Lecture 9 Probability Distributions - univet.hu · ©The McGraw-Hill Companies, Inc., 2000 6-2 Probability Distributions If a variable can assume only a specific number of values,

© The McGraw-Hill Companies, Inc., 2000

Lecture 9Lecture 9

Probability Probability DistributionsDistributions


OutlineOutline

� 6-1 Introduction� 6-2 Probability Distributions� 6-3 Mean, Variance, and

Expectation� 6-4 The Binomial Distribution


OutlineOutline

� 7-2 Properties of the Normal Distribution

� 7-3 The Standard Normal Distribution

� 7-4 Applications of the Normal Distribution

� 7-5 The Central Limit Theorem


66--2 Probability Distributions2 Probability Distributions

� A variablevariable is defined as a characteristic or attribute that can assume different values.

� A variable whose values are determined by chance is called a random variablerandom variable .



� If a variablevariable can assume only a specific number of values, such as the outcomes for the roll of a die or the outcomes for the toss of a coin, then the variable is called a discrete discrete variablevariable .

� Discrete variables have values that can be counted.



� If a variablevariable can assume all values in the interval between two given values then the variable is called a continuous continuous variable.variable. Example Example -- temperature between 68 0 to 780.

� Continuous random variables are obtained from data that can be measured rather than counted.


66--2 Probability Distributions 2 Probability Distributions --Tossing Two CoinsTossing Two Coins

First Toss T

H

H

T H

T

Second Toss


� From the tree diagram, the sample space will be represented by HH, HH, HT, TH, TT.HT, TH, TT.

� If XX is the random variable for the number of heads, then XX assumes the value 0, 1, or 20, 1, or 2 .




TTTT

THTH

HTHT

HHHH

00

11

22

Sample Space Number of Heads


OUTCOME X

PROBABILITYP(X)

0 1/4

1 2/4

2 1/4



� A probability distributionprobability distribution consists of the values a random variable can assume and the corresponding probabilities of the values. The probabilities are determined theoretically or by observation.



66--2 Probability Distributions 2 Probability Distributions ----Graphical RepresentationGraphical Representation

3210

1

0.5

.25

NUMBER OF HEADS

PR

OBA

BIL

ITY

Experiment: Toss Two Coins


Two requirements

� The sum of the probabilities of all the events in the sample space must equal 1

� The probability of each event in the sample space must be between 0 and 1.


66--3 Mean, Variance, and 3 Mean, Variance, and Expectation for Discrete VariableExpectation for Discrete Variable

=

The mean of the random of a

probability distribution is

X P X X P X X P X

X P X

where X X X are the outcomesand

P X P X P X are the corresponding

probabilities

n n

n

n

variable

µ1 1 2 2

1 2

1 2

⋅ + ⋅ + + ⋅= ⋅∑

( ) ( ) ... ( )

( )

, ,...,

( ), ( ), ... , ( )

.


� Find the mean of the number of spots that appear when a die is tossed. The probability distribution is given below.

66--3 Mean for Discrete Variable 3 Mean for Discrete Variable --Example

X 1 2 3 4 5 6

P(X) 1/6 1/6 1/6 1/6 1/6 1/6

X 1 2 3 4 5 6

P(X) 1/6 1/6 1/6 1/6 1/6 1/6



µ = ⋅∑

= ⋅ + ⋅ + ⋅ + ⋅+ ⋅ + ⋅

= =

X P X( )

( / ) ( / ) ( / ) ( / )

( / ) ( / )

/ .

1 1 6 2 1 6 3 1 6 4 1 6

5 1 6 6 1 6

21 6 35

That is, when a die is tossed many times,the theoretical mean will be 3.5.

That is, when a die is tossed many times,the theoretical mean will be 3.5.


� In a family with two children, find the mean number of children who will be girls. The probability distribution is given below.

X 0 1 2

P(X) 1/4 1/2 1/4

X 0 1 2

P(X) 1/4 1/2 1/4



= .

