =3,#

4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!

Zt = Z[1-[1- β

2   ]]

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Zt = 1 [1- [1-0,96

2   ]]Zt = 0,98

>t ; #,0

5) U(i keseragaman datai) Batas kontrol atas B)

B= x + ( Zt x σx )

B =166,09 + (2,06 x 3,#)

B ; 1"3,51ii) Batas kontrol ba-ah BB)

BB = x - ( Zt  x σx )

BB = 166,09-(2,06 x 3,62)BB ; 15!,1

) U(i ke&ukupan data

N'=[

Zt

α √N∑ i

2- (∑ i )

2

∑i]

2

N'=

[

2,06

0,04 √ 90(24 .922. 328 ,59) - (223 .356 .014,01)

14 .945,10

  ]

2

 2C ; 11,3

eterangan: 2C 2, data &ukup

1 # 3 4 5 " ! $150.00

155.00

10.00

15.00

1"0.00

1"5.00

11.0

1".53

1"#.00

1#.0 11.3"10.13

1"0.01"1."

1!.0$

U(i keseragaman data

B BB bar  

+ata subgroup

<ambar 3.4.10. eta kontrol u(i keseragaman data TBB

") Distribusi *rekuensii) angkauan +)

+ ; DmaM Dmin

+ ; 1!3,50 E 14",10

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+ ; 3,40

ii) Banyak kelas ; 1 F 3,3 log n ; 1 F 3,3 log 100

; ",0 K $

iii)   Interval  G)

I =R

I=36,40

9

I =  4,05

Tabel 3.4.54. Tabel distrbusi *rekuensi TBB

 Interval  &i ' ()( *)( ()B *)B )$ 'i.)$

14",10 151,14 ! ! 14",10 151,14 14",0$5 151,145 14$,1# 11$#,$

151,15 155,1$ 15 #3 151,15 155,1$ 151,145 155,1$5 153,1" ##$",55

155,#0 15$,#4 5 #! 155,#0 15$,#4 155,1$5 15$,#45 15",## "!,10

15$,#5 13,#$ 11 3$ 15$,#5 13,#$ 15$,#45 13,#$5 11,#" 1""3,$"

13,30 1",34 $ 4! 13,30 1",34 13,#$5 1",345 15,3# 14!",!!

1",35 1"1,3$ " 55 1",35 1"1,3$ 1",345 1"1,3$5 1$,3" 11!5,5$

1"1,40 1"5,44 13 ! 1"1,40 1"5,44 1"1,3$5 1"5,445 1"3,4# ##54,4

1"5,45 1"$,4$ $ "" 1"5,45 1"$,4$ 1"5,445 1"$,4$5 1"",4" 15$",#3

1"$,50 1!3,54 13 $0 1"$,50 1!3,54 1"$,4$5 1!3,545 1!1,5# #35$,"

+&i $0 +&i.)$ 14$35,50

!) Mean

´ x=∑fi.cm

∑fi

´ x=14935,50

90

´ x=  15,$5

$) %ediani) /etak median

/etak median ;1

2  N

/etak median ;1

2  90

/etak median ; 45

ii) 2ilai median

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 Median=%& $+ [1

2N- ( *$-1 )

"$]   I

 Median=163,295+ [12

  90 - 39

9   ]   4,05

 Median=163,295 + [0,66 ]   4,05

 Median=163,295 +  #,"0

 Median=   11,$5

10)%odus

)=%& $) + [   (" $) - " $)-1 )("$) - "$)+1) + ("$) -"$)-1 ) ]   I

=151,145 + [15 - 8

(15-8 )+ (15-10) ]   4,05

=151,145 + [ 7

7+5 ]   4,05

= 151,145 + #,3

=¿  15#,!1

11)ersentili) /etak 5

%tk 5  =I

100  N

%tk 5  =5

100  90

%tk 5  =4,5

ersentil 5 terletak pada data ke5 pada Interval  ke1

ii) /etak 50

%tk 50  =I

100  N

%tk 50  =50

100  90

%tk 50  = 45

ersentil 50 terletak pada data ke50 pada Interval  ke4

iii) /etak $5

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%tk 95 =I

100  N

%tk 95 =95

100

  90

%tk 95 = 85,5

ersentil $5 terletak pada data ke$5 pada Interval  ke!

i9) 2ilai 5

5  = %&5 + I [I

100N - *k5-1

"5]

5  = 147,095 + 4,05

[5

10090 - 0

8   ]5  =14$,3

9) 2ilai 50

50  = %& 50+ I [I

100N - *k50-1

"50]

50  = 163,295 + 4,05 [50100

90 - 39

9   ]50 =

1!,#5

9i) 2ilai $5

95  = %&95 + I [I

100N - *k95-1

" 95]

95  = 179,495+ 4,05 [95

10090 - 77

13   ]50 =

1!5,10

 b) Tinggi bahu berdiri thb)Tabel 3.4.55. data antrhopometri tinggi bahu berdiri thb)

n 

1 2 3 4 5 6 7 8 9 10   ´ x

115#,"

0

14#,3

0

14!,#

0

140,5

0

13!,"

0

14$,0

0

14#,3

0

1#$,

0

13,5

0

13",#

0 141,"0

2 143,30

133,"0

135,!0

14,30

1#5,00

14$,"0

150,50

13!,30

13,40

151,40 141,04

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>t ; #,0

5) U(i keseragaman datai) Batas kontrol atas B)

= + (Zt   σ)

= 139,56 + (2,06 3,01 )

= 145,"

ii) Batas kontrol ba-ah BB)

= - (Zt   σ)

= 139,56 + (2,06 3,01 )

= 133,35

) U(i ke&ukupan data dengan H ; 0,04

N'=[

Zt

α √N∑ i

2- (∑ i )

2

∑ i]

2

N'=[

2,06

0,04 √ 100(1956693) – (13955,7 )2

13955,7   ]2

N'=

[2,06

0,04 √ 195669261 – 194761562,5

13955,7   ]2

N'=[51,5 952,73

13955,7   ]2

N'= 12,36

eterangan: 2C 2, data &ukup

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1#5

130

135

140

145

150

141." 141.04 141.

13".13.43 13.05

14#.5#

13".5$13!.$"

14#.0"

U(i keseragaman data

B BB bar  

+ata subgroup

<ambar 3.4.11. peta kontrol u(i keseragaman data thb

") Distribusi *rekuensi

i) angkauan +)R = $- $i

R = 155,0 - 122,1

R = 32,9  

ii) Banyak kelas )

= 1 + 3,3

= 1 + 3,3 100

= 7,6  K !

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iii) Interval G)

I =R

I = 32,98

I = 4,12

Tabel 3.4.5. Tabel distribusi *rekuensi T'B)

 Interval  &i ' ()( *)( ()B *)B )$ &i.)$

1##,10 1#,#1 14 14 1##,10 1#,#1 1##,0$5 1#,#15 1#4,1 1"3!,1"

1#,## 130,33 " #1 1#,## 130,33 1#,#15 130,335 1#!,#! !$",$3

130,34 134,45 11 3# 130,34 134,45 130,335 134,455 13#,40 145,35

134,4 13!,5"15 4" 134,4 13!,5" 134,455 13!,5"5 13,5# #04","3

13!,5! 14#,$ 10 5" 13!,5! 14#,$ 13!,5"5 14#,$5 140,4 140,35

14#,"0 14,!1 1! "5 14#,"0 14,!1 14#,$5 14,!15 144," #05,5$

14,!# 150,$3 11 ! 14,!# 150,$3 14,!15 150,$35 14!,!! 13",3

150,$4 155,05 14 100 150,$4 155,05 150,$35 155,055 153,00 #141,$3

+&i 100 +&i.)$ 13$31,

!)   Mean

 Mean  =∑ " i &

∑ " i

 Mean  = 13931,66100

 Mean  = 13$,3#

$) %ediani) /etak median

/etak median  =1

2  N

/etak median  =1

2  100

/etak median  = 50

%edian terletak pada nilai ke50 pada Interval  ke5

ii) 2ilai median

%edian  = %&$+ [1

2N - *k$-1

"$]× G

%edian  = 138,53+ [50 - 47

10   ]× 4,12

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%edian  = 138,53 + [310 ]×4,12

