1 continuous probability distributions chapter 8

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Continuous Probability Continuous Probability DistributionsDistributions

Chapter 8Chapter 8

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在隨機實驗中，所觀察的隨機變數，如：人的在隨機實驗中，所觀察的隨機變數，如：人的身高、體重、家庭所得、地區的氣溫、股票的身高、體重、家庭所得、地區的氣溫、股票的價格、工人的工資、醫生看病的時間，都是連價格、工人的工資、醫生看病的時間，都是連續的隨機變數。這些連續隨機變數具有共同的續的隨機變數。這些連續隨機變數具有共同的特性，其機率分配的形態頗為類似，比如人的特性，其機率分配的形態頗為類似，比如人的身高，大部份的人都很接近，而且集中在平均身高，大部份的人都很接近，而且集中在平均身高左右，只有少部份的人較高或較矮。同樣身高左右，只有少部份的人較高或較矮。同樣的，氣溫、體重也類似此一形態。又如醫生看的，氣溫、體重也類似此一形態。又如醫生看病的時間，大部份的人看病時間不長，只有少病的時間，大部份的人看病時間不長，只有少數病人需要花較長的時間去診斷，而形成右偏數病人需要花較長的時間去診斷，而形成右偏的分配。 的分配。

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• 均等分佈均等分佈 ((Uniform Distribution)Uniform Distribution) • 常態分佈常態分佈 ((Normal Distribution)Normal Distribution) • 指數分佈指數分佈 ((Exponential Distribution)Exponential Distribution)

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• A continuous random variable has an uncountably A continuous random variable has an uncountably infinite number of values in the interval (a,b).infinite number of values in the interval (a,b).

8.2 8.2 Continuous Probability DistributionsContinuous Probability Distributions

• The probability that a continuous variable The probability that a continuous variable XX will will assume any particular value is zero. Why?assume any particular value is zero. Why?

0 11/21/3 2/3

1/2 + 1/2 = 11/3 + 1/3 + 1/3 = 11/4 + 1/4 + 1/4 + 1/4 = 1

The probability of each value

50 11/21/3 2/3

1/2 + 1/2 = 11/3 + 1/3 + 1/3 = 11/4 + 1/4 + 1/4 + 1/4 = 1

As the number of values increases the probability of each value decreases. This is so because the sum of all the probabilities remains 1.

When the number of values approaches infinity (because X is continuous) the probability of each value approaches 0.

The probability of each value

8.2 8.2 Continuous Probability DistributionsContinuous Probability Distributions

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• To calculate probabilities we define a probability To calculate probabilities we define a probability density function f(x).density function f(x).

• The density function satisfies the following conditionsThe density function satisfies the following conditions– f(x) is non-negative,f(x) is non-negative,– The total area under the curve representing f(x) equals 1.The total area under the curve representing f(x) equals 1.x1 x2

Area = 1P(x1<=X<=x2)

Probability Density Function (pdf)Probability Density Function (pdf)

• The probability that X falls between xThe probability that X falls between x11 and x and x22 is is found by calculating the area under the graph of f(x) found by calculating the area under the graph of f(x) between xbetween x11 and x and x22..

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– A random variable A random variable XX is said to be uniformly is said to be uniformly distributed if its density function isdistributed if its density function is

– The expected value and the variance areThe expected value and the variance are

12)ab(

)X(V2

baE(X)

.bxaab

1)x(f

2

Uniform DistributionUniform Distribution

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均等分佈均等分佈

均等分佈是指隨機變數在某一連均等分佈是指隨機變數在某一連續的區間，有同等的機率密度，續的區間，有同等的機率密度，如：等車的時間，顧客排隊買票如：等車的時間，顧客排隊買票時間。時間。

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• Example 8.1Example 8.1– The daily sale of gasoline is uniformly distributed between The daily sale of gasoline is uniformly distributed between

