8
Probability Distributions and Statistics
 Distributions of Random Variables
 Expected Value
 Variance and Standard Deviation
 The Binomial Distribution
 The Normal Distribution
 Applications of the Normal Distribution
8.1
Distributions of Random Variables
.03
.02
.01
0
1
2
3
4
5
6
7
8
x
Random Variable
 A random variable is a rule that assigns
a number to each outcome of a chance
experiment.
Example
 A coin is tossed three times.
 Let the random variable X denote the number of heads
that occur in the three tosses.
✦ List the outcomes of the experiment; that is, find the
domain of the function X.
✦ Find the value assigned to each outcome of the
experiment by the random variable X.
✦ Find the event comprising the outcomes to which a value
of 2 has been assigned by X.
This event is written (X = 2) and is the event consisting of
the outcomes in which two heads occur.
Example 1, page 418
Example
Solution
 As discussed in Section 7.1, the set of outcomes of this
experiment is given by the sample space
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
 The table below associates the outcomes of the
experiment with the corresponding values assigned to
each such outcome by the random variable X:
Outcome
X
HHH HHT HTH THH HTT THT TTH
3
2
2
2
1
1
1
TTT
0
 With the aid of the table, we see that the event (X = 2) is
given by the set
{HHT, HTH, THH}
Example 1, page 418
Applied Example: Product Reliability
 A disposable flashlight is turned on until its battery runs
out.
 Let the random variable Z denote the length (in hours)
of the life of the battery.
 What values can Z assume?
Solution
 The values assumed by Z can be any nonnegative real
numbers; that is, the possible values of Z comprise the
interval 0  Z < ∞.
Applied Example 3, page 419
Probability Distributions and Random Variables
 Since the random variable associated with an
experiment is related to the outcome of the experiment,
we can construct a probability distribution associated
with the random variable, rather than one associated
with the outcomes of the experiment.
 In the next several examples, we illustrate the
construction of probability distributions.
Example
 Let X denote the random variable that gives the sum of the





faces that fall uppermost when two fair dice are thrown.
Find the probability distribution of X.
Solution
The values assumed by the random variable X are
2, 3, 4, … , 12, correspond to the events E2, E3, E4, … , E12.
Next, the probabilities associated with the random variable
X when X assumes the values 2, 3, 4, … , 12, are precisely
the probabilities P(E2), P(E3), P(E4), … , P(E12),
respectively, and were computed as seen in Chapter 7.
Thus,
213
P( X …3)
2)
4)and
 P(soE324on.
)) 
36
36
Example 5, page 420
Applied Example: Waiting Lines
 The following data give the number of cars observed
waiting in line at the beginning of 2-minute intervals
between 3 p.m. and 5 p.m. on a given Friday at the Happy
Hamburger drive-through and the corresponding
frequency of occurrence.
 Find the probability distribution of the random variable X,
where X denotes the number of cars found waiting in line.
Cars
0
1
2
3
4
5
6
7
8
Frequency
2
9
16
12
8
6
4
2
1
Applied Example 6, page 420
Applied Example: Waiting Lines
Cars
0
1
2
3
4
5
6
7
8
Frequency
2
9
16
12
8
6
4
2
1
Solution
 Dividing each frequency number in the table by 60 (the
sum of all these numbers) give the respective probabilities
associated with the random variable X when X assumes the
values 0, 1, 2, … , 8.
 For example,
12
16
92
…and
P( X
1)
3)
0)
2) so on.
 .15
.20
.03
.27
60
60
Applied Example 6, page 420
Applied Example: Waiting Lines
Cars
0
1
2
3
4
5
6
7
8
Frequency
2
9
16
12
8
6
4
2
1
Solution
 The resulting probability distribution is
x
0
1
2
3
4
5
6
7
8
P(X = x)
.03
.15
.27
.20
.13
.10
.07
.03
.02
Applied Example 6, page 420
Histograms
 A probability distribution of a random variable may be
exhibited graphically by means of a histogram.
Examples
 The histogram of the probability distribution from the last
example is
.30
.20
.10
0
1
2
3
4
5
6
7
8
x
Histograms
 A probability distribution of a random variable may be
exhibited graphically by means of a histogram.
Examples
 The histogram of the probability distribution for the sum
of the numbers of two dice is
6/36
5/36
4/36
3/36
2/36
1/36
2
3
4
5
6
7
8
9 10 11 12
x
8.2
Expected Value
18
P( E )
 3818 
1  P( E ) 1  38
18
38
20
38
18 38 18 9
  

