Probability Distributionbs 2 - NYUpages.stern.nyu.edu/~igiddy/Probdistributions.pdfProbability...

Giddy/NYU Probability Distributions /1

Probability Distributions

Prof. Ian GiddyNew York University

New York UniversityStern School of Business

Copyright ©1999 Ian H. Giddy Probability Distributionbs 2

Source:

ELEMENTARY STATISTICS: A BriefVersion

Allan G. BlumanMcGraw-Hill, 2000ISBN: 0-07-237288-5


6-16-1

Chapter 6Chapter 6

ProbabilityProbabilityDistributionsDistributions


6-26-2

OutlineOutline

l 6-1 Introductionl 6-2 Probability Distributionsl 6-3 Mean, Variance, and

Expectation

l 6-4 The Binomial Distribution


6-36-3

ObjectivesObjectives

l Construct a probability distribution for arandom variable.

l Find the mean, variance, and expectedvalue for a discrete random variable.

l Find the exact probability for X successes inn trials of a binomial experiment.


6-46-4


l Find the mean, variance, and standarddeviation for the variable of a binomialdistribution.



6-56-5

6-2 Probability Distributions6-2 Probability Distributions

l A variablevariable is defined as acharacteristic or attribute that canassume different values.

l A variable whose values aredetermined by chance is called arandom variablerandom variable.


6-66-6


l If a variablevariable can assume only a specificnumber of values, such as the outcomesfor the roll of a die or the outcomes for thetoss of a coin, then the variable is called adiscrete variablediscrete variable.

l Discrete variables have values that can becounted.


6-76-7


l If a variablevariable can assume all values in theinterval between two given values then thevariable is called a continuous variable.continuous variable.Example -Example - temperature between 680 to780.

l Continuous random variables are obtainedfrom data that can be measured ratherthan counted.


6-86-8

6-2 Probability Distributions - 6-2 Probability Distributions - Tossing Two CoinsTossing Two Coins

First Toss T

H

H

T H

T

Second Toss


6-96-9

l From the tree diagram, the samplespace will be represented by HH,HH,HT, TH, TT.HT, TH, TT.

l If XX is the random variable for thenumber of heads, then XX assumesthe value 0, 1, or 20, 1, or 2.

6-2 Probability Distributions - 6-2 Probability Distributions - Tossing TwoTossing TwoCoinsCoins


6-106-10


TTTT

THTH

HTHT

HHHH

00

11

22

Sample Space Number of Heads



6-116-11

OUTCOME X

PROBABILITYP(X)

0 1/4

1 2/4

2 1/4



6-126-12

l A probability distributionprobability distribution consists of thevalues a random variable can assumeand the corresponding probabilities ofthe values. The probabilities aredetermined theoretically or byobservation.



6-136-13

6-2 Probability Distributions --6-2 Probability Distributions -- Graphical Representation Graphical Representation

3210

1

0.5

.25

NUMBER OF HEADS

PR

OBA

BILI

TY

Experiment: Toss Two Coins


6-146-14

6-3 Mean, Variance, and Expectation for Discrete6-3 Mean, Variance, and Expectation for DiscreteVariableVariable

=

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 outcomes and

P X P X P X are the correspond ing

probabilities

n n

n

n

variable

µ1 1 2 2

1 2

1 2

⋅ + ⋅ + + ⋅= ⋅∑

( ) ( ) ... ( )

( )

, ,...,

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

.


6-156-15

l Find the mean of the number of spots thatappear when a die is tossed. Theprobability distribution is given below.

6-3 Mean for Discrete Variable -6-3 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


6-166-16


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.



6-176-17

l In a family with two children, find the meannumber of children who will be girls. Theprobability 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



6-186-18

= .

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.



6-196-19

6-3 Formula for the Variance of a Probability Distribution

l The variance of a probability distribution isfound by multiplying the square of eachoutcome by its corresponding probability,summing these products, and subtractingthe square of the mean.


6-206-20

6-3 Formula for the Variance of a6-3 Formula for the Variance of a Probability Distribution Probability Distribution

[ ]

= 2

The formula for the of a


X P X

The standard deviation of a


variance

σ µ

σ σ

2 2 2= ⋅ −∑ ( ) .