%edian  = 138,53 + 1,24

%edian  =  13$,""

10)%odus

%odus = %&$ + [

"$ - "$-1

( "$ - "$-1 )+( "$ - "$+1 ) ]× G

%odus = 142,65 + [18 - 10

(18 - 10 )  + (18 - 11 ) ]   ×  4,1#

%odus = 142,65 +

[8

 8 + 7

 ]  ×  4,1#

%odus  = 142,65 + [815 ]× 4,1#

%odus  = 142,65 + 2,20

%odus  = 144,85

11)ersentili) /etak 5

%tk 5  =I

100   N

%(tk 5  =5

100  100

%tk 5  = 5

ersentil 5 terletak pada data ke5 pada Interval  ke1

ii) /etak 50

%tk 50  =I

100  N

%(tk 50  =50

100  100

%tk 50  = 50

ersentil 50 terletak pada data ke50 pada Interval  ke5

iii) /etak $5

%tk 95 =I

100  N

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%tk 95 =95

100  100

%tk 95 = 95

ersentil $5 terletak pada data ke$5 pada Interval  ke!

i9) 2ilai 5

5  = %&5 + [I

100N - *k5-1

"5]×  G

5  = 122,055 + [5 - 0

14   ]   ×  4,1#

5 = 122,055 +

[5

14

 ]×

 4,1#

5  = 122,055 + 1,47 

5  = 123,57 

9) 2ilai 50

50  = %& 50  + [I

100N - *k50-1

"50]×I

50  = 138,535 + [50 - 47

10   ]   ×  4,1#

50  = 138,535 + [310 ]×4,12

50  = 138,535 + 1,24

50  = 139,81

9i) 2ilai $5

95  = %&95 +[I

100N - *k95-1

" 95 ]×I

95  = 150,895 + [95 - 86

14   ]×4,12

95  = 150,895 +[914 ]×4,12

95  = 150,895 + 2,65

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95  = 153,58

&) /ebar bahu lb)Tabel 3.4.5" Data anthropometri lebar bahu lb)

n 

1 2 3 4 5 6 7 8 9 10   ´ x

1 40,0 41,40 3$,0 40,0 4,#0 3!,30 35,40 44,"0 3",!0 3,50 40,11

2 45,10 3",#0 3,30 3,#0 40,50 3,00 3!,!0 44,10 41,!0 45,00 40,10

3 44,40 3$,#0 41,"0 3!,10 4#,10 3!,0 35,!0 34,0 41,!0 34,$0 3$,1#

4 3!,30 41,10 3,10 3!,50 3$,10 3,40 3",#0 34,00 41,30 3!,"0 3!,0"

5 34,$0 3,10 41,40 4#,0 45,30 3$,$0 3,10 40,!0 35,30 45,00 3$,"4

6 3,40 3!,!0 3",0 35,0 40,#0 44,30 44,50 3!,"0 3",!0 3$,30 3$,3#

7 45,00 44,"0 3$,40 3,!0 40,00 40,#0 44,!0 34,!0 40,!0 35,00 40,15

8 3!,!0 34,10 43,10 3$,50 40,"0 3,#0 34,"0 4#,$0 3!,00 35,30 3!,33

9 4,#0 34,!0 4,00 3,$0 40,"0 3,#0 4#,"0 43,50 41,"0 3",50 40,#

10 35,#0 35,10 3",10 4#,50 4,00 4#,#0 3$,40 45,!0 44,50 3",50 40,53

L   ´ x 3$,0$

1) %enghitung ratarata subgroup

=∑ i

k

  =

396,09

10

  = 39,61

#) %enghitung standar de9iasi

σ =√∑(i- )

2

N

  = √(40,60 – 39,61 )2 + (41,40 – 39,61 )2 + / + (37,50 – 39,61 )2

100

  ; 3,4"

3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup

σ  =σ

√ k

 σ  =3,47

√ 10

 σ  = 1,10

4) 'itung nilai > tabel >t), dengan ? ; 0,$

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Zt = Z[1-[1- β

2   ]]

Zt = 1

[1-

[1-0,96

2

  ]]Zt = 0,98

>t ; #,0

5) U(i keseragaman dataa) Batas kontrol atas B)

= + (Zt   σ )

= 39,60 + (2,06 1,10 )

= 41,!"

 b) Batas kontrol ba-ah BB)

= - (Zt   σ)

= 39,60 + (2,06 1,10 )

= 37,35

) U(i ke&ukupan data dengan H ; 0,04

N

'

=

[

Zt

α √N∑ i

2- (∑ i )

2

∑ i

]

2

N'=[

2,06

0,04 √ 100(158093,8) – (3960,9 )2

3960,9   ]2

N'=[

2,06

0,04 √ 15809383 – 15688729

3960,9   ]2

N' =[51,5 347,353960,9   ]

2

N'= 20,42

eterangan: 2C 2, data &ukup

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1 # 3 4 5 " ! $ 103.00

3".00

3!.00

3$.00

40.00

41.00

4#.00

40.11 40.10

3$.1#

3!.0" 3!.0"

3$."43$.3#

40.15

3!.33

40.#

U(i keseragaman data

B BB bar  

+ata subgroup

<ambar 3.4.1#. eta kontrol u(i keseragaman data /B

") Distribusi *rekuensia) angkauan +)

R = $- $I

R = 46,2 - 34

R = 12,2  

 b) Banyak kelas )

= 1 + 3,3

= 1 + 3,3 100

= 7,6  K !

&)   Interval G)

I =R

I =12,2

8

I = 1,53

Tabel 3.4.5!. Tabel distribusi *rekuensi lebar bahu

 Interval  &i ' ()( *)( ()B *)B )$ &i.)$

34,00 35,5# 14 14 34,00 35,5# 33,$$5 35,5#5 34," 4!,4

35,53 3",05 15 #$ 35,53 3",05 35,5#5 3",055 3,#$ 544,35

3",0 3!,5! 13 4# 3",0 3!,5! 3",055 3!,5!5 3",!# 4$1,

3!,5$ 40,11 15 5" 3!,5$ 40,11 3!,5!5 40,115 3$,35 5$0,#5

40,1# 41,4 13 "0 40,1# 41,4 40,115 41,45 40,!! 531,44

41,5 43,1" 11 !1 41,5 43,1" 41,45 43,1"5 4#,41 4,51

43,1! 44,"0 ! !$ 43,1! 44,"0 43,1"5 44,"05 43,$4 351,5#

44,"1 4,#3 11 100 44,"1 4,#3 44,"05 4,#35 45,4" 500,1"

+&i 100 +&i.)$ 3$#,54

!)   Mean

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 Mean  =∑ " i &

∑ " i

 Mean =

3962,54

100

 Mean  = 39,63

$) %ediani) /etak median

/etak median  =1

2  N

/etak median  =1

2  100

/etak median  = 50

%edian terletak pada nilai ke50 pada Interval  ke4

ii) 2ilai median

%edian  = %&$()+ [1

2N - *k$()-1

"$()]×I

%edian  = 38,545 + [50 - 42

15   ]× 1,53

%edian  = 38,545 + [815 ]×1,53

%edian  = 38,545 + 0,82

%edian  =  3$,3

iii) %odus

%odus = %&$+ [

"$ - "$-1

( "$ - "$-1)+ ( "$ - "$+1 ) ]× I

%odus = 38,545 + [15 - 13

(15 - 13)  + (15 - 13 ) ]   ×  1,53

%odus  = 38,545 + [2 2 + 2 ]   ×  1,53

%odus  = 38,545 + [24 ]× 1,53

%odus  = 38,545 + 0,77

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%odus  = 39,31

10) ersentili) /etak 5

%tk 5  = I100

  N

%tk 5  =5

100  100

%tk 5  = 5

ersentil 5 terletak pada data ke5 pada Interval  ke1

ii) /etak 50

%tk 50  =I

100

  N

%tk 50  =50

100  100

%tk 50  = 50

ersentil 50 terletak pada data ke50 pada Interval  ke4

iii) /etak $5

%tk 95  =I

100  N

%tk 95  =95100

  100

%tk 95  = 95

ersentil $5 terletak pada data ke$5 pada Interval  ke!