38 20 20 10
1  P( E ) 1  18
 18 38 
P( E )
38
20
38
18
38
20 38 20 10
  

38 18 18 9
Average, or Mean
 The average, or mean, of the n numbers
x1 , x2 , … , x n
is x (read “x bar”), where
x1  x2    xn
x
n
Applied Example: Waiting Lines
 Find the average number of cars waiting in line at the
Happy Burger’s drive-through at the beginning of each
2-minute interval during the period in question.
Cars
0
1
2
3
4
5
6
7
8
Frequency
2
9
16
12
8
6
4
2
1
Solution
 Using the table above, we see that there are all together
2 + 9 + 16 + 12 + 8 + 6 + 4 + 2 + 1 = 60
numbers to be averaged.
 Therefore, the required average is given by
0  2  1  9  2  16  3  12  4  8  5  6  6  4  7  2  8  1
 3.1
60
Applied Example 1, page 428, refer Section 8.1
Expected Value of a Random Variable X
 Let X denote a random variable that assumes
the values x1, x2, … , xn with associated
probabilities p1, p2, … , pn, respectively.
 Then the expected value of X, E(X), is given by
E ( X )  x1 p1  x2 p2    xn pn
Applied Example: Waiting Lines
 Use the expected value formula to find the average
number of cars waiting in line at the Happy Burger’s
drive-through at the beginning of each 2-minute interval
during the period in question.
x
0
1
2
3
4
5
6
7
8
P(X = x)
.03
.15
.27
.20
.13
.10
.07
.03
.02
Solution
 The average number of cars waiting in line is given by the
expected value of X, which is given by
E ( X )  (0)(.03)  (1)(.15)  (2)(.27)  (3)(.20)  (4)(.13)
 (5)(.10)  (6)(.07)  (7)(.03)  (8)(.02)
 3.1
Applied Example 2, page 428, refer Section 8.1
Applied Example: Waiting Lines
 The expected value of a random variable X is a measure of
central tendency.
 Geometrically, it corresponds to the point on the base of
the histogram where a fulcrum will balance it perfectly:
.20
.10
x
0
1
4
2
3.1
E(X)
Applied Example 2, page 428
5
6
7
8
Applied Example: Raffles
 The Island Club is holding a fundraising raffle.
 Ten thousand tickets have been sold for $2 each.
 There will be a first prize of $3000, 3 second prizes of
$1000 each, 5 third prizes of $500 each, and 20
consolation prizes of $100 each.
 Letting X denote the net winnings (winnings less the cost
of the ticket) associated with the tickets, find E(X).
 Interpret your results.
Applied Example 5, page 431
Applied Example: Raffles
Solution
 The values assumed by X are (0 – 2), (100 – 2), (500 – 2),
(1000 – 2), and (3000 – 2).
 That is –2, 98, 498, 998, and 2998, which correspond,
respectively, to the value of a losing ticket, a consolation
prize, a third prize, and so on.
 The probability distribution of X may be calculated in
the usual manner:
x
–2
98
498
998
2998
P(X = x)
.9971
.0020
.0005
.0003
.0001
 Using the table, we find
E ( X )  ( 2)(.9971)  98(.0020)  498(.0005)
Applied Example 5, page 431
 998(.0003)  2998(.0001)
 0.95
Applied Example: Raffles
Solution
 The expected value of E(X) = –.95 gives the long-run
average loss (negative gain) of a holder of one ticket.
 That is, if one participated regularly in such a raffle by
purchasing one ticket each time, in the long-run, one
may expect to lose, on average, 95 cents per raffle.
Applied Example 5, page 431
Odds in Favor and Odds Against
 If P(E) is the probability of an event E
occurring, then
✦ The odds in favor of E occurring are
P( E )
P( E )

1  P( E ) P( E c )
[ P( E )  1]
✦ The odds against E occurring are
1  P( E ) P( E c )

P( E )
P( E )
[ P( E )  0]
Applied Example: Roulette
 Find the odds in favor of winning a bet on red in American
roulette.
 What are the odds against winning a bet on red?
Solution
 The probability that the ball lands on red is given by
18
P
38
 Therefore, we see that the odds in favor of winning a bet
on red are
18
P( E )
 3818 
1  P( E ) 1  38
Applied Example 8, page 433
18
38
20
38
18 38 18 9
 


38 20 20 10
Applied Example: Roulette
 Find the odds in favor of winning a bet on red in American
roulette.
 What are the odds against winning a bet on red?
Solution
 The probability that the ball lands on red is given by
18
P
38
 The odds against winning a bet on red are
1  P( E ) 1  18
 18 38 
P( E )
38
Applied Example 8, page 433
20
38
18
38
20 38 20 10

 

38 18 18 9
Probability of an Event (Given the Odds)
 If the odds in favor of an event E occurring are
a to b, then the probability of E occurring is
a
P( E ) 
ab
Example
 Consider each of the following statements.
✦ “The odds that the Dodgers will win the World Series
this season are 7 to 5.”
✦ “The odds that it will not rain tomorrow are 3 to 2.”
 Express each of these odds as a probability of the event
occurring.
Solution
 With a = 7 and b = 5, the probability that the Dodgers will
win the World Series is
a
7
7
P( E ) 


 .58
a  b 7  5 12
Example 9, page 434
Example
 Consider each of the following statements.
✦ “The odds that the Dodgers will win the World Series
this season are 7 to 5.”
✦ “The odds that it will not rain tomorrow are 3 to 2.”
 Express each of these odds as a probability of the event
occurring.
Solution
 With a = 3 and b = 2, the probability that it will not rain
tomorrow is
a
3
3
P( E ) 