.


6-216-21

6-3 Variance of a Probability6-3 Variance of a Probability Distribution - Distribution - Example

l The probability that 0, 1, 2, 3, or 4people will be placed on hold when theycall a radio talk show with four phonelines is shown in the distribution below.Find the variance and standarddeviation for the data.


6-226-22


X 0 1 2 3 4

P ( X ) 0 .18 0 .34 0 .23 0 .21 0 .04



6-236-23


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

µµ = 1.59 ΣΣX2⋅⋅P(X)=3.79

σσ2 = 3.79 – 1.592

= 1.26

σσ2 = 3.79 – 1.592

= 1.26


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

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

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

l σ 2 = 3.79 – 2.53 = 1.26l σ = = 1.12

6-246-24


1.26


6-256-25

6-3 Expectation6-3 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 ⋅∑


6-266-26

6-3 Expectation -6-3 Expectation - Example

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


6-276-27

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

6-3 Expectation -6-3 Expectation - Example

Profit, X 250,000 –70,000

P(X) 0.40 0.60


6-286-28

6-4 The Binomial Distribution6-4 The Binomial Distribution

l A binomial experimentbinomial experiment is a probabilityexperiment that satisfies the following fourrequirements:

l Each trial can have only two outcomes oroutcomes that can be reduced to twooutcomes. Each outcome can beconsidered as either a success ora failure.



6-296-29


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

independent of each other.l The probability of success must remain the

same for each trial.


6-306-30


l The outcomes of a binomial experimentand the corresponding probabilities ofthese outcomes are called a binomialbinomialdistribution.distribution.


6-316-31


l Notation for the Binomial Distribution:

ll PP((SS) = ) = pp, probability of a success

ll PP((FF) = 1 – ) = 1 – pp = = qq, probability of a failure

ll nn = number of trialsl XX = number of successes.


6-326-32

6-4 Binomial Probability Formula6-4 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,

( )!

( )! !=

− −


6-336-33

6-4 Binomial Probability -6-4 Binomial Probability - Example

l If a student randomly guesses at fivemultiple-choice questions, find theprobability that the student gets exactlythree correct. Each question has fivepossible choices.

ll 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.

≈Copyright ©1999 Ian H. Giddy Probability Distributionbs 36

6-346-34


l A survey from Teenage ResearchUnlimited (Northbrook, Illinois.) found that30% of teenage consumers received theirspending money from part-time jobs. Iffive teenagers are selected at random, findthe probability that at least three of themwill have part-time jobs.



6-356-35


ll 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.

ll NOTE:NOTE: You can use Table B in thetextbook to find the Binomialprobabilities as well.


6-366-36


l A report from the Secretary of Health andHuman Services stated that 70% of single-vehicle traffic fatalities that occur on weekendnights involve an intoxicated driver. If asample of 15 single-vehicle traffic fatalitiesthat occurred on a weekend night is selected,find the probability that exactly 12 involve adriver who is intoxicated.


6-376-37


ll Solution:Solution: n = 15, X = 12, andp = 0.7. From Table B,P(X =12) = 0.170


6-386-38

6-4 Mean, Variance, Standard Deviation for the Binomial6-4 Mean, Variance, Standard Deviation for the BinomialDistribution -Distribution - Example

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

ll Solution:Solution: n = 4, p = 1/2, and q = 1/2.

l µ = n⋅p = (4)(1/2) = 2.l σ 2 = n⋅p⋅q = (4)(1/2)(1/2) = 1.l σ = = 1.

1


7-17-1

Chapter 7Chapter 7

The NormalThe NormalDistributionDistribution


7-27-2

OutlineOutline

l 7-1 Introductionl 7-2 Properties of the Normal

Distributionl 7-3 The Standard Normal

Distributionl 7-4 Applications of the Normal

Distribution



7-37-3

OutlineOutline

l 7-5 The Central Limit Theoreml 7-6 The Normal Approximation to the

Binomial Distribution


7-47-4


l Identify distributions as symmetric orskewed.

l Identify the properties of the normaldistribution.

l Find the area under the standardnormal distribution given various zvalues.