µ = ⋅∑

= ⋅ + ⋅ + ⋅X P X( )

( / ) ( / ) ( / )0 1 4 1 1 2 2 1 4

1

That is, the average number of girls in a two-child family is 1.That is, the average number of girls in a two-child family is 1.



Variance of a Probability Distribution

� The mean describes the measure of the long-run or theoretical average, but it does not tell anything about the spread of the distribution.


6-3 Formula for the Variance of a Probability Distribution

� The variance of a probability distribution is found by multiplying the square of each outcome by its corresponding probability, summing these products, and subtracting the square of the mean.


66--3 Formula for the Variance of a 3 Formula for the Variance of a Probability DistributionProbability Distribution

[ ]

= 2

The formula for the of a


X P X

The standard deviation of a


variance

σ µ

σ σ

2 2 2= ⋅ −∑ ( ) .

.


66--3 Variance of a Probability 3 Variance of a Probability Distribution Distribution -- Example

� The probability that 0, 1, 2, 3, or 4 people will be placed on hold when they call a radio talk show with four phone lines is shown in the distribution below. Find the variance and standard deviation for the data.



X 0 1 2 3 4

P (X ) 0 .1 8 0 .3 4 0 .2 3 0 .2 1 0 .0 4



X P(X) X⋅⋅⋅⋅P(X) X2⋅⋅⋅⋅P(X)

0 0.18 0 0

1 0.34 0.34 0.34

2 0.23 0.46 0.92

3 0.21 0.63 1.89

4 0.04 0.16 0.64

µµµµ = 1.59 ΣΣΣΣX2⋅⋅⋅⋅P(X)=3.79

σσσσ2 = 3.79 – 1.592

= 1.26

σσσσ2 = 3.79 – 1.592

= 1.26


� Now, µµµµ = (0)(0.18) + (1)(0.34) + (2)(0.23) + (3)(0.21) + (4)(0.04) = 1.59.

� ΣΣΣΣ X 2 P(X) = (02)(0.18) + (12)(0.34) + (22)(0.23) + (32)(0.21) + (42)(0.04) = 3.79

� 1.592 = 2.53 (rounded to two decimal places).

� σσσσ 2 = 3.79 – 2.53 = 1.26� σσσσ = = 1.12


1.26


66--3 Expectation3 Expectation

= ( ) =

( )

.

The value of a discrete

random of a probability

distribution is the l average

of the The formula is

E X X P X

The symbol E X is used for the

value

expected

variable

theoretica

variable

expected

.

( )µ ⋅∑


66--3 Expectation 3 Expectation -- Example

� A ski resort loses $70,000 per season when it does not snow very much and makes $250,000 when it snows a lot. The probability of it snowing at least 75 inches (i.e., a good season) is 40%. Find the expected profit.


� The expected profit = ($250,000)(0.40) + (–$70,000)(0.60) = $58,000.

66--3 Expectation 3 Expectation -- Example

Profit, X 250,000 –70,000

P(X) 0.40 0.60


66--4 The Binomial Distribution4 The Binomial Distribution

� A binomial experimentbinomial experiment is a probability experiment that satisfies the following four requirements:

� Each trial can have only two outcomes or outcomes that can be reduced to two outcomes. Each outcome can be considered as either a success or a failure.



� There must be a fixed number of trials.� The outcomes of each trial must be

independent of each other.� The probability of success must remain

the same for each trial.



� The outcomes of a binomial experiment and the corresponding probabilities of these outcomes are called a binomial distribution.binomial distribution.



� Notation for the Binomial Distribution:

�� PP((SS) = ) = pp, probability of a success�� PP((FF) = 1 ) = 1 –– pp = = qq, probability of a

failure�� nn = number of trials� XX = number of successes.


66--4 Binomial Probability Formula4 Binomial Probability Formula

In a binomial the probability of

exactly X successes in n trials is

P Xn

n X Xp qX n X

experiment,

( )!

( )! !=

− −


66--4 Binomial Probability 4 Binomial Probability -- Example

� If a student randomly guesses at five multiple-choice questions, find the probability that the student gets exactly three correct. Each question has five possible choices.