  .6
a  b 3 2 5
Example 9, page 434
Median
 The median of a group of numbers arranged
in increasing or decreasing order is
a. The middle number if there is an odd
number of entries or
b. The mean of the two middle numbers if
there is an even number of entries.
Applied Example: Commuting Times
 The times, in minutes, Susan took to go to work on nine
consecutive working days were
46 42 49 40 52 48 45 43 50
 What is the median of her morning commute times?
Solution
 Arranging the numbers in increasing order, we have
40 42 43 45 46 48 49 50 52
 Here we have an odd number of entries, and the middle
number that gives us the required median is 46.
Applied Example 10, page 435
Applied Example: Commuting Times
 The times, in minutes, Susan took to return from work on
nine consecutive working days were
37 36 39 37 34 38 41 40
 What is the median of her evening commute times?
Solution
 If we include the number 44 for the tenth work day and
arrange the numbers in increasing order, we have
34 36 37 37 38 39 40 41
 Here we have an even number of entries so we calculate
the average of the two middle numbers 37 and 38 to find
the required median of 37.5.
Applied Example 10, page 435
Mode
 The mode of a group of numbers is the
number in the set that occurs most
frequently.
Example
 Find the mode, if there is one, of the given group of
numbers.
a. 1 2 3 4 6
b. 2 3 3 4 6 8
c. 2 3 3 3 4 4 4 8
Solution
a. The set has no mode because there isn’t a number that
occurs more frequently than the others.
b. The mode is 3 because it occurs more frequently than the
others.
c. The modes are 3 and 4 because each number occurs three
times.
Example 11, page 436
8.3
Variance and Standard Deviation
 x  (15.8)(.1)  (15.9)(.2)  (16.0)(.4)
 (16.1)(.2)  (16.2)(.1)  16
Var( X )  (.1)(15.8  16)  (.2)(15.9  16)  (.4)(16.0  16)
 (.2)(16.1  16)  (.1)(16.2  16)  0.012
 x  Var( X )  0.012  0.11
Variance of a Random Variable X
 Suppose a random variable has the probability
distribution
x
x1
x2
x3
···
xn
P(X = x)
p1
p2
p3
···
pn
and expected value
E(X ) = 
 Then the variance of the random variable X is
Var ( X )  p1 ( x1   )2  p2 ( x2   ) 2    pn ( xn   ) 2
Example
 Find the variance of the random variable X whose
probability distribution is
x
1
2
3
4
5
6
7
P(X = x)
.05
.075
.2
.375
.15
.1
.05
Solution
 The mean of the random variable X is given by
 x  1(.05)  2(.075)  3(.2)  4(.375)  5(.15)
 6(.1)  7(.05)
4
Example 1, page 442
Example
 Find the variance of the random variable X whose
probability distribution is
x
1
2
3
4
5
6
7
P(X = x)
.05
.075
.2
.375
.15
.1
.05
Solution
 Therefore, the variance of X is given by
Var ( X )  (.05)(1  4) 2  (.075)(2  4) 2  (.2)(3  4) 2
 (.375)(4  4) 2  (.15)(5  4) 2
 (.1)(6  4) 2  (.05)(7  4) 2
 1.95
Example 1, page 442
Standard Deviation of a Random Variable X
 The standard deviation of a random variable X,
 (pronounced “sigma”), is defined by
  Var( X )

p1 ( x1   )2  p2 ( x2   )2    pn ( xn   )2
where x1, x2, … , xn denote the values assumed by
the random variable X and
p1 = P(X = x1), p2 = P(X = x2), … , pn = P(X = xn).
Applied Example: Packaging
 Let X and Y denote the random variables whose values are
the weights of brand A and brand B potato chips,
respectively.
 Compute the means and standard deviations of X and Y
and interpret your results.
x
15.8
15.9
16.0
16.1
16.2
P(X = x)
.1
.2
.4
.2
.1
y
15.7
15.8
15.9
16.0
16.1
16.2
16.3
P(Y = y)
.2
.1
.1
.1
.2
.2
.1
Applied Example 3, page 443
Applied Example: Packaging
x
15.8
15.9
16.0
16.1
16.2
P(X = x)
.1
.2
.4
.2
.1
y
15.7
15.8
15.9
16.0
16.1
16.2
16.3
P(Y = y)
.2
.1
.1
.1
.2
.2
.1
Solution
 The means of X and Y are given by
 x  (.1)(15.8)  (.2)(15.9)  (.4)(16.0)
 (.2)(16.1)  (.1)(16.2)
 16
 y  (.2)(15.7)  (.1)(15.8)  (.1)(15.9)  (.1)(16.0)
 (.2)(16.1)  (.2)(16.2)  (.1)(16.3)
 16
Applied Example 3, page 443
Applied Example: Packaging
x
15.8
15.9
16.0
16.1
16.2
P(X = x)
.1
.2
.4
.2
.1
y
15.7
15.8
15.9
16.0
16.1
16.2
16.3
P(Y = y)
.2
.1
.1
.1
.2
.2
.1
Solution
 Therefore, the variance of X and Y are
Var ( X )  (.1)(15.8  16) 2  (.2)(15.9  16) 2  (.4)(16.0  16) 2
 (.2)(16.1  16) 2  (.1)(16.2  16) 2
 0.012
Var (Y )  (.2)(15.7  16) 2  (.1)(15.8  16) 2  (.1)(15.9  16) 2
 (.1)(16.0  16) 2  (.2)(16.1  16) 2  (.2)(16.2  16) 2
 (.1)(16.3  16) 2
 0.042
Applied Example 3, page 443
Applied Example: Packaging
x
15.8
15.9
16.0
16.1
16.2
P(X = x)
.1
.2
.4
.2
.1
y
15.7
15.8
15.9
16.0
16.1
16.2
16.3
P(Y = y)
.2
.1
.1
.1
.2
.2
.1
Solution
 Finally, the standard deviations of X and Y are
 x  Var ( X )  0.012  0.11
 y  Var (Y )  0.042  0.20
Applied Example 3, page 443
Applied Example: Packaging
x
15.8
15.9
16.0
16.1
16.2
P(X = x)
.1
.2
.4
.2
.1
y
15.7
15.8
15.9
16.0
16.1
16.2
16.3
P(Y = y)
.2
.1
.1
.1
.2
.2
.1
Solution
 The means of X and Y are both equal to 16.
✦ Therefore, the average weight of a package of potato
chips of either brand is the same.
 However, the standard deviation of Y is greater than that
of X.
✦ This tells us that the weights of the packages of brand B
potato chips are more widely dispersed than those of
brand A.
Applied Example 3, page 443
Chebychev’s Inequality
 Let X be a random variable with expected
value  and standard deviation .
 Then the probability that a randomly chosen
outcome of the experiment lies between
 – k and  + k is at least 1 – (1/k2) .
 That is,
1
P(   k  X    k )  1  2
k
Applied Example: Industrial Accidents
 Great Lumber Co. employs 400 workers in its mills.
 It has been estimated that X, the random variable
measuring the number of mill workers who have
industrial accidents during a 1-year period, is distributed
with a mean of 40 and a standard deviation of 6.
 Use Chebychev’s Inequality to find a bound on the
probability that the number of workers who will have an
industrial accident over a 1-year period is between 30 and
50, inclusive.
Applied Example 5, page 445
Applied Example: Industrial Accidents
Solution
 Here,  = 40 and  = 6.
 We wish to estimate P(30  X  50) .
 To use Chebychev’s Inequality, we first determine the
value of k from the equation
 – k = 30
or
 + k = 50
 Since  = 40 and  = 6, we see that k satisfies
40 – 6k = 30
and
40 + 6k = 50
from which we deduce that k = 5/3.
Applied Example 5, page 445
Applied Example: Industrial Accidents
Solution
 Thus, the probability that the number of mill workers who
will have an industrial accident during a 1-year period is
between 30 and 50 is given by
P(30  X  50)  1 
that is, at least 64%.
Applied Example 5, page 445
1
 