i9) 2ilai 5

5  = %&5 + [I

100N - *k5-1

"5]× I

5  = 33,955 + [5 - 014   ]   ×  1,53

5  = 33,955 + [514 ]×  1,53

5  = 33,955 + 0,55 

5  = 34,50 

9) 2ilai 50

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50 = %& 50  + [I

100N - *k50-1

"50]×I

50 = 38,545 + [50 - 4215   ]   ×  1,53

50 = 38,545 + [815 ]×1,53

50 = 38,545 + 0,82

50 = 39,36

9i) 2ilai $5

95  = %&95 +[I100

N - *k95-1

" 95 ]×I

95  = 44,665 + [95 - 89

11   ]×1,53

95  = 44,665 + [611 ]×1,53

95  = 44,665 + 0,83

95  = 45,49

d) an(ang (ari 3Tabel 3.4.5$. Data antrhopometri pan(ang (ari 3

  n

 1 2 3 4 5 6 7 8 9 10   ´ x

1 !,#0 !,#0 !,$0 ",10 !,0 !,10 ",30 $,00 !,30 ",0 !,13

2 ",10 ",!0 !,$0 ",10 ",30 ",40 !,00 $,00 ",$0 !,!0 ",$3

3 ",#0 ",10 !,30 !,50 !,40 ",50 !,40 ",10 ",#0 ",#0 ",$

4 !,30 !,!0 !,30 !,!0 !,0 $,00 ",30 ",10 ",00 ",#0 !,04

5 ",$0 ","0 !,#0 ",$0 $,00 ",!0 ",0 !,40 ",10 ",#0 ",!!

6 !,00 !,0 ",50 ",!0 ",!0 !,0 !,30 !,!0 !,10 !,40 !,1$

7 ",30 ",00 !,"0 ",!0 !,0 ","0 ",$0 ",#0 ",10 ",30 ",

8 ",40 ",!0 ",0 !,0 !,50 ",#0 !,30 ",$0 !,50 ",40 ",$#

9 !,30 ",0 ",0 ",50 !,40 !,#0 ",#0 ",!0 ",10 !,$0 ",!

10 !,10 !,$0 ",$0 ","0 !,10 !,00 ",00 ","0 !,10 ",30 ",!!

∑ ´ "$,1!

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1) %enghitung ratarata subgroup

=∑ i

k

  =79,18

10

  = 7,92

#) %enghitung standar de9iasi

σ =√∑(i- )

2

N

  =

√

(8,20 – 7,92 )2 + (8,20 – 7,92 )2 + / + (7,30 – 7,92 )2

100

  ; 0,5$

3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup

σ  =σ

√ k

σ  =0,59

√ 10

σ  = 0,19

4) 'itung nilai > tabel >t), dengan ? ; 0,$

Zt = Z[1-[1- β

2   ]]Zt = 1 [1- [1-0,96

2   ]]Zt = 0,98

>t ; #,0

5) U(i keseragaman dataa) Batas kontrol atas B)

= + (Zt   σ )

= 7,92 + (2,06 0,19 )

= !,31

 b) Batas kontrol ba-ah BB)

= - (Zt   σ)

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= 7,92 - (2,06 0,19 )

= 7,53

) U(i ke&ukupan data dengan H ; 0,04

N'=[

Zt

α √N∑ i

2- (∑ i )

2

∑ i]

2

N'=[

2,06

0,04 √ 100(6305,32) – (791,80)2

791,80   ]2

N' =

[2,06

0,04

 √ 630532 – 626947,24

791,80   ]2

N'=[51,5 59,87

791,80   ]2

N'= 15,16

eterangan: 2C 2, data &ukup

1 # 3 4 5 " ! $ 10".00

".#0

".40

".0

".!0

!.00

!.#0

!.40

!.13

".$3

".$

!.04 !.04

".!!

!.1$

".

".$#".!

U(i keseragaman data

B BB bar  +ata ,bgrup

<ambar 3.4.13. eta kontrol u(i keseragaman data 3

") Distribusi *rekuensii) angkauan +)

R = $- $I

R = 9 - 7

R = 2  

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ii) Banyak kelas )

= 1 + 3,3

= 1 + 3,3 100

= 7,6≈7

iii)   Interval G)

I =R

I =2

7

I = 0,29

Tabel 3.4.0. Tabel distribusi pan(ang (ari 3 Interval  &i ' ()( *)( ()B *)B )$ &i.)$

",00 ",#! #0 #0 ",00 ",#! ,$$5 ",#!5 ",14 14#,!0

",#$ ",5" 1# 3# ",#$ ",5" ",#!5 ",5"5 ",43 !$,1

",5! ",! 1 4! ",5! ",! ",5"5 ",!5 ","# 1#3,5#

",!" !,15 14 # ",!" !,15 ",!5 !,155 !,01 11#,14

!,1 !,44 1 "! !,1 !,44 !,155 !,445 !,30 13#,!0

!,45 !,"3 10 !! !,45 !,"3 !,445 !,"35 !,5$ !5,$0

!,"4 $,0# 1# 100 !,"4 $,0# !,"35 $,0#5 !,!! 10,5

+&i 100 +&i.)$ "$#,!!

8) Mean

 Mean  = ∑ " i &

∑ " i

 Mean  =792,88

100

 Mean  = 7,93

$) %ediani) /etak median

/etak median  =1

2   N

/etak median  =1

2  100

/etak median  = 50

%edian terletak pada nilai ke50 pada inter9al ke4

ii) 2ilai median

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%edian  = %&$()+ [1

2N - *k$()-1

"$()]×I

%edian  = 7,87 + [50 - 4814   ]× 0,29

%edian  = 7,87 + [214 ]×0,29

%edian  = 7,87 + 0,04

%edian  =  ",$1

10) %odus

%odus = %&$ +

["$ - "$-1

( "$ - "$-1 )+( "$ - "$+1 ) ]× I

%odus = 6,99 + [20 - 0

(20 - 0 ) + (20 - 12) ]   ×  0,#$

%odus  = 6,99 + [20

 20 + 8 ]   ×  0,#$

%odus  = 6,99 + [20

28 ]× 0,#$

%odus  = 6,99 + 0,20

%odus  = 7,20

11) ersentili) /etak 5

%tk 5  =I

100  N

%tk 5  =

5

100   100

%tk 5  = 5

ersentil 5 terletak pada data ke5 pada inter9al ke1

ii) /etak 50

%tk 50  =I

100  N

%tk 50  =50

100  100

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%tk 50  = 50

ersentil 50 terletak pada data ke50 pada inter9al ke4

iii) /etak $5

%tk 95 = I100

  N

%tk 95 =95

100  100

%tk 95 = 95

ersentil $5 terletak pada data ke$5 pada inter9al ke"

i9) 2ilai 5

5  = %&5 +

I

100N - *k5-1

"5

× I

5  = 6,99 + [5 - 0

20   ]   ×  0,#$

5  = 6,99 + [520 ]×  0,#$

5  = 6,99 + 0,07 

5

 = 7,06 

9) 2ilai 50

50  = %& 50  + [I

100N - *k50-1

"50]×I

50  = 7,87 + [50 - 48

14   ]   ×  0,#$

50  = 7,87+

[2

14

 ]×0,29

50  = 7,87 + (0.04)