�� Solution:Solution: n = 5, X = 3, and p = 1/5. Then, P(3) = [5!/((5 – 3)!3! )](1/5) 3(4/5)2 0.05.≈



� A survey from Teenage Research Unlimited (Northbrook, Illinois.) found that 30% of teenage consumers received their spending money from part-time jobs. If five teenagers are selected at random, find the probability that at least three of them will have part-time jobs.



�� Solution:Solution: n = 5, X = 3, 4, and 5, and p = 0.3. Then, P(X ≥ ≥ ≥ ≥ 3) = P(3) + P(4) + P(5) = 0.1323 + 0.0284 + 0.0024 = 0.1631.

�� NOTE:NOTE: You can use Table B in the textbook to find the Binomial probabilities as well.



� A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur on weekend nights involve an intoxicated driver. If a sample of 15 single-vehicle traffic fatalities that occurred on a weekend night is selected, find the probability that exactly 12 involve a driver who is intoxicated.



�� Solution:Solution: n = 15, X = 12, and p = 0.7. From Table B, P(X =12) = 0.170


66--4 Mean, Variance, Standard Deviation 4 Mean, Variance, Standard Deviation for the Binomial Distribution for the Binomial Distribution -- Example

� A coin is tossed four times. Find the mean, variance, and standard deviation of the number of heads that will be obtained.

�� Solution:Solution: n = 4, p = 1/2, and q = 1/2. � µµµµ = n⋅⋅⋅⋅p = (4)(1/2) = 2. � σ σ σ σ 2 = n⋅⋅⋅⋅p⋅⋅⋅⋅q = (4)(1/2)(1/2) = 1.� σσσσ = = 1.1


77--2 The Normal Distribution2 The Normal Distribution

� Many continuous variables have distributions that are bell-shaped and are called approximately normally distributed approximately normally distributed variables.variables.

� The theoretical curve, called the normal normal distribution curvedistribution curve , can be used to study many variables that are not normally distributed but are approximately normal.


77--2 Mathematical Equation for the 2 Mathematical Equation for the Normal DistributionNormal Distribution

The mathematical equation for the normal distribution:

where

e

population mean

population standard deviation

2 718

314

≈≈==

πµ

σ

.

.

πσ

σµ

2

222)( −−

=x

ey


77--2 Properties of the Normal 2 Properties of the Normal DistributionDistribution

� The shape and position of the normal distribution curve depend on two parameters, the mean and the standard deviation.

� Each normally distributed variable has its own normal distribution curve, which depends on the values of the variable’s mean and standard deviation.


77--2 Properties of the 2 Properties of the Theoretical Normal DistributionTheoretical Normal Distribution

� The normal distribution curve is bell-shaped.

� The mean, median, and mode are equal and located at the center of the distribution.

� The normal distribution curve isunimodal (single mode).



� The curve is symmetrical about the mean.

� The curve is continuous.� The curve never touches the x-axis.� The total area under the normal

distribution curve is equal to 1.



� The area under the normal curve that lies within � one standard deviation of the mean is

approximately 0.68 (68%). � two standard deviations of the mean is

approximately 0.95 (95%).� three standard deviations of the mean is

approximately 0.997 (99.7%).


77--2 Areas Under the Normal Curve2 Areas Under the Normal Curve

µµµµ −3−3−3−3σσσσ µµµµ −2−2−2−2σσσσ µµµµ −1−1−1−1σσσσ µµµµ µµµµ +1+1+1+1σσσσ µµµµ +2+2+2+2σσσσ µµµµ +3+3+3+3σσσσ

68%

95% 99.7%


77--3 The Standard Normal 3 The Standard Normal DistributionDistribution

� The standard normal distributionstandard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.

