Download Lecture 9 Probability Distributions

Outline
Lecture 9
Probability
Distributions
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© The McGraw-Hill Companies, Inc., 2000
Outline
6-1 Introduction
6-2 Probability Distributions
6-3 Mean, Variance, and
Expectation
6-4 The Binomial Distribution
6-2 Probability Distributions
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
A variable is defined as a
characteristic or attribute that can
assume different values.
A variable whose values are
determined by chance is called a
random variable.
variable
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6-2 Probability Distributions
6-2 Probability Distributions
If a variable 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
variable.
variable
Discrete variables have values that can
be counted.
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If a variable can assume all values in
the interval between two given values
then the variable is called a continuous
variable. Example - temperature
between 680 to 780.
Continuous random variables are
obtained from data that can be
measured rather than counted.
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6-2 Probability Distributions Tossing Two Coins
6-2 Probability Distributions Tossing Two Coins
H
H
T
Second Toss
H
T
First Toss
From the tree diagram, the sample
space will be represented by HH,
HT, TH, TT.
If X is the random variable for the
number of heads, then X assumes
the value 0, 1, or 2.
2
T
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6-2 Probability Distributions Tossing Two Coins
Sample Space
TT
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6-2 Probability Distributions Tossing Two Coins
Number of Heads
0
TH
OUTCOME
X
0
PROBABILITY
P(X)
1/4
1
2/4
2
1/4
1
HT
HH
2
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A probability 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.
6-2 Probability Distributions -Graphical Representation
Experiment: Toss Two Coins
1
PROBABILITY
6-2 Probability Distributions
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0.5
.25
0
1
2
3
NUMBER OF HEADS
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6-3 Mean, Variance, and
Expectation for Discrete Variable
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.
The mean of the random variable of a
probability distribution is
µ = 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 corresponding
probabilities.
1
1
1
1
2
2
2
2
n
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6-3 Mean for Discrete Variable Example
Find the mean of the number of spots
that appear when a die is tossed. The
probability distribution is given below.
XX
11
22
33
44
55
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
.
66
P(X)
P(X) 1/6
1/6 1/6
1/6 1/6
1/6 1/6
1/6 1/6
1/6 1/6
1/6
That is, when a die is tossed many times,
the theoretical mean will be 3.5.
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6-3 Mean for Discrete Variable Example
In a family with two children, find the
mean number of children who will be
girls. The probability distribution is
given below.
XX
00
11
n
n
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6-3 Mean for Discrete Variable Example
n
22
P(X)
P(X) 1/4
1/4 1/2
1/2 1/4
1/4
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6-3 Mean for Discrete Variable Example
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.
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Variance of a
Probability Distribution
6-3 Formula for the 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.
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.
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6-3 Formula for the Variance of a
Probability Distribution
The formula for the variance of a
probability distribution is
σ = ∑ [ X ⋅ P ( X )] − µ .
2
2
2
The standard deviation of a
probability distribution is
6-3 Variance of a Probability
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.
σ= σ .
2
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6-3 Variance of a Probability
Distribution - Example
X
0
1
2
3
4
P (X ) 0 .1 8 0 .3 4 0 .2 3 0 .2 1 0 .0 4
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6-3 Variance of a Probability
Distribution - Example
X⋅P(X) X2⋅P(X)
X
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
σ2 =
3.79 – 1.592
= 1.26
µ = 1.59 ΣX2⋅P(X)
=3.79
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6-3 Variance of a Probability
Distribution - Example
6-3 Expectation
The expected value of a discrete
random variable of a probability
distribution is the theoretical average
Now, µ = (0)(0.18) + (1)(0.34) + (2)(0.23) +
(3)(0.21) + (4)(0.04) = 1.59.
Σ 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).
σ 2 = 3.79 – 2.53 = 1.26
σ = 1.26 = 1.12
of the variable. The formula is
µ = E ( X ) = ∑ X ⋅ P( X )
The symbol E ( X ) is used for the
expected value.