50  = 7,91

9i) 2ilai $5

95  = %&95 +[I

100N - *k95-1

" 95]×I

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95  = 8,74 + [95 - 88

12   ]×0,29

95  = 8,74 +

[7

12

 ]×0,29

95  = 8,74 + 0,16

95  = 8,90

e) /ebar (ari #,3,4 dan 5Tabel 3.4.1. Data anthropometri lebar (ari #,3,4 dan 5

  n

 1 2 3 4 5 6 7 8 9 10   ´ x

1 ,4 , ,! ",1 ",! ",$ ,4 ,3 ,3 ,3 ,"$

2 ",5 ,5 ,# ",3 ",1 ,0 ", ",$ ", ,$ ",0

3 ",5 , ",4 ",1 ",# ,# ",4 "," ",3 ,5 ",0$

4 ", ,0 ",3 ",! ",0 ",3 , "," ,# ",1 ",0

5 "," ",# ",0 ," , ",$ ,$ ,1 ,! ", ",05

6 ",3 "," ",1 ",# ",$ ",3 ",0 ",4 ,$ ,0 ",1!

7 ," ,# ",# ,4 ,0 ",! ,$ ",# ,! ," ,"$

8 ",$ ,! ,5 ,0 ," , ",1 ," , ",! ,!"

9 ", ",! ,5 ,4 ,5 ",3 ,5 ",! ,3 ,3 ,$0

10 ",! , ," ,# ",4 ,1 , ," "," ," ,!5

∑ $.4

1) %enghitung ratarata subgroup

=∑ i

k

  =69,64

10

  = 6,97

#) %enghitung standar de9iasi

σ =√∑(i- )

2

N

= √(6,40 – 6,97 )2 + (6,60 – 6,97 )2  + / + (6,70 – 6,97 )2

100

; 0,5"

3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup

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σ  =σ

√ k

σ  =0,57

√ 10σ  = 0,18

4) 'itung nilai > tabel >t), dengan ? ; 0,$

Zt = Z[1-[1- β

2   ]]Zt = 1 [1- [1-0,96

2   ]]Zt = 0,98

>t ; #,0

5) %enentukan B dan BB untuk u(i keseragaman datai) Batas kontrol atas B)

= + (Zt   σ)

= 6,97 + (2,06 0,18 )

= ",33

ii) Batas kontrol ba-ah BB) = - (Zt   σ)

= 6,97 + (2,06 0,18 )

= 6,56

) U(i ke&ukupan data dengan H ; 0,04

N'=[

Zt

α √N∑ i

2- (∑ i )

2

∑ i]

2

N'=[

2,06

0,04 √ 100(4881,82) – (696,4 )2

696,4   ]2

N'=[

2,06

0,04 √ 488182 – 484973

696,4   ]2

N'=

[

51,5 56,65

696,4

  ]

2

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N'= 17,55

eterangan: 2C 2, data &ukup

1 # 3 4 5 " ! $ 10

.#0

.40

.0

.!0

".00

".#0

".40

."$

".0 ".0$ ".0 ".05

".1!

."$.!" .$0

.!5

U(i keseragaman data

B BB bar  

+ata subgroup

<ambar 3.4.14. eta kontrol u(i keseragaman data lebar (ari #, 3, 4 dan 5

") Distribusi *rekuensii) angkauan +)

R = $- $I

R = 7,90 - 6,00

R = 1,$0

ii) Banyak kelas )

= 1 + 3,3

= 1 + 3,3 100

  ; ", 8

iii)   Interval G)

I =R

I =1,9

8

I = 0,24

Tabel 3.4.#. Tabel distribusi *rekuensi lebar (ari #, 3, 4 dan 5

 Interval  &i ' ()( *)( ()B *)B )$ 'i.)$

,00 ,#3 1# 1# ,00 ,#3 5,$$5 ,#35 ,1# "3,3!

,#4 ,4" $ #1 ,#4 ,4" ,#35 ,4"5 ,3 5",#0

,4! ,"1 ## 43 ,4! ,"1 ,4"5 ,"15 ,0 145,0$

,"# ,$5 ! 51 ,"# ,$5 ,"15 ,$55 ,!4 54,!

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,$ ",1$ $ 0 ,$ ",1$ ,$55 ",1$5 ",0! 3,!

",#0 ",43 1 " ",#0 ",43 ",1$5 ",435 ",3# 11",04

",44 "," " !3 ",44 "," ",435 ","5 ",5 5#,!$

",! ",$1 1" 100 ",! ",$1 ","5 ",$15 ",!0 13#,5#

+&i 100 +&i.)$ $,4

!)   Mean

 Mean  =∑ " i &

∑ " i

 Mean =

696,46

100

 Mean  = 6,96

$) %ediani) /etak median

/etak median  =1

2  N

/etak median  =1

2  100

/etak median  = 50

%edian terletak pada nilai ke50 pada Interval  ke4

ii) 2ilai median

%edian  = %&$()+ [1

2N - *k$()-1

"$()]×I

%edian  = 6,715 + [50 - 43

8   ]× 0,24

%edian  = 6,715 + [78 ]×0,24

%edian  = 6,715 + 0,21

%edian  =  ,$3

10) %odus

%odus = %&$ + [

"$ - "$-1

( "$ - "$-1 )+ ( "$ - "$+1 ) ]× I

%odus = 6,475 + [22 - 9

(22 - 9 )  + (22 - 8 ) ]   ×  0,#4

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%odus  = 6,475 + [13

 13 + 14 ]   ×  0,#4

%odus

 = 6,475 +

[13

27

 ]×

0,#4

%odus  = 6,475 + 0,11

%odus  = 6,59

11) ersentil

i) /etak 5

%tk 5  =I

100  N

%tk 5

 =5

100  100

%tk 5  = 5

ersentil 5 terletak pada data ke5 pada Interval  ke1

ii) /etak 50

%tk 50  =I

100  N

%tk 50  =50

100  100

%tk 50  = 50

ersentil 50 terletak pada data ke50 pada Interval  ke4

iii) /etak $5

%tk 95 =I

100  N

%tk 95 =95

100  100

%tk 95 = 95

ersentil $5 terletak pada data ke$5 pada Interval  ke!

i9) 2ilai 5

5  = %&5 + [I

100N - *k5-1

"5]× I

5  = 5,995 + [5 - 0

12   ]   ×  0,#4

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5  = 5,995 + [512 ]×  0,#4

5  = 5,995 + 0,10 

5  = 6,10 

9) 2ilai 50

50  = %& 50  + [I

100N - *k50-1

"50]×I

50  = 6,715 + [50 - 43

8   ]   ×  0,#4

50  = 6,715 + [78 ]×0,24

50  = 6,715 + 0,21

50  = 6,93

9i) 2ilai $5

95  = %&95 +

[

I

100N - *k95-1

"95

]×I

95  = 7,675 + [95 - 83

17   ]×0,24

95  = 7,675 + [12

17 ]×0,24

95  = 7,675 + 0,17

95  = 7,84

4.2.1.1.4. endela

%eran&ang (endela yang sesuai data anthropometri praktikan mata kuliah IJ. Dataanthropometri yang digunakan untuk meran&ang (endela adalah rentang tangan, (angkauan tangandan tinggi badan berdiri. /alu perhitungannya adalah sebagai berikut:

a) Tinggi badan berdiri tbb)Tabel 3.4.3. data anthropometri tinggi badan berdiri tbb)