� All normally distributed variables can be transformed into the standard normally distributed variable by using the formula for the standard score: (see next slide)


77--3 The Standard Normal 3 The Standard Normal DistributionDistribution

z

or

zX

= −

= −

value mean

standard deviation

µσ



77--3 Area Under the Standard 3 Area Under the Standard Normal Curve Normal Curve -- Example

� Find the area under the standard normal curve between z = 0 and z = 2.34 ⇒⇒⇒⇒⇒⇒⇒⇒ PP(0(0 ≤≤≤≤≤≤≤≤ zz ≤≤≤≤≤≤≤≤ 2.34)2.34).

� Use your table at the end of the text to find the area.

� The next slide shows the shaded area.



00000000 2.342.342.342.342.342.342.342.34

0.49040.49040.49040.49040.49040.49040.49040.4904


77-- 3 Area Under the Standard 3 Area Under the Standard Normal Curve Normal Curve -- Example

� Find the area under the standard normal curve between z = 0 and z = –1.75 ⇒⇒⇒⇒⇒⇒⇒⇒ PP((––1.751.75 ≤≤≤≤≤≤≤≤ zz ≤≤≤≤≤≤≤≤ 0)0).

� Use the symmetric property of the normal distribution and your table at the end of the text to find the area.




00000000 1.751.751.751.751.751.751.751.75−1.75−1.75−1.75−1.75−1.75−1.75−1.75−1.75

0.45990.45990.45990.45990.45990.45990.45990.45990.45990.45990.45990.45990.45990.45990.45990.4599



� Find the area to the right of z = 1.11⇒⇒⇒⇒⇒⇒⇒⇒ PP((zz >>>>>>>> 1.11)1.11).





00000000 1.111.111.111.111.111.111.111.11

0.36650.36650.36650.36650.36650.36650.36650.3665

0.5000 0.5000 0.5000 0.5000−0.3665−0.3665−0.3665−0.36650.13350.13350.13350.1335



� Find the area to the left of z = –1.93⇒⇒⇒⇒⇒⇒⇒⇒ PP((zz <<<<<<<< ––1.93)1.93).


� The next slide shows the area.



−1.93−1.93−1.93−1.93 1.931.931.931.930000

0.47320.47320.47320.4732

0.50000.50000.50000.5000−0.4732−0.4732−0.4732−0.47320.02680.02680.02680.02680.02680.02680.02680.0268



� Find the area between z = 2 and z = 2.47 ⇒⇒⇒⇒⇒⇒⇒⇒ PP(2(2 ≤≤≤≤≤≤≤≤ zz ≤≤≤≤≤≤≤≤ 2.47)2.47).




77--3 Area Under the Standard Normal Curve 3 Area Under the Standard Normal Curve --Example

00000000 2.472.472.472.472.472.472.472.4722222222

0.47720.47720.47720.47720.47720.47720.47720.4772

0.49320.49320.49320.49320.49320.49320.49320.4932−0.4772−0.4772−0.4772−0.4772−0.4772−0.4772−0.4772−0.47720.01600.01600.01600.01600.01600.01600.01600.0160

0.49320.49320.49320.49320.49320.49320.49320.4932


� Find the area between z = 1.68 and z = –1.37 ⇒⇒⇒⇒⇒⇒⇒⇒ PP((––1.371.37 ≤≤≤≤≤≤≤≤ zz ≤≤≤≤≤≤≤≤ 1.68)1.68).





00000000 1.681.681.681.681.681.681.681.68

0.45350.45350.45350.45350.45350.45350.45350.4535

0.45350.45350.45350.45350.45350.45350.45350.4535+0.4147+0.4147+0.4147+0.4147+0.4147+0.4147+0.4147+0.41470.86820.86820.86820.86820.86820.86820.86820.8682

0.41470.41470.41470.41470.41470.41470.41470.4147

−1.37−1.37−1.37−1.37−1.37−1.37−1.37−1.37



� Find the area to the left of z = 1.99⇒⇒⇒⇒⇒⇒⇒⇒ PP((zz <<<<<<<< 1.99)1.99).





