6.continuous rv and prob dist

Chapter 6

Variabel Random Continu & Distribusi ProbabilitasStatistika

Statistics for Business and Economics, 6e 2007 Pearson Education, Inc.

Setelah mempelajari bab ini, anda akan mampu : Menjelaskan perbedaan antara variabel random diskrit dan kontinuMenerangkan karakteristik distribusi uniform dan normal Mentransformasi masalah dalam distribusi normal ke normal standardMencari probabilitas menggunakan tabel distribusi normalMenggunakan distribusi normal

Chapter Goals

After completing this chapter, you should be able to: Evaluate the normality assumptionUse the normal approximation to the binomial distribution Recognize when to apply the exponential distributionExplain jointly distributed variables and linear combinations of random variablesStatistics for Business and Economics, 6e 2007 Pearson Education, Inc.Chap 6-*Chapter Goals


Probability DistributionsContinuous Probability DistributionsBinomialPoissonHypergeometricProbability DistributionsDiscrete Probability DistributionsUniformNormalExponentialCh. 5Ch. 6


A continuous random variable is a variable that can assume any value in an intervalthickness of an itemtime required to complete a tasktemperature of a solutionheight, in inches

These can potentially take on any value, depending only on the ability to measure accurately.Continuous Probability Distributions

The cumulative distribution function, F(x), for a continuous random variable X expresses the probability that X does not exceed the value of x

Let a and b be two possible values of X, with a < b. The probability that X lies between a and b is

Cumulative Distribution Function

The probability density function, f(x), of random variable X has the following properties:f(x) > 0 for all values of xThe area under the probability density function f(x) over all values of the random variable X is equal to 1.0The probability that X lies between two values is the area under the density function graph between the two valuesThe cumulative density function F(x0) is the area under the probability density function f(x) from the minimum x value up to x0

where xm is the minimum value of the random variable xProbability Density Function

Probability as an Area abxf(x)Paxb()Shaded area under the curve is the probability that X is between a and bPaxb()

Distribusi Normal Distribusi Probabilitas kontinuDistribusi ProbabilitasUniformNormalExponential


Distribusi Normal Bell Shaped Simetris Mean= Median = ModusParameter lokasi : mean, Parameter skala : standard deviasi,

VR ini mempunyai nilai asal secara teori dari :

Formula pdf normal X~N(, 2) :Fungsi Densitas NormalWheree = bilangan alam 2.71828 = phi 3.14159 = mean populasi = standard deviasi populasix = nilai vr x, < x <

Untuk beberapa nilai parameter dan , kita peroleh beberapa kurva yang berbeda Bentuk Distribusi Normal


Chart3

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xpopulationsampling populationsigman

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Untuk X VR normal dengan mean dan variansi 2 , X~N(, 2), fungsi cdf nya adalah

Distribusi Normal Kumulatifxx0f(x)

xbaxbaxbaFinding Normal Probabilities

xbaPeluang nilai X berada diantara a dan b adalah Finding Normal Probabilities Menggunakan Integral Tidak Efisien

Normal StandardSebarang distribusi normal dapat ditransformasi ke distribusi normal standard (Z), dengan mean 0 dan variansi 1

X ditransformasi ke Z dengan transformasi :Zf(Z)01


ContohX berdistribusi normal dengan mean = 100 dan standard deviasi = 50, nilai Z untuk X = 200 adalah


Perbandingan X dan Z Z1002.00200XPerhatikan : Distribusi tetap sama, Normal, rata-rata dan skala berubah. Rata-rata baru menjadi nol, dan variansi menjadi 1( = 100, = 50)( = 0, = 1)

Tabel Normal StandardZ02.00.9772contoh P(X

The Standardized Normal TableZ0-2.00Example: P(X

Suppose X (Stat grade) is normal with mean 8.0 and standard deviation 5.0Find P(X < 8.6)Finding Normal ProbabilitiesX8.68.0

Z0.12 0X8.6 8 = 8 = 10 = 0 = 1Finding Normal ProbabilitiesP(X < 8.6)P(Z < 0.12)


Z0.12zF(z).10.5398.11.5438.12.5478.13.5517Solution: Finding P(Z < 0.12)F(0.12) = 0.5478Standardized Normal Probability Table (Portion)0.00= P(Z < 0.12)P(X < 8.6)


Suppose X is normal with mean 8.0 and standard deviation 5.0. Now Find P(X > 8.6)Upper Tail ProbabilitiesX8.68.0

Now Find P(X > 8.6)Upper Tail ProbabilitiesZ0.12 0Z0.120.5478 01.0001.0 - 0.5478 = 0.4522 P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z 0.12) = 1.0 - 0.5478 = 0.4522

Permasalahan dalam distribusi normal lainnya : Mencari nilai batas x, dengan diketahui luasnyaSteps to find the X value for a known probability:1. Find the Z value for the known probability2. Convert to X units using the formula:Finding the X value for a Known Probability

Example:Suppose X(stat grade) is normal with mean 8.0 and standard deviation 5.0. Now find the X value so that only 20% of all values are below this X (jika 20% nilai terendah dinilai D, tentukan batas nilai tertinggi untuk nilai D)Finding the X value for a Known ProbabilityX?8.0.2000Z? 0

20% area in the lower tail is consistent with a Z value of -0.84Find the Z value for 20% in the Lower TailStandardized Normal Probability Table (Portion)X?8.0.20Z-0.84 01. Find the Z value for the known probabilityzF(z).82.7939.83.7967.84.7995.85.8023.80

2. Convert to X units using the formula:Finding the X valueSo 20% of the values from a distribution with mean 8.0 and standard deviation 5.0 are less than 3.80

The Exponential DistributionContinuous Probability DistributionsProbability DistributionsNormalUniformExponential


Used to model the length of time between two occurrences of an event (the time between arrivals)

Examples: Time between trucks arriving at an unloading dockTime between transactions at an ATM MachineTime between phone calls to the main operatorThe Exponential Distribution

The Exponential DistributionThe exponential random variable T (t>0) has a probability density function

Where is the mean number of occurrences per unit timet is the number of time units until the next occurrencee = 2.71828T is said to follow an exponential probability distribution


Defined by a single parameter, its mean (lambda)

The cumulative distribution function (the probability that an arrival time is less than some specified time t) isThe Exponential Distributionwhere e = mathematical constant approximated by 2.71828 = the population mean number of arrivals per unitt = any value of the continuous variable where t > 0

Exponential Distribution ExampleExample: Customers arrive at the service counter at the rate of 15 per hour. What is the probability that the arrival time between consecutive customers is less than three minutes?The mean number of arrivals per hour is 15, so = 15Three minutes is .05 hoursP(arrival time < .05) = 1 e- X = 1 e-(15)(.05) = 0.5276So there is a 52.76% probability that the arrival time between successive customers is less than three minutes

6.continuous rv and prob dist

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