-o 1 2 3 4 5 6 7 8 9 10   ×

1154,00 1!,40 151,0 15,"0 1"#,0 1"1,"0 14!,10 14,!0 1",00 15#,10 11,0

21"$,#0 1#,10 1"!,$0 1"5,0 15#,$0 1"!,40 1"!,!0 10,00 155,10 154,30 1",53

31",#0 1$,30 1!0,"0 1!1,10 1"3,$0 1"0,"0 1$,00 1!0,30 1"3,00 154,!0 1"#,00

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410,50 1"5,00 153,00 13,30 151,40 14!,#0 14$,$0 1"$,50 14,00 1"5,!0 1#,0

514",10 15,00 15$,#0 155,10 1"#,$0 1!3,50 154,0 154,"0 15!,!0 1#,!0 11,3"

61"5,50 151,"0 10,$0 1"4,0 15!,10 1#,"0 153,0 15#,50 155,0 15,10 10,13

7 1!#,!0 1!,40 1"3,50 1!0,"0 1"3,30 1"4,00 15$,40 1"$,40 11,#0 14",$0 1"0,0

81"4,40 1#,40 150,0 1"3,00 15,#0 10,"0 1!3,50 1!3,#0 1!3,40 1!0,30 1"1,"

910,#0 1$,$0 1!0,$0 1!1,50 151,00 151,00 1"0,!0 1"1,$0 1"!,00 15,"0 1!,0$

+× 14$4,51

1) %enghitung rararata subgroup

= ∑ k

= ∑1494,51

9

=  1,0

#) %enghitung standar de9iasi

σ =√∑ ( x1−´ x )2

 N 

σ = √ (154,00- 14 9 ,45 )2 + .... + (165,70 – 149,45 )2

90

σ = 10,!5

3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup

σ =σ

√ k

σ =10,85

√ 9

σ =3,#

4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!

Zt = Z[1-[1- β

2   ]]Zt = 1 [1- [1-0,96

2   ]]Zt = 0,98

>t ; #,0

5) U(i keseragaman data

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i) Batas kontrol atas B)

B ¿ ´ x+(Z t  x σ  x)

B2,06 x

¿71,15+¿ 1,"!)

B ; "4,!1ii) Batas kontrol ba-ah BB)

BB ¿ ´ x−(Z t  x σ  x)

BB ¿71,15−(2,06  x1,78)

BB ; ",4!

) U(i ke&ukupan data

N

'

=

[

Zt

α √N∑ i

2- (∑ i )

2

∑i

]

2

N'=[

2,06

0,04 √ 50(5115,83) - (502539,21)

7089   ]2

 2C ; 1,0eterangan: 2C 2, data &ukup

1 # 3 4 5 " ! $150.00

155.00

10.00

15.00

1"0.00

1"5.00

11.0

1".53

1"#.00

1#.0 11.3"10.13

1"0.01"1."

1!.0$

U(i keseragaman data

B BB bar  

+ata subgroup

<ambar 3.4.14. eta kontrol u(i keseragaman data TBB

") Distribusi *rekuensi

i) angkauan +)+ ; DmaM Dmin

+ ; 1!3,50 E 14",10+ ; 3,40

ii) Banyak kelas

; 1 F 3,3 log n ; 1 F 3,3 log 100 ; ",0 K $

iii)   Interval  G)

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I =R

I=36,40

9I =  4,05

Tabel 3.4.4. Tabel distrbusi *rekuensi TBB

 Interval  &i & ()( *)( ()B *)B )$ 'i.)$

14",10 151,14 ! ! 14",10 151,14 14",0$5 151,145 14$,1# 11$#,$

151,15 155,1$ 15 #3 151,15 155,1$ 151,145 155,1$5 153,1" ##$",55

155,#0 15$,#4 5 #! 155,#0 15$,#4 155,1$5 15$,#45 15",## "!,10

15$,#5 13,#$ 11 3$ 15$,#5 13,#$ 15$,#45 13,#$5 11,#" 1""3,$"

13,30 1",34 $ 4! 13,30 1",34 13,#$5 1",345 15,3# 14!",!!

1",35 1"1,3$ " 55 1",35 1"1,3$ 1",345 1"1,3$5 1$,3" 11!5,5$

1"1,40 1"5,44 13 ! 1"1,40 1"5,44 1"1,3$5 1"5,445 1"3,4# ##54,4

1"5,45 1"$,4$ $ "" 1"5,45 1"$,4$ 1"5,445 1"$,4$5 1"",4" 15$",#3

1"$,50 1!3,54 13 $0 1"$,50 1!3,54 1"$,4$5 1!3,545 1!1,5# #35$,"

+&i $0 +&i.)$ 14$35,50

!)   Mean

´ x=∑fi.cm

∑fi

´ x=14935,50

90

´ x=  15,$5

$) %ediani) /etak median

/etak median ;1

2  N

/etak median ;1

2  90

/etak median ; 45

ii) 2ilai median

 Median=%& $()+ [1

2N- ( *$()-1 )

"$()]   I

 Median=163,295+ [1

2  90 - 39

9   ]   4,05

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 Median=163,295 + [0,66 ]   4,05

 Median=163,295 +  #,"0

 Median=   11,$5

10) %odus

)=%& $) + [   (" $) - " $)-1 )("$) - "$)+1) + ("$) -"$)-1 ) ]   I

=151,145 + [15 - 8

(15-8 )+ (15-10) ]   4,05

=151,145 + [ 7

7+5 ]   4,05

= 151,145 + #,3

=¿  15#,!1

11) ersentili) /etak 5

%tk 5  =I

100  N

%tk 5

 =5

100  90

%tk 5  =4,5

ersentil 5 terletak pada data ke5 pada Interval  ke1

ii) /etak 50

%tk 50  =I

100  N

%tk 50  =50

100  90

%tk 50  = 45

ersentil 50 terletak pada data ke50 pada Interval  ke4

iii) /etak $5

%tk 95 =I

100  N

%tk 95 =95

100  90

%tk 95 = 85,5

ersentil $5 terletak pada data ke$5 pada Interval  ke!

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i9) 2ilai 5

5  = %&5 + I

I

100N - *k5-1

"5

5  = 147,095 + 4,05 [5

10090 - 0

8   ]5  =

14$,3

9) 2ilai 50

50  = %& 50+ I

[I

100

N - *k50-1

"50 ]50  = 163,295 + 4,05 [

50

10090 - 39

9   ]50 =

1!,#5

9i) 2ilai $5

95  = %&95 + I [I

100 N - *k95-1

" 95 ]95  = 179,495+ 4,05 [

95

10090 - 77

13   ]50 =

1!5,10

 b) +entang tangan rt)Tabel 3.4.5. +entang tangan

% 

n1 2 3 4 5 6 7 8 9 10   ´ x  

1 1!5," 153, 1!, 11," 14$, 1$#, 1", 15!, 150, 14," 1,1

2 1!",1 15$, 10, 10,4 10 153, 154," 1!1, 11 1$0,$ 1,$3

3 1"! 1$3, 1!", 1!4,$ 1!5 1"#, 1!4,$ 1"#, 15", 1$," 1"!,3

4 1!#, 154, 1$#, 1$4 1"3, 1", 1, 1"4 1"$, 15"," 1"5,05

5 1"4,3 10, 1$0, 10,$ 1$#, 15$, 1"4, 1!, 15, 1"1,$ 1"#,!

6 1"3," 153, 14", 1"3,$ 1!3 15, 1"", 11, 1", 1$1, 1"0,5

714",# 14!, 1!3, 1"5,# 155, 14!, 1!, 1!", 154 1",5 1,33

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8 1"$ 15 1, 15$,$ 1!$, 13, 1!1 1"3 1$#, 1!3,4 1"4,4$

9 153,4 1$3, 1", 1!$,3 1!0, 1" 1!0,# 1!3 1!, 15$ 1"4,1!