7-57-5


l Find probabilities for a normallydistributed variable by transforming itinto a standard normal variable.

l Find specific data values for givenpercentages using the standard normaldistribution.


7-67-6


l Use the Central Limit Theorem to solveproblems involving sample means.

l Use the normal approximation tocompute probabilities for a binomialvariable.


7-77-7

7-2 Properties of the Normal7-2 Properties of the Normal Distribution Distribution

l Many continuous variables have distributionsthat are bell-shaped and are calledapproximately normally distributed variables.approximately normally distributed variables.

l The theoretical curve, called the normalnormaldistribution curvedistribution curve, can be used to study manyvariables that are not normally distributed butare approximately normal.


7-87-8

7-2 Mathematical Equation for the7-2 Mathematical Equation for the Normal Distribution Normal Distribution

The mathematical equation for the normal distribution:

where

e

population mean

population standard deviation

2 718

314

≈≈

=

=

π

µ

σ

.

.

πσ

σµ

2

222)( −−

=

xe

y



7-97-9

7-2 Properties of the Normal7-2 Properties of the Normal Distribution Distribution

l The shape and position of the normaldistribution curve depend on twoparameters, the mean and the standarddeviation.

l Each normally distributed variable has itsown normal distribution curve, whichdepends on the values of the variable’smean and standard deviation.


7-107-10

7-2 Properties of the7-2 Properties of the Theoretical Normal Distribution Theoretical Normal Distribution

l The normal distribution curve isbell-shaped.

l The mean, median, and mode are equaland located at the center of thedistribution.

l The normal distribution curve isunimodal (single mode).


7-117-11


l The curve is symmetrical about themean.

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

distribution curve is equal to 1.


7-127-12


l The area under the normal curve that lieswithinü one standard deviation of the mean is

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

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

approximately 0.997 (99.7%).


7-137-13

7-2 Areas Under the Normal Curve7-2 Areas Under the Normal Curve

µµ −3−3σσ µµ −2−2σσ µµ −1−1σσ µµ µµ +1+1σσ µµ +2+2σσ µµ +3+3σσ

68%

95% 99.7%


7-147-14

7-3 The Standard Normal7-3 The Standard Normal Distribution Distribution

l The standard normal distributionstandard normal distribution is anormal distribution with a mean of 0 and astandard deviation of 1.

l All normally distributed variables can betransformed into the standard normallydistributed variable by using the formulafor the standard score:(see next slide)



7-157-15

7-3 The Standard Normal7-3 The Standard Normal Distribution Distribution

z

or

zX

=−

=−

value mean

standard deviation

µσ


7-167-16

7-3 Area Under the Standard Normal Curve -7-3 Area Under the Standard Normal Curve -Example

l Find the area under the standardnormal curve between z = 0 andz = 2.34 ⇒⇒ PP(0(0 ≤≤ zz ≤≤ 2.34) 2.34).

l Use your table at the end of the text tofind the area.

l The next slide shows the shaded area.


7-177-17

7-3 Area Under the Standard 7-3 Area Under the Standard Normal Curve Normal Curve- - Example

0000 2.342.342.342.34

0.49040.49040.49040.4904


7-187-18

7- 3 Area Under the Standard 7- 3 Area Under the Standard Normal CurveNormal Curve-- Example

l Find the area under the standard normalcurve between z = 0 andz = –1.75 ⇒⇒ PP(–1.75(–1.75 ≤≤ zz ≤≤ 0) 0).

l Use the symmetric property of the normaldistribution and your table at the end of thetext to find the area.



7-197-19

7-3 Area Under the Standard 7-3 Area Under the Standard Normal CurveNormal Curve- - Example

0000 1.751.751.751.75−1.75−1.75−1.75−1.75

0.45990.45990.45990.45990.45990.45990.45990.4599


7-207-20


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





7-217-21

7-3 Area Under the Standard 7-3 Area Under the Standard Normal CurveNormal Curve- - Example

0000 1.111.111.111.11

0.36650.36650.36650.3665

0.5000 0.5000−0.3665−0.3665

0.13350.1335


7-227-22

7-3 Area Under the Standard 7-3 Area Under the Standard Normal CurveNormal Curve-- Example

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

l Use the symmetric property of thenormal distribution and your table at theend of the text to find the area.

l The next slide shows the area.