2,000 and 5,000 gallons. Find the probability that sales are:2,000 and 5,000 gallons. Find the probability that sales are:– Between 2,500 and 3,500 gallonsBetween 2,500 and 3,500 gallons– More than 4,000 gallonsMore than 4,000 gallons– Exactly 2,500 gallonsExactly 2,500 gallons

2000 5000

1/3000

f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]

x2500 3000

P(2500X3000) = (3000-2500)(1/3000) = .1667

Uniform DistributionUniform Distribution

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• Example 8.1Example 8.1– The daily sale of gasoline is uniformly distributed between The daily sale of gasoline is uniformly distributed between

2,000 and 5,000 gallons. Find the probability that sales are:2,000 and 5,000 gallons. Find the probability that sales are:– Between 2,500 and 3,500 gallonsBetween 2,500 and 3,500 gallons– More than 4,000 gallonsMore than 4,000 gallons– Exactly 2,500 gallonsExactly 2,500 gallons

2000 5000

1/3000

f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]

x4000

P(X4000) = (5000-4000)(1/3000) = .333

Uniform DistributionUniform Distribution

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• Example 8.1Example 8.1– The daily sale of gasoline is uniformly distributed between The daily sale of gasoline is uniformly distributed between

2,000 and 5,000 gallons. Find the probability that sales are:2,000 and 5,000 gallons. Find the probability that sales are:– Between 2,500 and 3,500 gallonsBetween 2,500 and 3,500 gallons– More than 4,000 gallonsMore than 4,000 gallons– Exactly 2,500 gallonsExactly 2,500 gallons

2000 5000

1/3000

f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]

x2500

P(X=2500) = (2500-2500)(1/3000) = 0

Uniform DistributionUniform Distribution

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常態分佈常態分佈

許多連續的隨機變數，如：某一年齡層的身高、體重許多連續的隨機變數，如：某一年齡層的身高、體重，人的智商，某公司的產品與標準規格的差異，測量，人的智商，某公司的產品與標準規格的差異，測量的誤差等。大都會集中於平均數的附近，其中特別大的誤差等。大都會集中於平均數的附近，其中特別大的數值或特別小的數值不多，而且對稱的分散於平均的數值或特別小的數值不多，而且對稱的分散於平均數的兩邊，亦即其次數分配係一個鐘形數的兩邊，亦即其次數分配係一個鐘形 ((bell-shaped)bell-shaped) 的的分配，而且數值大部份集中於離平均數三個標準差之分配，而且數值大部份集中於離平均數三個標準差之內，此種隨機變數，我們稱為常態隨機變數，其分佈內，此種隨機變數，我們稱為常態隨機變數，其分佈稱為常態分佈，該分配是由德國人稱為常態分佈，該分配是由德國人 Karl F. GaussKarl F. Gauss 所提所提出的，故又稱為高斯分佈出的，故又稱為高斯分佈 ((Gaussian Distribution)Gaussian Distribution) 。。常常態分佈是一個最重要、最常用的分配，因為社會及自態分佈是一個最重要、最常用的分配，因為社會及自然界的現象以常態分佈的情形最為普遍。然界的現象以常態分佈的情形最為普遍。

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常態分佈的重要性常態分佈的重要性

• 因為社會與自然界中，常態分佈是最為常見的因為社會與自然界中，常態分佈是最為常見的現象，因此可以做為我們統計推論程序中的基現象，因此可以做為我們統計推論程序中的基本模式。本模式。

• 間斷機率分佈在某些條件下可利用常態分佈，間斷機率分佈在某些條件下可利用常態分佈，求其近似值。求其近似值。

• 常態分佈在大樣本時，是推論統計的基礎。常態分佈在大樣本時，是推論統計的基礎。• 在統計推論上，許多樣本統計量的抽樣分佈，在統計推論上，許多樣本統計量的抽樣分佈，

如：如： tt 分佈，卡方分佈，分佈，卡方分佈， FF 分佈必須假設母體分佈必須假設母體為常態分佈方可獲得。為常態分佈方可獲得。

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8.3 8.3 Normal DistributionNormal Distribution

• This is the most important continuous distribution.This is the most important continuous distribution.