5
3
2

16
 .64
25
8.4
The Binomial Distribution
P(SFFF) = P(S)P(F)P(F)P(F) = p · q · q · q = pq3
P(FSFF) = P(F)P(S)P(F)P(F) = q · p · q · q = pq3
P(FFSF) = P(F)P(F)P(S)P(F) = q · q · p · q = pq3
P(FFFS) = P(F)P(F)P(F)P(S) = q · q · q · p = pq3
P( E )  pq3  pq3  pq3  pq3  4 pq3
3
 1  5 
 4      .386
 6  6 
Binomial Experiment
 A binomial experiment has the following
properties:
1. The number of trials in the experiment is
fixed.
2. There are two outcomes in each trial:
“success” and “failure.”
3. The probability of success in each trial is
the same.
4. The trials are independent of each other.
Example
 A fair die is thrown four times. Compute the probability of
obtaining exactly one 6 in the four throws.
Solution
 There are four trials in this experiment.
 Each trial consists of throwing the die once and observing
the face that lands uppermost.
 We may view each trial as an experiment with two
outcomes:
✦ A success (S) if the face that lands uppermost is a 6.
✦ A failure (F) if it is any of the other five numbers.
 Letting p and q denote the probabilities of success and
failure, respectively, of a single trial of the experiment, we
find that
1
1 5
p
and q  1  
6
6 6
Example 1, page 453
Example
 A fair die is thrown four times. Compute the probability of
obtaining exactly one 6 in the four throws.
Solution
 The trials of this experiment are independent, so we have a
binomial experiment.
 Using the multiplication principle, we see that the
experiment has 24 = 16 outcomes.
 The possible outcomes associated with the experiment are:
0 Successes
FFFF
Example 1, page 453
1 Success
SFFF
FSFF
FFSF
FFFS
2 Successes
SSFF
SFSF
SFFS
FSSF
FSFS
FFSS
3 Successes
SSSF
SSFS
SFSS
FSSS
4 Successes
SSSS
Example
 A fair die is thrown four times. Compute the probability of
obtaining exactly one 6 in the four throws.
Solution
 From the table we see that the event of obtaining exactly
one success in four trials is given by
E = {SFFF, FSFF, FFSF, FFFS}
 The probability of this event is given by
P(E) = P(SFFF) + P(FSFF) + P(FFSF) + P(FFFS)
Success
00Successes
FFFF
Example 1, page 453
1 Success
SFFF
FSFF
FFSF
FFFS
2 Successes
SSFF
SFSF
SFFS
FSSF
FSFS
FFSS
3 Successes
SSSF
SSFS
SFSS
FSSS
4 Successes
SSSS
Example
 A fair die is thrown four times. Compute the probability of
obtaining exactly one 6 in the four throws.
Solution
 Since the trials (throws) are independent, the probability of
each possible outcome with one success is given by
P(SFFF) = P(S)P(F)P(F)P(F) = p · q · q · q = pq3
P(FSFF) = P(F)P(S)P(F)P(F) = q · p · q · q = pq3
P(FFSF) = P(F)P(F)P(S)P(F) = q · q · p · q = pq3
P(FFFS) = P(F)P(F)P(F)P(S) = q · q · q · p = pq3
 Therefore, the probability of obtaining exactly one 6 in four
throws is
P( E )  pq3  pq3  pq3  pq3  4 pq3
3
Example 1, page 453
 1  5 
 4      .386
 6  6 
Computation of Probabilities in Bernoulli Trials
 In general, experiments with two outcomes
are called Bernoulli trials, or binomial trials.
 In a binomial experiment in which the
probability of success in any trial is p, the
probability of exactly x successes in n
independent trials is given by
C ( n, x ) p x q n  x
Binomial Distribution
 If we let X be the random variable that gives the
number of successes in a binomial experiment, then
the probability of exactly x successes in n
independent trials may be written
P( X  x)  C (n, x) p x qn x
( x  0,1,2,..., n)
 The random variable X is called a binomial
random variable, and the probability distribution
of X is called a binomial distribution.
Example
 A fair die is thrown five times.
 If a 1 or a 6 lands uppermost in a trial, then the throw is
considered a success.
 Otherwise, the throw is considered a failure.
 Find the probabilities of obtaining exactly 0, 1, 2, 3, 4, and 5
successes, in this experiment.
 Using the results obtained, construct the binomial
distribution for this experiment and draw the histogram
associated with it.
Example 2, page 455
Example
Solution
 This is a binomial experiment with X taking on each of the
values 0, 1, 2, 3, 4, and 5 corresponding to exactly 0, 1, 2, 3,
4, and 5 successes, respectively, in five trials.
 Since the die is fair, the probability of a 1 or a 6 landing
uppermost in any trial is given by
2 1
p 
6 3
from which it also follows that
1 2
q  1 p  1 
3 3
 Finally, n = 5 since there are five trials (throws) in this
experiment.
Example 2, page 455
Example
Solution
 Using the formula for the binomial random variable, we
find that the required probabilities are
P( X  x )  C (n, x) p x qn x
0
1  2
P( X  0)  C (5, 0)    
3  3
x
0
P(X = x)
.132
Example 2, page 455
1
2
5 0
5! 32