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6-3 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.
6-3 Expectation - Example
Profit, X 250,000
P(X)
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A binomial 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.
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0.60
The expected profit = ($250,000)(0.40)
+ (–$70,000)(0.60) = $58,000.
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6-4 The Binomial Distribution
0.40
–70,000
6-4 The Binomial Distribution
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.
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6-4 The Binomial Distribution
The outcomes of a binomial
experiment and the corresponding
probabilities of these outcomes are
called a binomial distribution.
6-4 The Binomial Distribution
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6-4 Binomial Probability Formula
6-4 Binomial Probability - Example
In a binomial experiment, the probability of
exactly X successes in n trials is
P( X ) =
n!
p Xq n − X
−
( n X )! X !
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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.
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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: n = 5, X = 3, and p = 1/5.
Then,
P(3) = [5!/((5 – 3)!3! )](1/5)3(4/5)2 ≈ 0.05.
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6-4 Binomial Probability - Example
Notation for the Binomial
Distribution:
P(S) = p, probability of a success
P(F) = 1 – p = q, probability of a
failure
n = number of trials
X = number of successes.
6-4 Binomial Probability - Example
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: You can use Table B in the
textbook to find the Binomial
probabilities as well.
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6-4 Binomial Probability - Example
A report from the Secretary of Health and
Human Services stated that 70% of singlevehicle 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.
6-4 Binomial Probability - Example
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6-4 Mean, Variance, Standard Deviation
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: n = 4, p = 1/2, and q = 1/2.
µ = n⋅⋅p = (4)(1/2) = 2.
σ 2 = n⋅⋅p⋅⋅q = (4)(1/2)(1/2) = 1.
σ = 1 = 1.
7-2 The Normal Distribution
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y=
e
−( x−µ )2 2σ
7-2 Properties of the Normal
Distribution
2
σ 2π
where
e ≈ 2.718
π ≈ 314
.
µ = population mean
σ = population standard deviation
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Many continuous variables have
distributions that are bell-shaped and are
called approximately normally distributed
variables.
The theoretical curve, called the normal
distribution curve,
curve can be used to study
many variables that are not normally
distributed but are approximately normal.
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7-2 Mathematical Equation for the
Normal Distribution
The mathematical equation for the normal distribution:
Solution: n = 15, X = 12, and
p = 0.7. From Table B,
P(X =12) = 0.170
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.
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7-2 Properties of the
Theoretical 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 is
unimodal (single mode).
7-2 Properties of the
Theoretical Normal Distribution
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.
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7-2 Properties of the
Theoretical Normal Distribution
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7-2 Areas Under the Normal Curve
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%).
68%
95%
µ −3σ
µ −2σ µ −1σ µ µ +1σ µ +2σ µ +3σ
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7-3 The Standard Normal
Distribution
The standard 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)
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99.7%
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7-3 The Standard Normal
Distribution
z=
value − mean
standard deviation
or
z=
X −µ
σ
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7-3 Area Under the Standard
Normal Curve - Example
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7-3 Area Under the Standard
Normal Curve - Example
7- 3 Area Under the Standard
Normal Curve - Example
0.4904
0
2.34
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−1.75
0
The next slide shows the shaded area.
7-3 Area Under the Standard
Normal Curve - Example
0.4599
Find the area under the standard
normal curve between z = 0 and
z = –1.75 ⇒ P(–1.75 ≤ z ≤ 0).
0)
Use the symmetric property of the
normal distribution and your table at
the end of the text to find the area.
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7-3 Area Under the Standard
Normal Curve - Example
0.4599
Find the area under the standard
normal curve between z = 0 and
z = 2.34 ⇒ P(0 ≤ z ≤ 2.34).
2.34)
Use your table at the end of the text
to find the area.
The next slide shows the shaded
area.
Find the area to the right of z = 1.11
⇒ P(z > 1.11).