– Many distributions can be approximated by a normal Many distributions can be approximated by a normal distribution.distribution.

– The normal distribution is the cornerstone distribution The normal distribution is the cornerstone distribution of statistical inference.of statistical inference.

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• A random variable A random variable XX with mean with mean and and variancevariance is normally distributed if its is normally distributed if its probability density function is given byprobability density function is given by

...71828.2eand...14159.3where

xe2

1)x(f

2x

)2/1(

...71828.2eand...14159.3where

xe2

1)x(f

2x

)2/1(

Normal DistributionNormal Distribution

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The Shape of the Normal DistributionThe Shape of the Normal Distribution

The normal distribution is bell shaped, and symmetrical around

Why symmetrical? Let = 100. Suppose x = 110.

2210

)2/1(100110

)2/1(e

21

e2

1)110(f

Now suppose x = 9022

10)2/1(

10090)2/1(

e2

1e

21

)90(f

11090

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The effects of and The effects of The effects of and and

How does the standard deviation affect the shape of f(x)?

= 2

=3 =4

= 10 = 11 = 12How does the expected value affect the location of f(x)?

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標準常態分佈標準常態分佈 ((Standard Normal Standard Normal

Distribution)Distribution)

不同的常態分佈因其平均數與標準差互不相同，因此不同的常態分佈因其平均數與標準差互不相同，因此有不同的常態曲線，而要計算某一常態分佈，其在某有不同的常態曲線，而要計算某一常態分佈，其在某一區間的機率，必須利用積分的方法。一區間的機率，必須利用積分的方法。

求出該區間常態曲線以下的面積。這種計算方法相當求出該區間常態曲線以下的面積。這種計算方法相當麻煩且費時，有沒有比較容易省事的方法呢？麻煩且費時，有沒有比較容易省事的方法呢？

2

221?

2

xb

ap a X b e dx

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設設 X~ X~ ，，令令 ZZ ＝ ， ＝ ，則則ZZ 為一標準常態變數，一般以為一標準常態變數，一般以 Z~ N(0,1)Z~ N(0,1)來表示。標準常態分佈的任何值域之機來表示。標準常態分佈的任何值域之機率可查標準常態機率值表而獲得，如：率可查標準常態機率值表而獲得，如：求求 ZZ 落在落在 ((a,b)a,b) 間的機率 間的機率 可查表而獲得。 可查表而獲得。

2,N X

dzzfbZapb

a

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• Two facts help calculate normal probabilities:Two facts help calculate normal probabilities:– The normal distribution is symmetrical.The normal distribution is symmetrical.– Any normal distribution can be transformed into a Any normal distribution can be transformed into a

specific normal distribution called…specific normal distribution called…• ““STANDARD NORMAL DISTRIBUTION”STANDARD NORMAL DISTRIBUTION”

ExampleExampleThe amount of time it takes to assemble a computer is normally The amount of time it takes to assemble a computer is normally

distributed, with a mean of 50 minutes and a standard distributed, with a mean of 50 minutes and a standard deviation of 10 minutes. What is the probability that a deviation of 10 minutes. What is the probability that a computer is assembled in a time between 45 and 60 minutes?computer is assembled in a time between 45 and 60 minutes?

Finding Normal ProbabilitiesFinding Normal Probabilities

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• SolutionSolution– If If X X denotes the assembly time of a computer, we denotes the assembly time of a computer, we

seek the probability P(45<X<60).seek the probability P(45<X<60).– This probability can be calculated by creating a new This probability can be calculated by creating a new

normal variable the normal variable the standard normal variable.standard normal variable.

x

xXZ

x

xXZ

E(Z) = = 0 V(Z) = 2 = 1

Every normal variablewith some and, canbe transformed into this Z.