10 1"!,1 1!4, 1!0, 10,3 1!1, 1", 1$3,4 15$, 1$0, 1$# 1"$,

  ∑ ´ x 1."#4,"3

1) %enghitung ratarata subgroup

=∑ i

k

  =1726,73

10

  = 172,67

#) %enghitung standar de9iasi

σ =√∑(i- )

2

N

= √(185,7 – 172,67 )2 + (153,6 – 172,67 )2 + / + (192,0 – 172,67 )2

100

; 13,4"

3) %enghitung standar de9iasi dari distribusi harga ratatata subgroup

σ  =σ√ k

σ  =13,47

√ 10

σ  = 4,26

4) 'itung nilai > tabel >t), dengan ? ; 0,$

Zt = Z[1-[1- β

2   ]]Zt = 1 [1- [1-0,96

2   ]]Zt = 0,98

>t ; #,0

5) U(i keseragaman datai) Batas kontrol atas B)

= + (Zt   σ )

= 172,67+ (2,06 4,26 )

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= 1!1,45

ii) Batas kontrol ba-ah BB)

= - (Zt   σ)

= 172,67- (2,06 4,26 )

= 163,90

1 # 3 4 5 " ! $ 10155.00

10.00

15.00

1"0.00

1"5.00

1!0.00

1!5.00

1.11".$3

1"!.31"5.05

1"5.05 1"#.!01"0.50

1.33

1"4.4$1"5.1!

U(i keseragaman data

B BB bar  

+ata sbgrup

<ambar 3.4.15. eta kontrol u(i keseragaman data +T

) U(i ke&ukupan data dengan H ; 0,04

N'=[

Zt

α √N∑ i

2- (∑ i )

2

∑ i]

2

N'=[

2,06

0,04 √ 100(2999751,95) – (1726,73 )2

1726,73   ]2

N'=[

2,06

0,04 √ 299975195 – 298159649,29

1726,73   ]2

N'=[51,5 1347,40

1726,73   ]2

N'= 40,20

eterangan: 2C 2, data &ukup

") Distribusi *rekuensi

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i) angkauan +)

R = $- $I

R = 194 - 147,20

R = 46,80  

ii) Banyak kelas )

= 1 + 3,3

= 1 + 3,3 90

= 7,4 7

iii) Interval G)

I =R

I =46,80

7

I = 6,69

Tabel 3.4.. Tabel distribusi *rekuensi rentang tangan +T)

 Interval  &i & ()( *)( ()B *)B )$ &i.)$

14",#0 153,!! ! ! 14",#0 153,!! 14",1$5 153,!!5 150,54 1#04,3#

153,!$ 10,5" 1$ #" 153,!$ 10,5" 153,!!5 10,5"5 15",#3 #$!",3"

10,5! 1",# 11 3! 10,5! 1",# 10,5"5 1",#5 13,$# 1!03,1#

1",#" 1"3,$5 1# 50 1",#" 1"3,$5 1",#5 1"3,$55 1"0,1 #04",3#

1"3,$ 1!0,4 1 1"3,$ 1!0,4 1"3,$55 1!0,45 1"",30 #!3,!0

1!0,5 1!",33 1" !3 1!0,5 1!",33 1!0,45 1!",335 1!3,$$ 31#",!3

1!",34 1$4,0# 1" 100 1!",34 1$4,0# 1!",335 1$4,0#5 1$0,! 3#41,5

+&i 100 +&i.)$ 1"#4!,3#

!)   Mean

 Mean  =∑ " i &

∑ " i

 Mean  =17248,32

100

 Mean  = 172,48

$) %ediani) /etak median

/etak median  =1

2  N

/etak median  =1

2  100

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/etak median  = 50

%edian terletak pada nilai ke50 pada Interval  ke4

ii) 2ilai median

%edian  = %&$+ [12

N - *k$-1

"$ ]×I

%edian  = 167,26 + [50 - 38

12   ]× 6,69

%edian  = 167,26 + [12

12 ]×6,69

%edian  = 167,26 + 6,69

%edian  =  1"3,$5

10) %odus

%odus = %&$) + [

"$) - "$)-1

( "$) - "$)-1 )+ ( "$) - "$)+1 ) ]× I

%odus = 167,26 + [19 - 8

(19 - 8 )  + (19 - 11) ]   ×  ,$

%odus  = 167,26 +

[11

 11 + 8 ]  ×  ,$

%odus  = 167,26 + [11

19 ]× ,$

%odus  = 167,26 + 3,87

%odus  = 171,14

11) ersentil

i) /etak 5

%tk 5  =I

100  N

%(tk 5  =5

100  100

%tk 5  = 5

ersentil 5 terletak pada data ke5 pada Interval  ke1

ii) /etak 50

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%tk 50  =I

100  N

%(tk 50  =50

100

  100

%tk 50  = 50

ersentil 50 terletak pada data ke50 pada Interval  ke4

iii) /etak $5

%tk 95 =I

100  N

%tk 95 =95

100  100

%tk 95 = 95

ersentil $5 terletak pada data ke$5 pada Interval  ke"

i9) 2ilai 5

5  = %&5 + [I

100N - *k5-1

"5]× I

5  = 147,195 + [5 - 0

8   ]   ×  ,$

5  = 147,195 + [5

8 ]×  ,$

5  = 147,195 + 4,18 

5  = 151,38 

9) 2ilai 50

50  = 167,26 +

[

I

100N - *k50-1

" 50

]×I

50   = 167,26 + [50 - 38

12   ]   ×  ,$

50  = 167,26 + [12

12 ]×6,69

50  = 167,26 + 6,69

50  = 173,96

9i) 2ilai $5

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95  = %&95 +[I

100N - *k95-1

" 95]×I

95  = 187,33 + [95 - 10017   ]×6,69

95  = 187,33 + [-517 ]×6,69

95  = 187,33 + (-1,97)

95  = 185,37

&) angkauan tangan (t)Tabel 3.4." angkauan tangan

% n

1 2 3 4 5 6 7 8 9 10   ´ x

1 "",$ $,3 "4 4,3 5," ",4 !,$ $,3 "4,$ !0,! "1,#5

2 5, 5 5 "#,! 3," 5,4 "5,3 !0,# ", !0 "0,$

3 "0 3," 3," "4,4 "!,3 "!,4 "!," ""," "$ $,$ "3,3!

4 "4,4 "1,$ "1,$ "3," ",# 3,4 3,# "!,# "5,! "0,5"

5 "3," 1, 1, !0,1 ",5 $ "5 !1 ,3 5,3 "1,01

6 "!,1 1," 1," "",! 1,$ #,3 "1,! "1,3 !,! ",5 $,1$

7 $,1 $," $," ",5 "1 $,3 #,3 "",# "4 !,! $,!

8 !, "4,! "4,! "1, #,3 "#,! ",5 ",5 "3 "$,! "1,#"

9 "",5 ",$ ",$ $," "3, "0,1 "!," 1,4 "0,$ 5,$ "#,1

10 #,$ ",1 ",1 !0," $ ,1 "1,5 !0 5,$ #,# $,#5

  ∑ ´ x "0!,$

1) %enghitung rararata subgroup

x = ∑  xk

x = ∑ 708,9

10x =70,89  

#) %enghitung standar de9iasi

σ =√∑ ( x1−´ x )2

 N 

σ = √ (77,9-70,89 )2  + .... + (62,2 – 70,89 )2

100

σ = 5,79

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3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup

σx =σ

√ k

σx =5,79

√ 10

σx = 1,83

4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!

Z = 1 – [1 - β

2   ]

Z = 1 –

[1 – 0, 98

2   ]Z = 1 – 0, 01

Z = 0,99

Berdasarkan tabel distribusi normal, nilai yang paling mendekati untuk harga > ; 0,$$

adalah #,0 maka >t ; #,0

5) U(i keseragaman datai) Batas kontrol atas B)

B

= x+(Z t  x σx )

B =70,89+(2,06 x 1,!3)

B ; "4,ii) Batas kontrol ba-ah BB)

BB= x-(Zt  x σ x )

BB = x-(2,06 x 1,83)

BB ; ",1#

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1 # 3 4 5 " ! $ 10".1#

!.1#

$.1#

"0.1#

"1.1#

"#.1#

"3.1#

"4.1#

"1.#5"0.$

"3.3!