7-237-237-3 Area Under the Standard 7-3 Area Under the Standard NormalNormalCurve -Curve - Example

−1.93−1.93 1.931.9300

0.47320.4732

0.50000.5000−0.4732−0.4732

0.02680.02680.02680.0268


7-247-24


l Find the area between z = 2 andz = 2.47 ⇒⇒ PP(2(2 ≤≤ zz ≤≤ 2.47) 2.47).




7-257-25

7-3 Area Under the Standard Normal Curve -7-3 Area Under the Standard Normal Curve - Example

0000 2.472.472.472.472222

0.47720.47720.47720.4772

0.49320.49320.49320.4932−0.4772−0.4772−0.4772−0.4772

0.01600.01600.01600.0160

0.49320.49320.49320.4932


7-267-26

l Find the area between z = 1.68 andz = –1.37 ⇒⇒ PP(–1.37(–1.37 ≤≤ zz ≤≤ 1.68) 1.68).






7-277-27

0000 1.681.681.681.68

0.45350.45350.45350.4535

0.45350.45350.45350.4535+0.4147+0.4147+0.4147+0.4147

0.86820.86820.86820.8682

0.41470.41470.41470.4147

−1.37−1.37−1.37−1.37



7-287-28

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





7-297-29

0000 1.991.991.991.99

0.47670.47670.47670.4767

0.50000.50000.50000.5000+0.4767+0.4767+0.4767+0.4767

0.97670.97670.97670.9767

0.50000.50000.50000.5000



7-307-30

l Find the area to the right ofz = –1.16 ⇒⇒ PP((zz >> –1.16) –1.16).





7-317-31

−1.16−1.16−1.16−1.16 0000

0.50000.50000.50000.5000

0.50000.50000.50000.5000+ 0.3770+ 0.3770+ 0.3770+ 0.3770

0.87700.87700.87700.8770

0.3770.3770.3770.377



7-327-32

RECALL:RECALL: The Standard Normal Distribution

z

or

zX

=−

=−

value mean

standard deviation

µσ



7-337-33

7-4 Applications of the Normal7-4 Applications of the Normal Distribution - Distribution - Example

l Each month, an American householdgenerates an average of 28 pounds ofnewspaper for garbage or recycling.Assume the standard deviation is 2pounds. Assume the amount generated isnormally distributed.

l If a household is selected at random, findthe probability of its generating:


7-347-34

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

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

selected household will generate more than30.2 lbs. of newspapers is 0.1357 or 13.57%.



7-357-35

1.11.11.11.10000

0.50000.50000.50000.5000−0.3643−0.3643−0.3643−0.3643

0.13570.13570.13570.1357



7-367-36

ll Between 27 and 31 pounds per month.Between 27 and 31 pounds per month.l First find the z-value for 27 and 31. z1

= [X –µ]/σ = [27 – 28]/2 = –0.5; z2= [31 – 28]/2 = 1.5

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



7-377-37

−0.5−0.5−0.5−0.5 0000 1.51.51.51.5

0.19150.19150.19150.1915++++ 0.4332 0.4332 0.4332 0.4332

0.62470.62470.62470.62470.19150.19150.19150.1915

0.43320.43320.43320.4332



7-387-38


l The American Automobile Association reportsthat the average time it takes to respond to anemergency call is 25 minutes. Assume thevariable is approximately normally distributedand the standard deviation is 4.5 minutes. If80 calls are randomly selected, approximatelyhow many will be responded to in less than15 minutes?



7-397-39

l First find the z-value for 15 is z =[X –µ]/σ = [15 – 25]/4.5 = –2.22.

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

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



7-407-40

0000 2.222.222.222.22−2.22−2.22−2.22−2.22


0.50000.50000.50000.5000− − − − 0.48680.48680.48680.4868

0.01320.01320.01320.01320.01320.01320.01320.0132


7-417-41


l An exclusive college desires to accept onlythe top 10% of all graduating seniorsbased on the results of a nationalplacement test. This test has a mean of500 and a standard deviation of 100. Findthe cutoff score for the exam. Assume thevariable is normally distributed.