Therefore, once probabilities for Zare calculated, probabilities of any normal variable can be found.

Finding Normal ProbabilitiesFinding Normal Probabilities

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• Example - continuedExample - continued

P(45<X<60) = P( < < )45 X 60- 50 - 5010 10

= P(-0.5 < Z < 1)

To complete the calculation we need to compute the probability under the standard normal distribution

Finding Normal ProbabilitiesFinding Normal Probabilities

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z 0 0.01 ……. 0.05 0.060.0 0.0000 0.0040 0.0199 0.02390.1 0.0398 0.0438 0.0596 0.0636

. . . . .

. . . . .1.0 0.3413 0.3438 0.3531 0.3554

. . . . .

. . . . .1.2 0.3849 0.3869 ……. 0.3944 0.3962

. . . . . .

. . . . . .

Standard normal probabilities have been calculated and are provided in a table .

The tabulated probabilities correspondto the area between Z=0 and some Z = z0 >0 Z = 0 Z = z0

P(0<Z<z0)

Using the Standard Normal TableUsing the Standard Normal Table

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• Example - continuedExample - continued

P(45<X<60) = P( < < )45 X 60- 50 - 5010 10

= P(-.5 < Z < 1)

z0 = 1z0 = -.5

We need to find the shaded area

Finding Normal ProbabilitiesFinding Normal Probabilities

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P(-.5<Z<0)+ P(0<Z<1)

P(45<X<60) = P( < < )45 X 60- 50 - 5010 10

z 0 0.1 ……. 0.05 0.060.0 0.0000 0.0040 0.0199 0.02390.1 0.0398 0.0438 0.0596 0.636

. . . . .

. . . . .1.0 0.3413 0.3438 0.3531 0.3554

. . . . .

P(0<Z<1

• Example - continuedExample - continued

= P(-.5<Z<1) =

z=0 z0 = 1z0 =-.5

.3413

Finding Normal ProbabilitiesFinding Normal Probabilities

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• The symmetry of the normal distribution makes it The symmetry of the normal distribution makes it possible to calculate probabilities for negative possible to calculate probabilities for negative values of Z using the table as follows:values of Z using the table as follows:

-z0 +z00

P(-z0<Z<0) = P(0<Z<z0)

Finding Normal ProbabilitiesFinding Normal Probabilities

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z 0 0.1 ……. 0.05 0.060.0 0.0000 0.0040 0.0199 0.02390.1 0.0398 0.0438 0.0596 0.636

. . . . .

. . . . .0.5 0.1915 …. …. ….

. . . . .

• Example - continuedExample - continued

Finding Normal ProbabilitiesFinding Normal Probabilities

.3413

.5-.5

.1915

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z 0 0.1 ……. 0.05 0.060.0 0.0000 0.0040 0.0199 0.02390.1 0.0398 0.0438 0.0596 0.636

. . . . .

. . . . .0.5 0.1915 …. …. ….

. . . . .

• Example - continuedExample - continued

Finding Normal ProbabilitiesFinding Normal Probabilities

.1915.1915.1915.1915

.3413

.5-.5

P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) = .1915 + .3413 = .5328

1.0

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10%0%

20-2

(i) P(X< 0 ) = P(Z< ) = P(Z< - 2)0 - 10

5

=P(Z>2) =Z

X

• Example 8.2Example 8.2– The rate of return (X) on an investment is normally distributed The rate of return (X) on an investment is normally distributed

with mean of 10% and standard deviation of (i) 5%, (ii) 10%.with mean of 10% and standard deviation of (i) 5%, (ii) 10%.– What is the probability of losing money?What is the probability of losing money?