"0.5""1.01

$.1$

$.!

"1.#"

"#.1

$.#5

U(i keseragaman data

A bar B BB # bar

<ambar 3.4.1". U(i keseragaman data

) U(i ke&ukupan data

N'=[

Zt

α √N∑ x i

2- (∑ x i )

2

∑ xi]2

N'=

[

2,06

0,04 √ 505115,83 - 502539,21

7089

  ]

2

 2C ; 13,0eterangan: 2C 2, data &ukup

") Distribusi *rekuensii) angkauan +)

+ ; DmaA Dmin

+ ; !1,00 E 1,40+ ; 1$,0

ii) Banyak kelas ; 1 F 3,3 log n

; 1 F 3,3 log 100 ; ",Dibulatkan ke atas, sehingga ; $

iii) Gnter9al G)

I=R

I=19,60

9

I=2,18

Tabel 3.4.!. Distribusi *rekuensi

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!) %ean

x =!"i.#$

!"i

x =7079,080100

x =70,79

$) osisi median

osisi median ;1

2 x N

osisi median ;1

2 x 100

osisi median ; 50

10) %edian

x =%&$+ [1

2N- (*$-1 )

"$]  x I

~x =67,935+[1

2 x 100- (35 )

51   ]  x 2,18

x =67,935+[50-35

51   ]x 2,18

x = 70,05

11) %odus

x =%&$) +[   ( "$) - "$)-1 )( "$)- "$)+1 )+ (" $) - "$)-1) ] x I

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x = 67,935+[17-9

(17-10) + (17-9) ]  x 2,18

x =67,935+

[8

7+8

 ] x 2,18

x =69,30

1#) ersentili) /etak 5

%tk 5  =I

100  N

%(tk 5  =5

100  100

%tk 5  = 5

ersentil 5 terletak pada data ke5 pada interval  ke1

ii) /etak 50

%tk 50  =I

100  N

%(tk 50  =50

100  100

%tk 50  = 50

ersentil 50 terletak pada data ke50 pada interval  ke4

iii) /etak $5

%tk 95  =I

100  N

%tk 95  =95

100  100

%tk 95  = 95

ersentil $5 terletak pada data ke$5 pada interval  ke$

i9) 2ilai 5

5  = %&5 + I [I

100N - *k5-1

"5]

5  = 61,395, + 2,18 [5

100100 - 0

13   ]5  = 61,89

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9) 2ilai 50

50 = %&50 + I

I

100N - *k50-1

" 50

50  = 67,935 + 2,18 [50

100100 - 35

16   ]50 = 70,05

9i) 2ilai $5

95  = %&95 + I

[I

100

N - *k95-1

" 95 ]95  = 78,835 + 2,18 [

95

100100 - 91

9   ]50  = 80,01

4.2.3 Peran#angan Produ 

4.2.3.1. Peran#angan Produ dan /uangan seretariat !urusan

eran&angan usulan tata letak ruangan merupakan pengembangan atau perbaikan dari ran&anganyang sudah ada dan masih bersi*at usulan, hal ini bertu(uan untuk membuat tata letak ruanganyang lebih e*ekti* dan nyaman sehingga segala ma&am kegiatan yang ada didalamnya bisadilakukan dengan lebih baik kedepannya. Berikut adalah desain usulan tata letak ruang sekretariat (urusan:

<ambar 3.4.#3. Usulan ruangan sekretariat (urusan

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Gambar 3.4.24. Desain produk meja

 Tabel 3.4.69. Data anthropometri meja

Data antropometri Persentil 5 Persentil 50 Persentil 95

angkauan tangan #,30 "0,$5 !0,01

+entang tangan 150,!# 1"3,$5 1$#,$0

Tinggi siku duduk 1",#0 ##,00 #,#1

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<ambar 3.4.#5 Desain produk lemari kabinet

Tabel 3.4."0. Desain produk lemari kabinet

Data antropometri Persentil 5 Persentil 50 Persentil 95

Tinggi bahu berdiri 1#3,5" 13$,!1 153,5!

/ebar bahu 34,50 3$,3 45,4$

an(ang (ari 3 ",0 ",$1 !,$0

/ebar (ari #,3,4 dan 5 ,10 ,$3 ",!4

angkauan tangan #,30 "0,$5 !0,01

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<ambar 3.4.#. Desain produk pintu

 Tabel 3.4.71. Data anthropometri meja

Data antropometri Persentil 5 Persentil 50 Persentil 95

Tinggi badan berdiri 14$,3 1!,#5 1!5,10

Tinggi bahu berdiri 1#3,5" 13$,!1 153,5!

/ebar bahu 34,50 3$,3 45,4$

an(ang (ari 3 ",0 ",$1 !,$0

/ebar (ari #,3,4 dan 5 ,10 ,$3 ",!4

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P95 (LB)=45,75

P95 (TBB)=182,44

P5 (JT)=80,72

P50 (TSB)= 122,172

<ambar 3.4.#". Desain produk (endela

 Tabel 3.4.72. Data anthropometri meja

Data antropometri Persentil 5 Persentil 50 Persentil 95Tinggi badan berdiri 14$,3 1!,#5 1!5,10

+entang tangan 151,3! 1"3,$ 1!3,3"

angkauan tangan #,30 "0,$5 !0,01

4.2.4. ata pen#ahaaan dan ebisingan

4.2.4.1. Perhitungan data pen#ahaaan

a) 'ari pertamaTabel 3.4."3. lux meter hari pertama

 n

  1 2 3 4 5 6 7 8 9 10  x

1 0,#$ 0,#! 0,30 0,## 0,30 0,#3 0,#$ 0,#3 0,30 0,30 0,#"

2 0,30 0,# 0,30 0,30 0,30 0,#! 0,30 0,30 0,#3 0,#$ 0,#$

3 0,30 0,30 0,30 0,## 0,# 0,#! 0,30 0,30 0,## 0,3# 0,#!

4 0,30 0,#" 0,#3 0,30 0,31 0,3# 0,## 0,30 0,30 0,#4 0,#!

5 0,# 0,#! 0,30 0,#$ 0,30 0,30 0,3# 0,#5 0,30 0,## 0,#!

∑ : 140

1) %enghitung rararata subgroup

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x =∑ x

k

x = 1,405

x =0,28

#) %enghitung standar de9iasi

σ =√∑ (x1- x )2

N

σ =

√

 (0,29-0,28 )2 + .... + (0,28-0,28 )

2

50

σ = 0,03

3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup

σx =σ

√ k

σx =0,03

√ 5

σx = 0,01

4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!