7-427-42

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

under the normal curve for acceptedstudents.

l Find the z value that corresponds to anarea of 0.4000 by looking up 0.4000 in thearea portion of Table E. Use the closestvalue, 0.3997.

7- 4 Applications of the Normal7- 4 Applications of the Normal Distribution - Distribution - Example


7-437-43

l Substitute in the formulaand solve for X.

l The z-value for the cutoff score (X) is z =[X –µ]/σ = [X – 500]/100 = 1.28. (See nextslide).

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

cutoff score.

7- 4 Applications of the Normal7- 4 Applications of the Normal Distribution - Distribution - Example

σµ−= Xz


7-447-44

0000 X X = 1.28= 1.28= 1.28= 1.28


0.10.10.10.1

0.40.40.40.4



7-457-45

ll NOTE:NOTE: To solve for X, use the followingformula: XX = = zz⋅⋅σσ + + µµ..

l Example: For a medical study, aresearcher wishes to select people in themiddle 60% of the population based onblood pressure. (Continued on the nextslide).



7-467-46

l (Continued)-- If the mean systolic bloodpressure is 120 and the standard deviationis 8, find the upper and lower readings thatwould qualify people to participate in thestudy.



7-477-47

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

l X = (z)(σ) + µ = (0.84)(8) + 120 = 126.72.The other X = (–0.84)(8) + 120 = 113.28.See next slide.

l i.e. the middle 60% of BP readings is between113.28 and 126.72.


(continued)


7-487-48

0000


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

0.20.20.20.2 0.30.30.30.3

0.20.20.20.20.30.30.30.3


7-497-49

7-5 Distribution of Sample Means7-5 Distribution of Sample Means

ll Distribution of Sample means:Distribution of Sample means: Asampling distribution of samplemeans is a distribution obtained byusing the means computed fromrandom samples of a specific sizetaken from a population.


7-507-50

ll Sampling errorSampling error is the differencebetween the sample measure and thecorresponding population measuredue to the fact that the sample is nota perfect representation of thepopulation.

7-5 Distribution of Sample Means7-5 Distribution of Sample Means



7-517-51

l The mean of the sample means will be thesame as the population mean.

l The standard deviation of the samplemeans will be smaller than the standarddeviation of the population, and it will beequal to the population standard deviationdivided by the square root of the samplesize.

7-5 Properties of the Distribution of7-5 Properties of the Distribution of Sample Means Sample Means


7-527-52

l Suppose a professor gave an 8-point quizto a small class of four students. Theresults of the quiz were 2, 6, 4, and 8.Assume the four students constitute thepopulation.

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

7-5 Properties of the Distribution of7-5 Properties of the Distribution of Sample Means - Sample Means - Example


7-537-53

l The standard deviation of the population is

=2.236.l The graph of the distribution of the scores is

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

taken with replacement.

7-5 Properties of the Distribution of7-5 Properties of the Distribution of Sample Means - Sample Means - Example

( ) ( ) ( ) ( ) /4}258254256252{ó

−+−+−+−=


7-547-54

7-5 Graph of the Original 7-5 Graph of the Original DistributionDistribution


7-557-55 7-5 Properties of the Distribution of7-5 Properties of the Distribution of Sample Means - Sample Means - Example

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


7-567-567-5 Frequency Distribution of the7-5 Frequency Distribution of the Sample Means - Sample Means - Example

X-bar(mean)

2 3 4 5 6 7 8

f 1 2 3 4 3 2 1



7-577-57

7-5 Graph of the Sample Means7-5 Graph of the Sample Means

8765432

4

3

2

1

0

SAM PLE MEANS

Fre

quen

cy

(APPROXIMAT ELY NORMAL)DISTRIBUT ION OF SAMPLE MEANS


7-587-58

7-5 Mean and Standard Deviation of 7-5 Mean and Standard Deviation of the Samplethe SampleMeansMeans

Mean of Sample Means

which is the same as the

population mean Thus

X

X

µ

µ µ

=+ + +

= =

=

2 3 816

8016

5...