.4772

0.5 - P(0<Z<2) = 0.5 - .4772 = .0228

Finding Normal ProbabilitiesFinding Normal Probabilities

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10%0%

-1

(ii) P(X< 0 ) = P(Z< ) 0 - 10

10

= P(Z< - 1) = P(Z>1) =Z

X

• Example 8.2Example 8.2– The rate of return (X) on an investment is normally distributed The rate of return (X) on an investment is normally distributed

with mean of 10% and standard deviation of (i) 5%, (ii) 10%.with mean of 10% and standard deviation of (i) 5%, (ii) 10%.– What is the probability of losing money?What is the probability of losing money?

.3413

0.5 - P(0<Z<1) = 0.5 - .3413 = .1587

Finding Normal ProbabilitiesFinding Normal Probabilities

Find Normal Probabilities

1

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• Sometimes we need to find the value of Z for a Sometimes we need to find the value of Z for a given probabilitygiven probability

• We use the notation zWe use the notation zA A to express a Z value for to express a Z value for which P(Z > zwhich P(Z > zAA) = A) = A

Finding Values of ZFinding Values of Z

zA

A

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• Example 8.3 & 8.4Example 8.3 & 8.4– Determine z exceeded by 5% of the populationDetermine z exceeded by 5% of the population– Determine z such that 5% of the population is belowDetermine z such that 5% of the population is below

• SolutionSolutionzz.05.05 is defined as the z value for which the area on its right under the is defined as the z value for which the area on its right under the

standard normal curve is .05.standard normal curve is .05.

0.05

Z0.050

0.45

1.645

Finding Values of ZFinding Values of Z

0.05

-Z0.05

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二項分佈，波氏分佈與常態分佈二項分佈，波氏分佈與常態分佈

• nn 很大，很大， pp 很小時，二項分佈很小時，二項分佈波氏分佈。波氏分佈。• nn 很大，很大， pp 不是很小時，二項分佈不是很小時，二項分佈常態分佈常態分佈

。 。 • npnp5 and nq5 and nq55 時，則可以常態分佈代替時，則可以常態分佈代替二項分佈，求其機率值， 二項分佈，求其機率值，

＝＝ npnp ，，

2 2 2

2 npq

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指數分佈指數分佈 ((Exponential Distribution)Exponential Distribution)

指數分佈是另一個常用的連續機率分佈，通常指數分佈是另一個常用的連續機率分佈，通常應用於間隔或等待的時間，如：燈泡被使用至應用於間隔或等待的時間，如：燈泡被使用至壞掉的時間，顧客排隊等待服務的時間，機器壞掉的時間，顧客排隊等待服務的時間，機器發生故障的間隔時間等。發生故障的間隔時間等。

指數分佈的隨機變數為非負之值，且通常該隨指數分佈的隨機變數為非負之值，且通常該隨機變數超過其平均數的機率較小，如服務時間機變數超過其平均數的機率較小，如服務時間、發生事件間隔時間等，通常具有這樣的性質、發生事件間隔時間等，通常具有這樣的性質。 。

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Exponential DistributionExponential Distribution

• The exponential distribution can be used to modelThe exponential distribution can be used to model– the length of time between telephone callsthe length of time between telephone calls– the length of time between arrivals at a service stationthe length of time between arrivals at a service station– the life-time of electronic components.the life-time of electronic components.

• When the number of occurrences of an event When the number of occurrences of an event follows the Poisson distribution, the time between follows the Poisson distribution, the time between occurrences follows the exponential distribution.occurrences follows the exponential distribution.

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A random variable is exponentially distributed A random variable is exponentially distributed if its probability density function is given byif its probability density function is given by

f(x) = f(x) = ee--xx, x>=0., x>=0.is the distribution parameter (is the distribution parameter (>0).>0).