1.11)
Use your table at the end of the text
to find the area.
The next slide shows the shaded
area.
1.75
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7-3 Area Under the Standard
Normal Curve - Example
7-3 Area Under the Standard
Normal Curve - Example
0.5000
−0.3665
0.1335
0.3665
0
1.11
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7-3 Area Under the Standard
Normal Curve - Example
7-3 Area Under the Standard
Normal Curve - Example
0.5000
−0.4732
0.0268
0.0268
0.4732
−1.93
0
1.93
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7-3 Area Under the Standard
Normal Curve - Example
0.4932
0.4932
−0.4772
0.0160
0.4772
Find the area between z = 2 and
z = 2.47 ⇒ P(2 ≤ z ≤ 2.47).
2.47)
Use the symmetric property of the
normal distribution and your table at
the end of the text to find the area.
The next slide shows the area.
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7-3 Area Under the Standard Normal Curve Example
0
Find the area to the left of z = –1.93
⇒ P(z < –1.93).
1.93)
Use the symmetric property of the
normal distribution and your table at
the end of the text to find the area.
The next slide shows the area.
Find the area between z = 1.68 and
z = –1.37 ⇒ P(–1.37 ≤ z ≤ 1.68).
1.68)
Use the symmetric property of the
normal distribution and your table at
the end of the text to find the area.
The next slide shows the area.
2 2.47
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7-3 Area Under the Standard Normal Curve Example
0.4535
+0.4147
7-3 Area Under the Standard
Normal Curve - Example
0.8682
0.4147
0.4535
−1.37
0
Find the area to the left of z = 1.99
⇒ P(z < 1.99).
1.99)
Use your table at the end of the text
to find the area.
The next slide shows the area.
1.68
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7-3 Area Under the Standard Normal Curve Example
0.5000
+0.4767
0.9767
0.5000
0.4767
0
7-3 Area Under the Standard
Normal Curve - Example
Find the area to the right of
z = –1.16 ⇒ P(z > –1.16).
1.16)
Use your table at the end of the text
to find the area.
The next slide shows the area.
1.99
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7-3 Area Under the Standard Normal Curve Example
0.5000
+ 0.3770
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RECALL: The Standard Normal
Distribution
z=
value − mean
standard deviation
0.8770
or
0.377 0.5000
−1.16
0
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z=
X −µ
σ
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7-4 Applications of the Normal
Distribution - Example
7-4 Applications of the Normal
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:
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7-4 Applications of the Normal
Distribution - Example
7-4 Applications of the Normal
Distribution - Example
0.5000
−0.3643
0.1357
0
Between 27 and 31 pounds per month.
First find the z-value for 27 and 31.
z1 = [X –µ]/σ = [27 – 28]/2 = –0.5;
z2 = [31 – 28]/2 = 1.5
Thus, P(–0.5 ≤ z ≤ 1.5) = 0.1915 + 0.4332
= 0.6247.
1.1
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7-4 Applications of the Normal
Distribution - Example
7-4 Applications of the Normal
Distribution - Example
0.4332
0.1915
0.1915
+ 0.4332
0.6247
−0.5 0
More than 30.2 pounds per month.
First find the z-value for 30.2.
z =[X –µ]/σ = [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%.
1.5
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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?
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7-4 Applications of the Normal
Distribution - Example
7-4 Applications of the Normal
Distribution - Example
First find the z-value for 15 is
z = [X –µ]/σ = [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.
0.5000
− 0.4868
0.0132
−2.22
0.0132
0
2.22
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7-4 Applications of the Normal
Distribution - Example
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.
7- 4 Applications of the Normal
Distribution - Example
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.
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7- 4 Applications of the Normal
Distribution - Example
Substitute in the formula z = X − µ
σ
and solve for X.
The z-value for the cutoff score (X) is
z = [X –µ]/σ = [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.