 .132
0!5! 243
3
4
5
Example
Solution
 Using the formula for the binomial random variable, we
find that the required probabilities are
P( X  x )  C (n, x) p x qn x
1
1  2
P( X  1)  C (5,1)    
 3  3
x
0
1
P(X = x)
.132
.329
Example 2, page 455
51
2
5! 16


 .329
1!4! 243
3
4
5
Example
Solution
 Using the formula for the binomial random variable, we
find that the required probabilities are
P( X  x )  C (n, x) p x qn x
2
1  2
P( X  2)  C (5, 2)    
 3  3
5 2
x
0
1
2
P(X = x)
.132
.329
329
Example 2, page 455
5!
8


 .329
2!3! 243
3
4
5
Example
Solution
 Using the formula for the binomial random variable, we
find that the required probabilities are
P( X  x )  C (n, x) p x qn x
3
1  2
P( X  3)  C (5,3)    
 3  3
53
5!
4


 .165
3!2! 243
x
0
1
2
3
P(X = x)
.132
.329
329
.165
Example 2, page 455
4
5
Example
Solution
 Using the formula for the binomial random variable, we
find that the required probabilities are
P( X  x )  C (n, x) p x qn x
4
1  2
P( X  4)  C (5, 4)    
 3  3
5 4
5! 2


 .041
4!1! 243
x
0
1
2
3
4
P(X = x)
.132
.329
329
.165
.041
Example 2, page 455
5
Example
Solution
 Using the formula for the binomial random variable, we
find that the required probabilities are
P( X  x )  C (n, x) p x qn x
5
1  2
P( X  5)  C (5,5)    
 3  3
55
5!
1


 .004
5!0! 243
x
0
1
2
3
4
5
P(X = x)
.132
.329
329
.165
.041
.004
Example 2, page 455
Example
Solution
 We can now use the probability distribution table to
construct a histogram for this experiment:
.4
.3
.2
.1
0
1
2
3
4
x
5
x
0
1
2
3
4
5
P(X = x)
.132
.329
329
.165
.041
.004
Example 2, page 455
Applied Example: Quality Control
 A division of Solaron manufactures photovoltaic cells to
use in the company’s solar energy converters.
 It estimates that 5% of the cells manufactured are
defective.
 If a random sample of 20 is selected from a large lot of
cells manufactured by the company, what is the
probability that it will contain at most 2 defective cells?
Applied Example 5, page 457
Applied Example: Quality Control
Solution
 We may view this as a binomial experiment with n = 20
trials that correspond to 20 photovoltaic cells.
 There are two possible outcomes of the experiment:
defective (“success”) and non-defective (“failure”).
 The probability of success in each trial is p = .05 and the
probability of failure in each trial is q = .95.
✦ Since the lot from which the sample is selected is large,
the removal of a few cells will not appreciably affect the
percentage of defective cells in the lot in each successive
trial.
 The trials are independent of each other.
✦ Again, this is because of the large lot size.
Applied Example 5, page 457
Applied Example: Quality Control
Solution
 Letting X denote the number of defective cells, we find that
the probability of finding at most 2 defective cells in the
sample of 20 is given by
P( X  0)  P ( X  1)  P ( X  2)
 C (20,0)(.05)0 (.95) 20  C (20,1)(.05)1 (.95)19
 C (20, 2)(.05) 2 (.95)18
 .3585  .3774  .1887  .9246
 Thus, approximately 92% of the sample will have at most 2
defective cells.
 Equivalently, approximately 8% of the sample will contain
more than 2 defective cells.
Applied Example 5, page 457
Mean, Variance, and Standard Deviation
of a Random Variable
 If X is a binomial random variable associated with
a binomial experiment consisting of n trials with
probability of success p and probability of failure
q, then the mean (expected value), variance, and
standard deviation of X are
  E ( X )  np
Var ( X )  npq
 X  npq
Applied Example: Quality Control
 PAR Bearings manufactures ball bearings packaged in lots of
100 each.
 The company’s quality-control department has determined
that 2% of the ball bearings manufactured do not meet
specifications imposed by a buyer.
 Find the average number of ball bearings per package that
fail to meet the buyer’s specification.
Solution
 Since this is a binomial experiment, the average number of
ball bearings per package that fail to meet the specifications
is given by the expected value of the associated binomial
random variable.
 Thus, we expect to find
  E ( X )  np  (100)(.02)  2
substandard ball bearings in a package of 100.
Applied Example 7, page 459
8.5
The Normal Distribution
y
y  f ( x)
Area is
.9545
 – 2