E(X) = 1/E(X) = 1/ V(X) = (1/ V(X) = (1/

Exponential DistributionExponential Distribution

37

0

0.5

1

1.5

2

2.5f(x) = 2e-2x

f(x) = 1e-1x

f(x) = .5e-.5x

0 1 2 3 4 5

Exponential distribution for = .5, 1, 2

0

0.5

1

1.5

2

2.5

a bP(a<x<b) = e-a - e-b

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• Finding exponential probabilities is relatively Finding exponential probabilities is relatively easy:easy:– P(X > a) = eP(X > a) = e––aa..– P(X < a) = 1 – e P(X < a) = 1 – e ––aa

– P(aP(a11 < X < a < X < a22) = e ) = e – – a1)a1) – e – e – – a2)a2)

Exponential DistributionExponential Distribution

39

• Example 8.5Example 8.5– The lifetime of an alkaline battery is exponentially The lifetime of an alkaline battery is exponentially

distributed with distributed with = .05 per hour.= .05 per hour.– What is the mean and standard deviation of the What is the mean and standard deviation of the

battery’s lifetime?battery’s lifetime?– Find the following probabilities:Find the following probabilities:

• The battery will last between 10 and 15 hours.The battery will last between 10 and 15 hours.• The battery will last for more than 20 hours?The battery will last for more than 20 hours?

Exponential DistributionExponential Distribution

40

• SolutionSolution– The mean = standard deviation = The mean = standard deviation =

1/1/1/.05 = 20 hours.1/.05 = 20 hours.– Let X denote the lifetime.Let X denote the lifetime.

• P(10<X<15) = eP(10<X<15) = e-.05(10)-.05(10) – e – e-.05(15)-.05(15) = .1341 = .1341• P(X > 20) = eP(X > 20) = e-.05(20)-.05(20) = .3679 = .3679

Exponential DistributionExponential Distribution

41

• Example 8.6Example 8.6– The service rate at a supermarket checkout is 6 The service rate at a supermarket checkout is 6

customers per hour. customers per hour. – If the service time is exponential, find the following If the service time is exponential, find the following

probabilities:probabilities:• A service is completed in 5 minutes,A service is completed in 5 minutes,• A customer leaves the counter more than 10 minutes after A customer leaves the counter more than 10 minutes after

arrivingarriving• A service is completed between 5 and 8 minutes.A service is completed between 5 and 8 minutes.

Exponential DistributionExponential Distribution

42

• SolutionSolution– A service rate of 6 per hour = A service rate of 6 per hour =

A service rate of .1 per minute (A service rate of .1 per minute ( = .1/minute). = .1/minute).– P(X < 5) = 1-eP(X < 5) = 1-e--xx = 1 – e = 1 – e-.1(5)-.1(5) = .3935 = .3935– P(X >10) = eP(X >10) = e--xx = e = e-.1(10)-.1(10) = .3679 = .3679– P(5 < X < 8) = eP(5 < X < 8) = e-.1(5)-.1(5) – e – e-.1(8)-.1(8) = .1572 = .1572

Exponential DistributionExponential Distribution

Compute Exponential probabilities

43

8.5 8.5 Other Continuous DistributionOther Continuous Distribution

• Three new continuous distributions:Three new continuous distributions:– Student t distributionStudent t distribution– Chi-squared distributionChi-squared distribution– F distributionF distribution

44

The Student t DistributionThe Student t Distribution

• The Student t density functionThe Student t density function

is the parameter of the student t distributionis the parameter of the student t distribution

E(t) = 0 V(t) =E(t) = 0 V(t) =((– 2)– 2)

2/)1(2t1

)]!2[()]!1[(

)t(f

2/)1(2t1

)]!2[()]!1[(

)t(f

((for n > 2)for n > 2)

45

The Student t DistributionThe Student t Distribution

0

0.05

0.1

0.15

0.2

-6 -4 -2 0 2 4 6

0

0.05

0.1

0.15

0.2

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

= 3

= 10

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Determining Student t Values Determining Student t Values

• The student t distribution is used extensively in The student t distribution is used extensively in statistical inference.statistical inference.

• Thus, it is important to determine values of tThus, it is important to determine values of tAA associated with a given number of degrees of freedom.associated with a given number of degrees of freedom.