00000000 1.991.991.991.991.991.991.991.99

0.47670.47670.47670.47670.47670.47670.47670.4767

0.50000.50000.50000.50000.50000.50000.50000.5000+0.4767+0.4767+0.4767+0.4767+0.4767+0.4767+0.4767+0.47670.97670.97670.97670.97670.97670.97670.97670.9767

0.50000.50000.50000.50000.50000.50000.50000.5000



� Find the area to the right of z = –1.16 ⇒⇒⇒⇒⇒⇒⇒⇒ PP((zz >>>>>>>> ––1.16)1.16).





−1.16−1.16−1.16−1.16−1.16−1.16−1.16−1.16 00000000

0.50000.50000.50000.50000.50000.50000.50000.5000

0.50000.50000.50000.50000.50000.50000.50000.5000++++++++ 0.37700.37700.37700.37700.37700.37700.37700.3770

0.87700.87700.87700.87700.87700.87700.87700.8770

0.3770.3770.3770.3770.3770.3770.3770.377



RECALL:RECALL: The Standard Normal Distribution

z

or

zX

= −

= −

value mean

standard deviation

µσ


77--4 Applications of the Normal 4 Applications of the Normal Distribution Distribution -- Example

� Each month, an American household generates an average of 28 pounds of newspaper for garbage or recycling. Assume the standard deviation is 2 pounds. Assume the amount generated is normally distributed.

� If a household is selected at random, find the probability of its generating:


�� More than 30.2 pounds per month.More than 30.2 pounds per month.� First find the z-value for 30.2.

z =[X –µµµµ]/σσσσ = [30.2 – 28]/2 = 1.1.� Thus, P(z > 1.1) = 0.5 – 0.3643 = 0.1357.� That is, the probability that a randomly

selected household will generate more than 30.2 lbs. of newspapers is 0.1357 or 13.57%.



1.11.11.11.11.11.11.11.100000000

0.50000.50000.50000.50000.50000.50000.50000.5000−0.3643−0.3643−0.3643−0.3643−0.3643−0.3643−0.3643−0.36430.13570.13570.13570.13570.13570.13570.13570.1357



�� Between 27 and 31 pounds per month.Between 27 and 31 pounds per month.� First find the z-value for 27 and 31.

z1 = [X –µµµµ]/σσσσ = [27 – 28]/2 = –0.5; z2 = [31 – 28]/2 = 1.5

� Thus, P(–0.5 ≤ ≤ ≤ ≤ z ≤≤≤≤ 1.5) = 0.1915 + 0.4332 = 0.6247.



−0.5−0.5−0.5−0.5−0.5−0.5−0.5−0.5 00000000 1.51.51.51.51.51.51.51.5

0.19150.19150.19150.19150.19150.19150.19150.1915++++++++ 0.4332 0.4332 0.4332 0.4332 0.4332 0.4332 0.4332 0.4332

0.62470.62470.62470.62470.62470.62470.62470.62470.19150.19150.19150.19150.19150.19150.19150.1915

0.43320.43320.43320.43320.43320.43320.43320.4332




� The American Automobile Association reports that the average time it takes to respond to an emergency call is 25 minutes. Assume the variable is approximately normally distributed and the standard deviation is 4.5 minutes. If 80 calls are randomly selected, approximately how many will be responded to in less than 15 minutes?


� First find the z-value for 15 is z = [X –µµµµ]/σσσσ = [15 – 25]/4.5 = –2.22.

� Thus, P(z < < < < –2.22) = 0.5000 – 0.4868 = 0.0132.

� The number of calls that will be made in less than 15 minutes = (80)(0.0132) = 1.056 ≈≈≈≈ 1.



00000000 2.222.222.222.222.222.222.222.22−2.22−2.22−2.22−2.22−2.22−2.22−2.22−2.22


0.50000.50000.50000.50000.50000.50000.50000.5000−−−−−−−− 0.48680.48680.48680.48680.48680.48680.48680.4868

0.01320.01320.01320.01320.01320.01320.01320.01320.01320.01320.01320.01320.01320.01320.01320.0132



� An exclusive college desires to accept only the top 10% of all graduating seniors based on the results of a national placement test. This test has a mean of 500 and a standard deviation of 100. Find the cutoff score for the exam. Assume the variable is normally distributed.