 + 2
x
Probability Density Functions
 In this section we consider probability distributions
associated with a continuous random variable:
✦ A random variable that may take on any value lying in
an interval of real numbers.
 Such probability distributions are called continuous
probability distributions.
 A continuous probability distribution is defined by a
function f whose domain coincides with the interval of
values taken on by the random variable associated with
the experiment.
 Such a function f is called the probability density function
associated with the probability distribution.
Properties of a Probability Density Function
 The properties of a probability density function are:
✦ f(x) is nonnegative for all values of x.
✦ The area of the region between the graph of f and
the x-axis is equal to 1.
For example:
y
y  f ( x)
Area = 1
x
Properties of a Probability Density Function
 Given a continuous probability distribution defined by a
probability density function f, the probability that the
random variable X assumes a value in an interval
a < x < b is given by the area of the region between the
graph of f and the x-axis, from x = a to x = b:
y
y  f ( x)
Area = P(a < x < b)
a
b
x
Properties of a Probability Density Function
 The mean  and the standard deviation  of a continuous
probability distribution have roughly the same meaning as
the mean and standard deviation of a finite probability
distribution.
 The mean of a continuous probability distribution is a
measure of the central tendency of the probability
distribution, and the standard deviation measures its
spread about the mean.
Normal Distributions
 A special class of continuous probability distributions is
known as normal distributions.
 The normal distributions are without doubt the most
important of all the probability distributions.
 There are many phenomena with probability
distributions that are approximately normal:
✦ For example, the heights of people, the weights of
newborn infants, the IQs of college students, the actual
weights of 16-ounce packages of cereal, and so on.
 The normal distribution also provides us with an accurate
approximation to the distributions of many random
variables associated with random-sampling problems.
Normal Distributions
y
y  f ( x)

x
 The graph of a normal distribution is bell shaped and is
called a normal curve.
 The curve has a peak at x = .
 The curve is symmetric with respect to x = .
Normal Distributions
y
y  f ( x)
Area = 1
x
 The curve always lies above the x-axis but approaches the
x-axis as x extends indefinitely in either direction.
 The area under the curve is 1.
Normal Distributions
y
y  f ( x)
Area is
.6827
–

+
 For any normal curve, 68.27% of the area under the
curve lies within 1 standard deviation.
x
Normal Distributions
y
y  f ( x)
Area is
.9545
 – 2

 + 2
 For any normal curve, 95.45% of the area under the
curve lies within 2 standard deviations.
x
Normal Distributions
y
y  f ( x)
Area is
.9973
 – 3

 + 3
 For any normal curve, 99.73% of the area under the
curve lies within 3 standard deviations.
x
Normal Distributions
y
1
2
3
4
 A normal distribution is completely described by the
x
mean  and the standard deviation  :
✦ The mean  of a normal distribution determines where
the center of the curve is located.
Normal Distributions
1
2
3
4
y

 A normal distribution is completely described by the
mean  and the standard deviation  :
✦ The standard deviation  of a normal distribution
determines the sharpness (or flatness) of the curve.
x
Normal Distributions
 There are infinitely many normal curves corresponding




to different means  and standard deviations  .
Fortunately, any normal curve may be transformed into
any other normal curve, so in the study of normal curves
it is enough to single out one such particular curve for
special attention.
The normal curve with mean  = 0 and standard
deviation  = 1 is called the standard normal curve.
The corresponding distribution is called the standard
normal distribution.
The random variable itself is called the standard normal
variable and is commonly denoted by Z.
Examples
 Let Z be the standard normal variable. Make a sketch of
the appropriate region under the standard normal curve,
and find the value of P(Z < 1.24).
Solution
 The region under the standard normal curve associated
with the probability P(Z < 1.24) is
y
0
Example 1, page 465
1.24
z
Examples
 Let Z be the standard normal variable. Make a sketch of
the appropriate region under the standard normal curve,
and find the value of P(Z < 1.24).
Solution
 Use Table 2 in Appendix B to find the area of the required
region:
✦ We find the z value of 1.24 in the table by first locating
the number 1.2 in the column and then locating the
number 0.04 in the row, both headed by z.
✦ We then read off the number 0.8925 appearing in the
body of the table, on the found row and column that
correspond to z = 1.24.
Example 1, page 465
Examples
 Let Z be the standard normal variable. Make a sketch of
the appropriate region under the standard normal curve,
and find the value of P(Z < 1.24).
Solution
 Thus, the area of the required region under the curve
is .8925, and we find that P(Z < 1.24) = 0.8925.
y
Area is
.8925
0
Example 1, page 465
1.24
z
Examples
 Let Z be the standard normal variable. Make a sketch of
the appropriate region under the standard normal curve,
and find the value of P(Z > 0.5).
Solution
 The region under the standard normal curve associated
with the probability P(Z > 0.5) is
y
0 0.5
Example 1, page 465
z
Examples
 Let Z be the standard normal variable. Make a sketch of
the appropriate region under the standard normal curve,
and find the value of P(Z > 0.5).
Solution
 Since the standard normal curve is symmetric, the
required area is equal to the area to the left of z = – 0.5, so
P(Z > 0.5) = P(Z < – 0.5)
y
– 0.5 0
Example 1, page 465
z
Examples
 Let Z be the standard normal variable. Make a sketch of
the appropriate region under the standard normal curve,
and find the value of P(Z > 0.5).
Solution
 Using Table 2 in Appendix B as before to find the area of
the required region we find that
P(Z > 0.5) = P(Z < – 0.5) = .3085
y
Area is
.3085
– 0.5 0
Example 1, page 465
z
Examples
 Let Z be the standard normal variable. Make a sketch of
the appropriate region under the standard normal curve,
and find the value of P(0.24 < Z < 1.48).
Solution
 The region under the standard normal curve associated
with the probability P(0.24 < Z < 1.48) is
y
P(0.24 < Z < 1.48)
0 0.24 1.48
Example 1, page 465
z
Examples
 Let Z be the standard normal variable. Make a sketch of
the appropriate region under the standard normal curve,
and find the value of P(0.24 < Z < 1.48).
Solution
 This area is obtained by subtracting the area under the
curve to the left of z = 0.24 from the area under the curve
to the left of z = 1.48:
y
P(Z < 1.48)
0 0.24 1.48
Example 1, page 465
z
Examples
 Let Z be the standard normal variable. Make a sketch of
the appropriate region under the standard normal curve,
and find the value of P(0.24 < Z < 1.48).
Solution
 This area is obtained by subtracting the area under the
curve to the left of z = 0.24 from the area under the curve
to the left of z = 1.48:
y
P(Z < 0.24)
0 0.24 1.48
Example 1, page 465
z
Examples
 Let Z be the standard normal variable. Make a sketch of
the appropriate region under the standard normal curve,
and find the value of P(0.24 < Z < 1.48).
Solution
 Thus, the required area is given by
y
P(0.24  Z  1.48)  P(Z  1.48)  P(Z  0.24)
 .9306  .5948  .3358
Area is
.3358
0 0.24 1.48
Example 1, page 465
z
Transforming into a Standard Normal Curve
 When dealing with a non-standard normal curve, it is
possible to transform such a curve into a standard
normal curve.
If X is a normal random variable with mean
 and standard deviation , then it can be
transformed into the standard normal
random variable Z by substituting in
Z
X 