• We can do this usingWe can do this using– t tables t tables – ExcelExcel– MinitabMinitab

47

Degrees of Freedom1 3.078 6.314 12.706 31.821 63.6572 1.886 2.92 4.303 6.965 9.925. . . . . .. . . . . .

10 1.372 1.812 2.228 2.764 3.169. . . . . .. . . . . .

200 1.286 1.653 1.972 2.345 2.6011.282 1.645 1.96 2.326 2.576

tA

t.100 t.05 t.025 t.01 t.005

A = .05A = .05

-tA

The t distribution issymmetrical around 0

=1.812=-1.812

• The table provides the t values (tThe table provides the t values (tAA) for which P(t) for which P(t > t > tAA) = A) = AUsing the t TableUsing the t Table

tttttttt

48

The Chi – Squared Distribution The Chi – Squared Distribution

• The Chi – Squared density function:The Chi – Squared density function:

• The parameter The parameter is the number of degrees of is the number of degrees of freedom.freedom.

0)(2]!1)2/[(

1)( 221)2/(2

2/2 2

ef 0)(

2]!1)2/[(

1)( 221)2/(2

2/2 2

ef

49

The Chi – Squared Distribution The Chi – Squared Distribution

00.00020.00040.00060.0008

0.0010.00120.00140.00160.0018

0 5 10 15 20 25 30 35

= 5

= 10

50

• Chi squared values can be found from the chi squared Chi squared values can be found from the chi squared table, from Excel, or from Minitab.table, from Excel, or from Minitab.

• The The 22-table entries are the -table entries are the values of the right hand values of the right hand tail probability (A), for which P(tail probability (A), for which P(

= A.= A.

Determining Chi-Squared ValuesDetermining Chi-Squared Values

0 5 10 15 20 25 30 35

A

2A

51

Degrees offreedom

1 0.0000393 0.0001571 . 3.84146 6.6349 7.87944..

10 2.15585 2.55821 . 18.307 23.2093 25.1882. . . . .. . . . . . . .

Degrees offreedom

1 0.0000393 0.0001571 . 3.84146 6.6349 7.87944..

10 2.15585 2.55821 . 18.307 23.2093 25.1882. . . . .. . . . . . . .

0 5 10 15 20 25 30 35

=.05A =.99

Using the Chi-Squared TableUsing the Chi-Squared Table

A

To find 2 for which P(2

<2)=.01, lookup the column labeled2

1-.01 or 2.99

52

The F DistributionThe F Distribution

• The density function of the F distribution:The density function of the F distribution:

11 and and 2 2 are the numerator and denominator are the numerator and denominator degrees of freedom.degrees of freedom.

0FF

1

F

22

22

22

)F(f2

2

1

22

2

2

1

21

21

21

11

0FF

1

F

22

22

22

)F(f2

2

1

22

2

2

1

21

21

21

11

!!

!

53

0

0.002

0.004

0.006

0.008

0.01

0 1 2 3 4 5

• This density function generates a rich family of This density function generates a rich family of distributions, depending on the values of distributions, depending on the values of 11 and and 22

The F DistributionThe F Distribution

1 = 5, 2 = 10

1 = 50, 2 = 10

00.0010.0020.0030.0040.0050.0060.0070.008

0 1 2 3 4 5

1 = 5, 2 = 10

1 = 5, 2 = 1

54

Determining Values of FDetermining Values of F

• The values of the F variable can be found in the The values of the F variable can be found in the F table, Excel, or from Minitab.F table, Excel, or from Minitab.

• The entries in the table are the values of the F The entries in the table are the values of the F variable of the right hand tail probability (A), for variable of the right hand tail probability (A), for which P(Fwhich P(F1,1,22>F>FAA) = A.) = A.

55

Homework for Chapter 8Homework for Chapter 8

Chapter 8Chapter 8 5, 11, 13, 21, 37, 53, 83, 5, 11, 13, 21, 37, 53, 83, 85, 85,

93, 99, 105.93, 99, 105.