jawaban latihan soal continuous probability distribution

7/23/2019 Jawaban Latihan Soal Continuous Probability Distribution

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1. Kode soal 6.5

a. P(0 < z < 1.96) = 0.4750.

b. P(z > 1.645) = 0.5 -2

4505.04495.0 = 0.5 - 0.4500 = 0.05.

c.

P(1.28< z 2.33) = P(0 < z 2.33) - P(0 < z 1.28) = 0.49010.3997 = 0.0904d. P(-2 z 3) = P(-2 < z 0) + P(0 < z 3) = P(0 < z 2) + P(0 < z 3) =

0.4772 + 0.49865 = 0.97585.

e. P(z > -1) = P(-1 < z 0) + P(Z 0) = P(0 < z 1) + P(z 0) = 0.3413 + 0.5000 =

0.8513.

2. Kode soal 6.6

a. P(z > 1.645) = 0.05. xo= + zo; so xo= 13.6 + 1.645(2.90) = 18.37.

b.

P(0 z 1.96) = 0.4750. So P(z 1.96) = 0.05 + 0.475. xo= + zo; so xo= 13.6

+ 1.96(2.90) = 19.284.c. P(0 z 1.96) = 0.4750. So P(-1.96 z 1.96) = 0.4750 + 0.4750 = 0.95. xo=

+ zo; so - xo= zo= 1.96(2.90) = 5.684. Therefore, xo= - 5.684 = 13.6

5.684 = 7.916.

3. Kode soal 6.9

a. The following steps are used to compute the desired value of x

Step 1: Determine the mean and standard deviation

The mean and standard deviation are 5.5 and .50

Step 2: Determine the event of interest.

We are interested in determining the value of x such that the probability of a

value exceeding x is at most 0.10.

Step 3: Determine the z-value corresponding to the known probability.

The area in the upper tail of the distribution above x is defined to be 0.10.

That means that the area between x and the population mean of 5.5 is 0.40. In

Appendix D, we go to the inside of the table and locate the value 0.40 or just

larger and determine the z-value associated with this probability. The closest

probability is 0.4015. The z-value corresponding to this probability is z =

1.29.

Step 4: Substitute the known values into the following equation:

x

z

5.51.29

.50

x

Step 5: Solve for x:

1.29(.50) 5.5 6.145x

b.

We are asked to determine what the population mean must be if we want the

following:

( 6.145) 0.05P x


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The following steps can be used to solve for the new population mean:

Step 1: Determine the mean and standard deviation

The mean and standard deviation are 5.5 and .50

Step 2: Determine the event of interest.

We are interested in determining the value of such that the probability of avalue exceeding 6.145 is at most 0.05: P(x > 6.145) 0.05.

Step 3: Determine the z-value that corresponds to an upper tail area equal to

0.05.

From the standard normal distribution table, we look for a probability on the

inside of the table equal to 0.45 (or slightly larger) and determine the

corresponding z-value. The closest probability is .4505. The z-value

corresponding to this probability is 1.65. (Note, students could interpolate

between 0.4495 and 0.4505 giving a z=1.645.)

Step 4: Substitute the known values into the following equation:

6.1451.65

.50

Step 5: Solve for

6.145 (1.65)(.50) 5.32

Thus, the population mean must be reduced from 5.5 to 5.32 in order for the

probability of a value exceeding 6.145 to be reduced from 0.10 to 0.05.

4.

Kode soal 6.16

a.

Since P(0 z 1.67) = 0.4525, P(z 1.67) = P(z > 1.67) = 0.50.4525 =

0.0475 = P(x > 3.5).

b. Since P(-0.83 z 0) = 0.2967, P(z > -0.83) = P(-0.83 z 0) + 0.5 P(z > -

0.83) = 0.2967 + 0.5 = 0.7967.

c. If P(z > z0) = 0.10, then P(0 < z < z0) = 0.5 - 0.10 = 0.40. From the standard

normal table, P(0 < z < 1.28) = 0.40. So z0= 1.28.

Since x0= + z0, x0= 3 + 1.28(0.3) = 3.384 years. Therefore, the length of life

values for the ten percent of the watches batteries that last the longest are those

greater than 3.384 years.

5. Kode soal 6.34

a. P(c x d) = f(x)(dc) = 0.25(85.5) = 0.625.

b.

P(x > 7) = P(7 < x 9) = f(x)(dc) =

0.25(97) = 0.50.

c.

= 72

59

2

ab = 333.1

12

)59(

12

)( 22

ab= 1.155


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d. (2) = [7 2(1.155)] = 7 2.31 = (4.69, 9.31). Since the distributions limits

are 5 and 9, P(4.69 x 9.31) = 1 or 100%.

6. Kode soal 6.37

a.

Lambda is equal to 1/mean, so = 1/1.5 = 0.6667. The probability that the timebetween the next two calls is 45 seconds or less is computed by converting the 45

seconds to minutes (45/60) = 0.75. Therefore, a = 0.75. Probability (x < 0.75) = 1

- e-(0.6667)(0.75)= 1-0.6065 = 0.3935

b. In this case a = 112.5 seconds which is 112.5/60 = 1.875 minutes. Probability (x

> 112.5 seconds) = e-a= e-(0.0.6667)(1.875)= e-1.25= 0.2865.

7. Kode soal 6.41

Students can use Excels EXPONDIST function to solve this problem.

a.

= 1/4000 = .00025; P(x

jawaban latihan soal continuous probability distribution

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