. .


7-597-59

7-5 Mean and Standard Deviation of 7-5 Mean and Standard Deviation of the Samplethe SampleMeansMeans

The standard deviation of the sample

means is

This is the same as

X

σ

σ

=− + − + + −

=

( ) ( ) ... ( )

. .

.

2 5 3 5 8 5

161581

2

2 2 2


7-607-60

7-5 The Standard Error of the 7-5 The Standard Error of the MeanMean

The standard deviation of the sample

means is called the standard error of

the mean Hence

nX

.

σσ

= .


7-617-61

7-5 The Central Limit Theorem7-5 The Central Limit Theorem

l As the sample size nn increases, the shapeof the distribution of the sample meanstaken from a population with mean µµ andstandard deviation of σσ will approach anormal distribution. As previously shown,this distribution will have a mean µµ andstandard deviationσ σ // √√n n ..


7-627-62

.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−

7-5 The Central Limit Theorem7-5 The Central Limit Theorem



7-637-63

l A.C. Neilsen reported that children between theages of 2 and 5 watch an average of 25 hours ofTV per week. Assume the variable is normallydistributed and the standard deviation is 3 hours.If 20 children between the ages of 2 and 5 arerandomly selected, find the probability that themean of the number of hours they watch TV isgreater than 26.3 hours.

7-5 The Central Limit Theorem -7-5 The Central Limit Theorem - Example


7-647-64

l The standard deviation of the samplemeans is σ/ √n = 3/ √20 = 0.671.

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

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



7-657-65

0000 1.941.941.941.94

0.50000.50000.50000.5000 − 0.4738 − 0.4738 − 0.4738 − 0.4738

0.02620.02620.02620.0262



7-667-66

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



7-67 7-67

l The standard deviation of the samplemeans is σ/ √n = 16/ √36 = 2.6667.

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

l ThusP(–2.25 ≤ z ≤ 1.50) = 0.4878 + 0.4332= 0.921 or 92.1%.



7-687-68


0000 1.501.501.501.50−2.25−2.25−2.25−2.25

0.48780.48780.48780.4878 0.43320.43320.43320.4332



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7-6 The Normal Approximation to 7-6 The Normal Approximation to the Binomialthe BinomialDistributionDistribution

l The normal distribution is often used tosolve problems that involve the binomialdistribution since when n is large (say,100), the calculations are too difficult to doby hand using the binomial distribution.


7-707-70


l The normal approximation to thebinomial is appropriate when npnp ≥≥ 5 and 5 andnqnq ≥≥ 5. 5.

l In addition, a correction for continuitymay be used in the normalapproximation.


7-717-71


l A correction for continuitycorrection for continuity is a correctionemployed when a continuous distribution isused to approximate a discrete distribution.

l The continuity correction means that for anyspecific value of X, say 8, the boundaries of Xin the binomial distribution (in this case 7.5and 8.5) must be used.


7-727-72

7-6 The Normal Approximation to the Binomial7-6 The Normal Approximation to the BinomialDistribution -Distribution - Example

l Prevention magazine reported that 6% ofAmerican drivers read the newspaperwhile driving. If 300 drivers are selected atrandom, find the probability that exactly 25say they read the newspaper while driving.


7-737-73

l Here p = 0.06, q = 0.94, and n = 300.

l Check for normal approximation: np =(300)(0.06) = 18 andnq = (300)(0.94) = 282. Since bothvalues are at least 5, the normalapproximation can be used.



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l (continued) µ = np = (300)(0.06) = 18andσ = √npq = √ (300)(0.06)(0.94) = 4.11.

l So P(X = 25) = P(24.5 ≤ X ≤ 25.5).l z1 = [24.5 – 18]/4.11 = 1.58 and

z2= [25.5 – 18]/4.11 = 1.82.




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l (continued) P(24.5 ≤ X ≤ 25.5) =P(1.58 ≤ z ≤ 1.82)= 0.4656 – 0.4429 = 0.0227.

l Hence, the probability that exactly 25people read the newspaper whiledriving is 2.27%.



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