Transforming into a Standard Normal Curve
 The area of the region under the normal curve
between x = a and x = b is equal to the area of the
region under the standard normal curve between
z
a

and
z
b

 Thus, in terms of probabilities we have
b
a
P ( a  X  b)  P 
Z





Transforming into a Standard Normal Curve
 Similarly, we have
a

P( X  a )  P  Z 




and
b

P ( X  b)  P  Z 




Transforming into a Standard Normal Curve
 This transformation can be seen graphically as well.
 The area of the region under a nonstandard normal curve
between a and b is equal to the area of the region under a
standard normal curve between z = (a – )/ and z = (b – )/ :
y
Nonstandard
Normal Curve
Same Area
a

b
x
Transforming into a Standard Normal Curve
 This transformation can be seen graphically as well.
 The area of the region under a nonstandard normal curve
between a and b is equal to the area of the region under a
standard normal curve between z = (a – )/ and z = (b – )/ :
y
Standard
Normal Curve
Same Area
a

0
b

x
Example
 If X is a normal random variable with
 = 100 and  = 20,
find the values of P(X < 120), P(X > 70), and P(75 < X <110).
Solution
 For the case of P(X < 120), we use the formula
b

P ( X  b)  P  Z 
 

with  = 100,  = 20, and b = 120, which gives us
120  100 

P( X  120)  P  Z 

20


 P( Z  1)
=.8413
Example 3, page 468
Example
 If X is a normal random variable with
 = 100 and  = 20,
find the values of P(X < 120), P(X > 70), and P(75 < X <110).
Solution
 For the case of P(X > 70), we use the formula
a

P( X  a )  P  Z 




with  = 100,  = 20, and a = 70, which gives us
70  100 

P( X  70)  P  Z 

20 

 P( Z  1.5)
 P( Z  1.5)
=.9332
Example 3, page 468
Example
 If X is a normal random variable with
 = 100 and  = 20,
find the values of P(X < 120), P(X > 70), and P(75 < X <110).
Solution
 Finally, for the case of P(75 < X <110), we use the formula
b
a
P ( a  X  b)  P 
Z
 
 
with  = 100,  = 20, a = 75, and b = 110, which gives us
110  100 
 75  100
P(75  X  110)  P 
Z

20
20


 P( 1.25  Z  0.5)
 P( Z  0.5)  P( Z  1.25)
 .6915  .1056  .5859
Example 3, page 468
8.6
Applications of the Normal Distribution
.20
.15
.10
.05
9.5
0
1
2 3 4 5 6 7 8 9 10
12
14
16
x  10
18
20
x
Applied Example: Birth Weights of Infants
 The medical records of infants delivered at the Kaiser
Memorial Hospital show that the infants’ birth weights in
pounds are normally distributed with a mean of 7.4 and a
standard deviation of 1.2.
 Find the probability that an infant selected at random
from among those delivered at the hospital weighed more
than 9.2 pounds at birth.
Applied Example 1, page 471
Applied Example: Birth Weights of Infants
Solution
 Let X be the normal random variable denoting the birth
weights of infants delivered at the hospital.
 Then, we can calculate the probability that an infant selected
at random has a birth weight of more than 9.2 pounds by
setting  = 7.4,  = 1.2, and a = 9.2 in the formula
a