� Work backward to solve this problem.� Subtract 0.1 (10%) from 0.5 to get the

area under the normal curve for accepted students.

� Find the z value that corresponds to an area of 0.4000 by looking up 0.4000 in the area portion of Table E. Use the closest value, 0.3997.

77-- 4 Applications of the Normal 4 Applications of the Normal Distribution Distribution -- Example


� Substitute in the formulaand solve for X.

� The z-value for the cutoff score ( X) is z = [X –µµµµ]/σσσσ = [X – 500]/100 = 1.28. (See next slide).

� Thus, X = (1.28)(100) + 500 = 628.� The score of 628 should be used as a

cutoff score.

77-- 4 Applications of the Normal 4 Applications of the Normal Distribution Distribution -- Example

σµ−= Xz


00000000 X X = 1.28= 1.28= 1.28= 1.28= 1.28= 1.28= 1.28= 1.28


0.10.10.10.10.10.10.10.1

0.40.40.40.40.40.40.40.4


�� NOTE:NOTE: To solve for X, use the following formula: XX = = zz⋅⋅⋅⋅⋅⋅⋅⋅σσσσσσσσ + + µµµµµµµµ..

� Example: For a medical study, a researcher wishes to select people in the middle 60% of the population based on blood pressure. (Continued on the next slide).



� (Continued)-- If the mean systolic blood pressure is 120 and the standard deviation is 8, find the upper and lower readings that would qualify people to participate in the study.



� Note that two values are needed, one above the mean and one below the mean. The closest zvalues are 0.84 and – 0.84 respectively.

� X = (z)(σσσσ) + µµµµ = (0.84)(8) + 120 = 126.72. The other X = (–0.84)(8) + 120 = 113.28.See next slide.

� i.e. the middle 60% of BP readings is between 113.28 and 126.72 .


(continued)


00000000


−0.84 −0.84 −0.84 −0.84 −0.84 −0.84 −0.84 −0.84 0.84 0.84 0.84 0.84 0.84 0.84 0.84 0.84

0.20.20.20.20.20.20.20.20.30.30.30.30.30.30.30.3

0.20.20.20.20.20.20.20.20.30.30.30.30.30.30.30.3


77--5 Distribution of Sample Means5 Distribution of Sample Means

�� Distribution of Sample means:Distribution of Sample means: A sampling distribution of sample means is a distribution obtained by using the means computed from random samples of a specific size taken from a population.


�� Sampling errorSampling error is the difference between the sample measure and the corresponding population measure due to the fact that the sample is not a perfect representation of the population.

77--5 Distribution of Sample Means5 Distribution of Sample Means


� The mean of the sample means will be the same as the population mean.

� The standard deviation of the sample means will be smaller than the standard deviation of the population, and it will be equal to the population standard deviation divided by the square root of the sample size.

77--5 Properties of the Distribution of 5 Properties of the Distribution of Sample MeansSample Means


� Suppose a professor gave an 8-point quiz to a small class of four students. The results of the quiz were 2, 6, 4, and 8. Assume the four students constitute the population.

� The mean of the population isµµµµµµµµ = ( 2 + 6 + 4 + 8)/4 = 5= ( 2 + 6 + 4 + 8)/4 = 5.

77--5 Properties of the Distribution of 5 Properties of the Distribution of Sample Means Sample Means -- Example


� The standard deviation of the population is

=2.236.� The graph of the distribution of the scores

is uniform and is shown on the next slide.� Next we will consider all samples of size 2

taken with replacement.