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7-4 Applications of the Normal
Distribution - Example
0.1
0.4
0
X = 1.28
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7-4 Applications of the Normal
Distribution - Example
NOTE: To solve for X, use the following
formula: X = z⋅σ + µ.
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).
7-4 Applications of the Normal
Distribution - Example
(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.
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7-4 Applications of the Normal
Distribution - Example
7-4 Applications of the Normal
Distribution - Example
(continued)
Note that two values are needed, one above the
mean and one below the mean. The closest z
values are 0.84 and – 0.84 respectively.
X = (z)(σ) + µ = (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.
0.2
−0.84
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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.
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0
0.2
0.84
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7-5 Distribution of Sample Means
0.3 0.3
7-5 Distribution of Sample Means
Sampling 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.
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7-5 Properties of the 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.
7-5 Properties of the Distribution of
Sample Means - Example
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 is
µ = ( 2 + 6 + 4 + 8)/4 = 5.
5
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7-5 Properties of the Distribution of
Sample Means - Example
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7-5 Graph of the Original
Distribution
The standard deviation of the population
is
2
2
2
2
σ = { (2 − 5 ) + (6 − 5 ) + (4 − 5 ) + (8 − 5 )  /4}


=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.
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7-5 Properties of the Distribution of
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
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7-5 Frequency Distribution of the
Sample Means - Example
X-bar 2
(mean)
f
1
3
4
5
6
7
8
2
3
4
3
2
1
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7-5 Graph of the Sample Means
DIST RIBUT ION O F SAMPLE MEANS
(APPROXIMAT ELY NORMAL)
Mean of Sample Means
2 + 3+...+8 80
µ =
= =5
16
16
which is the same as the
population mean. Thus µ = µ.
4
Frequency
7-5 Mean and Standard Deviation of
the Sample Means
3
X
2
1
0
2
3
4
5
6
SAMPLE MEANS
7
8
X
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7-5 Mean and Standard Deviation of
the Sample Means
The standard deviation of the sample
means is
(2 − 5) + (3 − 5) +...+ (8 − 5)
σ =
16
= 1581
. .
2
2
2
X
This is the same as
σ
2
© The McGraw-Hill Companies, Inc., 2000
7-5 The Standard Error of the
Mean
The standard deviation of the sample
means is called the standard error of
the mean. Hence
σ =
X
.
© The McGraw-Hill Companies, Inc., 2000
7-5 The Central Limit Theorem
As the sample size n increases, the
shape of the distribution of the sample
means taken from a population with
mean µ and standard deviation of σ will
approach a normal distribution. As
previously shown, this distribution will
have a mean µ and standard deviation
σ / √n .
n
.
© The McGraw-Hill Companies, Inc., 2000
7-5 The Central Limit Theorem
The central limit theorem can be used
to answer questions about sample means
in the same manner that the normal distributi on
can be used to answer questions about
individual values . The only difference is that
a new formula must be used for the z - values .
It is
z=
© The McGraw-Hill Companies, Inc., 2000
σ
X −µ
.
σ/ n
© The McGraw-Hill Companies, Inc., 2000
7-5 The Central Limit Theorem Example
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.
7-5 The Central Limit Theorem Example
The standard deviation of the sample
means is σ/ √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%.
© The McGraw-Hill Companies, Inc., 2000
© The McGraw-Hill Companies, Inc., 2000
7-5 The Central Limit Theorem Example
7-5 The Central Limit Theorem Example
0.5000
− 0.4738
0.0262
0
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.
1.94
© The McGraw-Hill Companies, Inc., 2000
7-5 The Central Limit Theorem Example
The standard deviation of the sample
means is σ/ √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%.
© The McGraw-Hill Companies, Inc., 2000
© The McGraw-Hill Companies, Inc., 2000
7-5 The Central Limit Theorem Example
0.4878
−2.25
0.4332
0
1.50
© The McGraw-Hill Companies, Inc., 2000