P( X  a )  P  Z 




to find
9.2  7.4 

P ( X  9.2)  P  Z 
  P ( Z  1.5)
1.2 

 P ( Z  1.5)  .0668
 Thus, the probability that an infant delivered at the hospital
weighs more than 9.2 pounds is .0668.
Applied Example 1, page 471
Approximating Binomial Distributions
 One important application of the normal distribution is
that it can be used to accurately approximate other
continuous probability distributions.
 As an example, we will see how a binomial distribution
may be approximated by a suitable normal distribution.
 This provides a convenient and simple solution to certain
problems involving binomial distributions.
Approximating Binomial Distributions
 Recall that a binomial distribution is a probability
distribution of the form
P( X  x )  C (n, x) p x qn x
 For small values of n, the arithmetic computations may be
done with relative ease. However, if n is large, then the
work involved becomes overwhelming, even when tables of
P(X = x) are available.
Approximating Binomial Distributions
 To see how a normal distribution can help in such
situations, consider a coin-tossing experiment.
 Suppose a fair coin is tossed 20 times and we wish to
compute the probability of obtaining 10 or more heads.
 The solution to this problem may be obtained, of course,
by laboriously computing
P( X  10)  P( X  10)  P( X  11)    P( X  20)
 As an alternative solution, let’s begin by interpreting the
solution in terms of finding the areas of rectangles in the
histogram.
Approximating Binomial Distributions
 We may calculate the probability of
obtaining exactly x heads in 20 coin
tosses with the formula
P( X  x )  C (n, x) p x qn x
 The results lead to the binomial
distribution shown in the table to
the right.
x
0
1
2
3
4
5
6
7
8
9
10
11
12

20
P(X = x)
.0000
.0000
.0002
.0011
.0046
.0148
.0370
.0739
.1201
.1602
.1762
.1602
.1201

.0000
Approximating Binomial Distributions
 Using the data from the table, we may construct a
histogram for the distribution:
.20
.15
.10
.05
0
1
2 3 4 5 6 7 8 9 10
12
14
16
18
20
x
Approximating Binomial Distributions
 The probability of obtaining 10 or more heads in 20 coin
tosses is equal to the sum of the areas of the blue shaded
rectangles of the histogram of the binomial distribution:
.20
.15
.10
.05
0
1
2 3 4 5 6 7 8 9 10
12
14
16
18
20
x
Approximating Binomial Distributions
 Note that the shape of the histogram suggests that the
binomial distribution under consideration may be
approximated by a suitable normal distribution.
.20
.15
.10
.05
0
1
2 3 4 5 6 7 8 9 10
12
14
16
18
20
x
Approximating Binomial Distributions
 The mean and standard deviation of the binomial
distribution in this problem are given, respectively, by
  np
 (20)(.5)
 10
  npq
 (20)(.5)(.5)
 2.24
 Thus, we should choose a normal curve for this purpose
with a mean of 10 and a standard deviation of 2.24.
Approximating Binomial Distributions
 Superimposing on the histogram a normal curve with a
mean of 10 and a standard deviation of 2.24 clearly gives
us a good fit:
.20
.15
.10
.05
0
1
2 3 4 5 6 7 8 9 10
12
14
16
18
20
x
Approximating Binomial Distributions
 The good fit suggests that the sum of the areas of the
rectangles representing P(X = x), the probability of obtaining
10 or more heads in 20 coin tosses, may be approximated by
the area of an appropriate region under the normal curve.
.20
.15
.10
.05
0
1
2 3 4 5 6 7 8 9 10
12
14
16
18
20
x
Approximating Binomial Distributions
 To determine this region, note that the base of the portion of
the histogram representing the required probability extends
from x = 9.5 on, since the base of the leftmost rectangle is
centered on 10:
.20
.15
.10
.05
9.5
0
1
2 3 4 5 6 7 8 9 10
12
14
16
x  10
18
20
x
Approximating Binomial Distributions
 Therefore, the required region under the normal curve
should also have x  9.5.
 Letting Y denote the continuous normal variable, we obtain
P( X  10)  P(Y  9.5)
 P(Y  9.5)
9.5  10 

 PZ 

2.24


 P( Z  0.22)
= P ( Z  0.22)
 .5871
Approximating Binomial Distributions
 The exact value can be found by computing
P( X  10)  P( X  10)  P( X  11)    P( X  20)
in the usual (time-consuming) fashion.
 Using this method yields a probability of .5881, which is not
very different from the approximation of .5871 obtained
using the normal distribution.
Theorem 1
 Suppose we are given a binomial distribution
associated with a binomial experiment
involving n trials, each with a probability of
success p and a probability of failure q.
 Then, if n is large and p is not close to 0 or 1,
the binomial distribution may be
approximated by a normal distribution with
  np
and
  npq
Applied Example: Quality Control
 An automobile manufacturer receives the
microprocessors used to regulate fuel consumption in its
automobiles in shipments of 1000 each from a certain
supplier.
 It has been estimated that, on the average, 1% of the
microprocessors manufactured by the supplier are
defective.
 Determine the probability that more than 20 of the
microprocessors in a single shipment are defective.
Applied Example 4, page 475
Applied Example: Quality Control
Solution
 Let X denote the number of defective microprocessors in a
single shipment.
 Then X has a binomial distribution with n = 1000, p = .01,
and q = .99, so
  np
 (1000)(.01)  10
  npq
 (1000)(.01)(.99)  3.15
Applied Example 4, page 475
Applied Example: Quality Control
Solution
 Approximating the binomial distribution by a normal
distribution with a mean of 10 and a standard deviation of
3.15, we find that the probability that more than 20
microprocessors in a shipment are defective is given by
P( X  20)  P(Y  20.5)
20.5  10 

 PZ 

3.15


 P( Z  3.33)
 P( Z  3.33)
 .0004
 Thus, approximately 0.04% of the shipments with 1000
microprocessors each will have more than 20 defective units.
Applied Example 4, page 475
End of
Chapter