Z = 1 – [1 - β

2   ]Z = 1 – [1 – 0,98

2   ]Z = 1 – 0,01

Z = 0,99

Berdasarkan tabel distribusi normal, nilai yang paling mendekati untuk harga > ; 0,$$adalah #,0 maka >t ; #,0

5) U(i keseragaman datai) Batas kontrol atas B)

B= x+(Z t  x σx¿

B =0,28+ (2,06 x 0,01)

B ;0,31

ii) Batas kontrol ba-ah BB)

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BB= x-(Zt  x σ x )

BB = 0,28-(2,06 x 0,01)

BB ; 0,#5

1 # 3 4 50.#5

0.#"

0.#$

0.31

0.27

0.290.28 0.28 0.28

U(i eseragaman Data

bka bkb bar

rata sub!roup

<ambar 3.4.#!. eta kontrol u(i keseragaman data

) U(i ke&ukupan data

N'=[

Zt

α √N∑ x i

2- (∑ x i )

2

∑ x i]2

N'=[

2,06

0,04 √ 50 x 3,97 - 196,28

14,01   ]2

 2C ; 31,!eterangan: 2C 2, data &ukup

 b) 'ari kedua

Tabel 3.4."4. lux meter hari kedua

  1 2 3 4 5 6 7 8 9 10   x

1 0,34 0,34 0,34 0,3# 0,35 0,34 0,35 0,35 0,35 0,35 0,34

2 0,35 0,35 0,34 0,35 0,35 0,35 0,3# 0,35 0,35 0,35 0,35

3 0,35 0,3 0,35 0,34 0,34 0,35 0,35 0,35 0,3# 0,3 0,35

4 0,3 0,35 0,35 0,35 0,3" 0,3 0,3 0,3" 0,35 0,3# 0,35

5 0,35 0,35 0,35 0,35 0,31 0,35 0,3" 0,34 0,35 0,3$ 0,35

∑ : 174

1) %enghitung rararata subgroup

x =∑ x

k

x =1,74

5

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x =0,35

#) %enghitung standar de9iasi

σ =

√∑ (

x1

- x

)

2

N

σ = √ (0,34-0,35)2 + .... + (0,39-0,35 )2

50

σ = 0,01

3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup

σx =σ

√ k

σx =0,01

√ 5

σx = 0,004

4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!

Z = 1 – [1 - β

2   ]Z = 1 –

[

1 – 0,98

2

  ]Z = 1 – 0,01

Z = 0,99

Berdasarkan tabel distribusi normal, nilai yang paling mendekati untuk harga > ; 0,$$adalah #,0 maka >t ; #,0

5) U(i keseragaman datai) Batas kontrol atas B)

B= x+(Z t  x σx¿

B =0,01+ (2,06 x 0,004)B ;0,3

ii) Batas kontrol ba-ah BB)

BB= x-(Zt  x σ x )

BB = 0,01-(2,06 x 0,00 4)

BB ; 0,34

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1 2 3 4 "0.33

0.34

0.34

0.3"0.3"

0.36

0.36

0.34300.340 0.34"0

0.35400.3510

U(i eseragaman Data

bka bkb bar

rata sub!roup

<ambar 3.4.#$. eta kontrol u(i keseragaman data

) U(i ke&ukupan data

N'=[

Zt

α √N∑ x i

2- (∑ x i )

2

∑ x i]2

N'=[

2,06

0,04 √ 50 x 6,07 - 303,11

17,41   ]2

 2C ; 4,00eterangan: 2C 2, data &ukup

4.2.4.2 Perhitungan data ebisingan

a) 'ari pertamaTabel 3.4."5. sound level  hari pertama

  1 2 3 4 5 6 7 8 9 10   x

1 ",00 3,30 0,#0 1,0 1,$0 ","0 4,10 3,0 3,50 0,0 3,35

2 4,50 3,!0 ,50 5,!0 0,30 1,!0 3,10 0,40 ","0 1,$0 3,5!

3 0,#0 3,50 ,50 5,!0 4,50 3,10 3,40 ","0 3,0 1,!0 4,01

4 0,40 5,!0 ,50 0,#0 1,!0 3,0 4,50 5,!0 ,50 1,!0 3,$

5 5,!0 0,30 3,50 1,!0 ,50 0,30 5,!0 4,50 5,!0 4,"0 3,$0

∑ :

31853

1) %enghitung rararata subgroup

x =∑ x

k

x =318,53

5

7/23/2019 Ergonomi - Bab 4 Pengolahan Data - Aldi Dwi Apriliana - Universitas Komputer Indonesia

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x =63,71

#) %enghitung standar de9iasi

σ =

√∑ (

x1

- x

)

2

N

σ = √ (67,00-63,71 )2 + .... + (64,70-63,71)2

50

σ = 63,47

3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup

σx =σ

√ k

σx =63,47

√ 5

σx = 28,38

4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!

Z = 1 – [1 - β

2   ]Z = 1 –

[

1 – 0,98

2

  ]Z = 1 – 0,01

Z = 0,99

Berdasarkan tabel distribusi normal, nilai yang paling mendekati untuk harga > ; 0,$$adalah #,0 maka >t ; #,0

5) U(i keseragaman datai) Batas kontrol atas B)

B= x+(Z t  x σx¿

B =63,47+ (2,06 x #!,3!)B ;1##,1!

ii) Batas kontrol ba-ah BB)

BB= x-(Zt  x σ x )

BB = 63,47-(2,06 x 28,38)

BB ; 0,#5

7/23/2019 Ergonomi - Bab 4 Pengolahan Data - Aldi Dwi Apriliana - Universitas Komputer Indonesia

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1 # 3 4 55$.00

0.00

1.00

#.00

3.004.00

5.00

.00

".00

3.35 3.5! 4.01 3.$ 3.$0

U(i eseragaman Data

bka bkb bar

rata sub!roup

<ambar 3.4.#$. eta kontrol u(i keseragaman data

) U(i ke&ukupan data

N'=[

Zt

α √N∑ x i

2- (∑ x i )

2

∑ x i ]2

N'=[

2,06

0,04 √ 50 x 203188,29- 10146136,09

3185,30   ]2

 2C ; 3,4"eterangan: 2C 2, data &ukup

 b) 'ari keduaTabel 3.4.". sound level hari kedua

1 2 3 4 5 6 7 8 9 10   x

1 ","0 $,#0 5,10 "3,#0 !,30 4,0 #,50 4,50 5,40 4,!0 ,53

2 ,10 ,!0 4,#0 #,30 3,00 ",30 4,00 4,40 ,50 !,$0 5,35

3 !,$0 3,30 ",50 5,30 ,50 4,50 !,"0 ,"0 5,40 3,50 ,03

4 ,!0 !,"0 "#,10 4,0 $,30 4,50 #,30 4,10 1,!0 ",!0 ,#0

5 3,"0 !,40 5,50 4,30 ,#0 ,$0 $,!0 !,30 5,10 #,$0 ,11

∑ : 33022

1) %enghitung rararata subgroup

x =∑ x

k

x =330,22

5

x =66,04

#) %enghitung standar de9iasi

7/23/2019 Ergonomi - Bab 4 Pengolahan Data - Aldi Dwi Apriliana - Universitas Komputer Indonesia

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σ =√∑ (x1- x )2

N

σ =

√ (67,70-66,04 )

2 + .... + (62,90-66,04 )

2

50

σ = 2,52

3) %enghitung standar de9iasi dan distribusi harga ratarata subgroup

σx =σ

√ k

σx =2,52

√ 5

σx = 1,13

4) %enghitung nilai > tabel >t) dengan ? ; $!@ ; 0,$!

Z = 1 – [1 - β

2   ]Z = 1 – [1 – 0,98

2   ]Z = 1 – 0,01

Z = 0,99Berdasarkan tabel distribusi normal, nilai yang paling mendekati untuk harga > ; 0,$$adalah #,0 maka >t ; #,0

5) U(i keseragaman datai) Batas kontrol atas B)

B= x+(Z t  x σx¿

B =66,04+ (2,06 x 1,13)

B ;!,3"

ii) Batas kontrol ba-ah BB)

BB= x-(Zt  x σ x )

BB = 66,04-(2,06 x 1,13)

BB ; 3,"#

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1 # 3 4 51.00

#.00

3.00

4.00

5.00

.00

".00

!.00

$.00

.53

5.35

.03 .#0 .11

U(i eseragaman Data

bka bkb bar

rata sub!roup

<ambar 3.4.30. eta kontrol u(i keseragaman data

) U(i ke&ukupan data

N'=[

Zt

α √N∑ x i

2- (∑ x i )

2

∑ x i]2

N' =

[2,06

0,04

 √ 50 x 218402,16 - 10904524,84

3302,20

  ]

2

 2C ; 3,"$  eterangan: 2C 2, data &ukup