( ) ( ) ( ) ( ) /4}258254256252{σ

−+−+−+−=


77--5 Graph of the Original 5 Graph of the Original DistributionDistribution



Sample Mean Sample Mean

2, 2 2 6, 2 4

2, 4 3 6, 4 5

2, 6 4 6, 6 6

2, 8 5 6, 8 7

4, 2 3 8, 2 5

4, 4 4 8, 4 6

4, 6 5 8, 6 7

4, 8 6 8, 8 8


77--5 Frequency Distribution of the5 Frequency Distribution of theSample Means Sample Means -- Example

X-bar(mean)

2 3 4 5 6 7 8

f 1 2 3 4 3 2 1


77--5 Graph of the Sample Means5 Graph of the Sample Means

8765432

4

3

2

1

0

SAMPLE MEANS

Fre

quen

cy

(APPROXIMATELY NORMAL)DISTRIBUTION OF SAMPLE MEANS


77--5 Mean and Standard Deviation of 5 Mean and Standard Deviation of the Sample Meansthe Sample Means

Mean of Sample Means

which is the same as the

population mean Thus

X

X

µ

µ µ

= + + + = =

=

2 3 816

8016

5...

. .


77--5 Mean and Standard Deviation of 5 Mean and Standard Deviation of the Sample Meansthe Sample Means

The standard deviation of the sample

means is

This is the same as

X

σ

σ

= − + − + + −

=

( ) ( ) ... ( )

. .

.

2 5 3 5 8 516

1581

2

2 2 2


77--5 The Standard Error of the 5 The Standard Error of the MeanMean

The standard deviation of the sample

means is called the standard error of

the mean Hence

nX

.

σ σ= .


77--5 The Central Limit Theorem5 The Central Limit Theorem

� As the sample size nn increases, the shape of the distribution of the sample means taken from a population with mean µµµµµµµµ and standard deviation of σσσσσσσσ will approach a normal distribution. As previously shown, this distribution will have a mean µµµµµµµµ and standard deviation σσσσσσσσ // √√√√√√√√n n ..


.n/

X=z

is It

.values-z the for used be must formula new a

that is difference only The .values individual

about questions answer to used be can

ondistributi normal the that manner same the in

means sample about questions answer to

used be can theorem itlim central The

σµ−

77--5 The Central Limit Theorem5 The Central Limit Theorem


� A.C. Neilsen reported that children between the ages of 2 and 5 watch an average of 25 hours of TV per week. Assume the variable is normally distributed and the standard deviation is 3 hours. If 20 children between the ages of 2 and 5 are randomly selected, find the probability that the mean of the number of hours they watch TV is greater than 26.3 hours.

77--5 The Central Limit Theorem 5 The Central Limit Theorem --Example


� The standard deviation of the sample means is σσσσ/ √√√√n = 3/ √√√√20 = 0.671.

� The z-value is z = (26.3 - 25)/0.671= 1.94.� Thus P(z > 1.94) = 0.5 – 0.4738 = 0.0262.

That is, the probability of obtaining a sample mean greater than 26.3 is 0.0262 = 2.62%.



00000000 1.941.941.941.941.941.941.941.94

0.50000.50000.50000.50000.50000.50000.50000.5000 −−−−−−−− 0.4738 0.4738 0.4738 0.4738 0.4738 0.4738 0.4738 0.4738

0.02620.02620.02620.02620.02620.02620.02620.0262



� The average age of a vehicle registered in the United States is 8 years, or 96 months. Assume the standard deviation is 16 months. If a random sample of 36 cars is selected, find the probability that the mean of their age is between 90 and 100 months.



� The standard deviation of the sample means is σσσσ/ √√√√n = 16/√√√√36 = 2.6667.

� The two z-values are z1 = (90 – 96)/2.6667 = –2.25 and z2 = (100 – 96)/2.6667 = 1.50.

� Thus P(–2.25 ≤ ≤ ≤ ≤ z ≤≤≤≤ 1.50) = 0.4878 + 0.4332 = 0.921 or 92.1%.




00000000 1.501.501.501.501.501.501.501.50−2.25−2.25−2.25−2.25−2.25−2.25−2.25−2.25

0.48780.48780.48780.48780.48780.48780.48780.4878 0.43320.43320.43320.43320.43320.43320.43320.4332

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Lecture 9 Probability Distributions - univet.hu · ©The McGraw-Hill Companies, Inc., 2000 6-2 Probability Distributions If a variable can assume only a specific number of values,