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Chapter 6:
CONTINUOUS RANDOM
VARIABLES AND THE NORMAL
DISTRIBUTION
CONTINUOUS PROBABILITY
DISTRIBUTION
Table 6.1 Frequency and Relative Frequency Distributions of Heights of Female Students
Height of a Female Student
(inches)
f
x
60
61
62
63
64
65
66
67
68
69
70
to
to
to
to
to
to
to
to
to
to
to
less
less
less
less
less
less
less
less
less
less
less
than
than
than
than
than
than
than
than
than
than
than
61
62
63
64
65
66
67
68
69
70
71
Relative Frequency
90
170
460
750
970
760
640
440
320
220
180
.018
.034
.092
.150
.194
.152
.128
.088
.064
.044
.036
N = 5000
Sum = 1.0
2
Figure 6.1
Histogram and polygon for Table 6.1.
3
Figure 6.2
Probability distribution curve for heights.
4
CONTINUOUS PROBABILITY
DISTRIBUTION
Two characteristics
1. The probability that
x assumes a value in any
interval lies in the range 0 to 1
2. The total probability of all the (mutually
exclusive) intervals within which x can
assume a value of 1.0
5
Figure 6.3
Area under a curve between two points.
Shaded area is
between 0 and 1
x=a
x=b
x
6
Figure 6.4
Total area under a probability distribution
curve.
Shaded area is 1.0 or
100%
x
7
Figure 6.5
Area under the curve as probability.
Shaded area gives the
probability P (a ≤ x ≤ b)
a
b
x
8
Figure 6.6
Probability that x lies in the interval 65 to
68 inches.
9
Figure 6.7
The probability of a single value of x is
zero.
10
Figure 6.8
Probability “from 65 to 68” and “between
65 and 68”.
11
THE NORMAL DISTRIBUTION
Normal Probability Distribution
A normal probability distribution , when
plotted, gives a bell-shaped curve such that
The total area under the curve is 1.0.
2. The curve is symmetric about the mean.
3. The two tails of the curve extend indefinitely.
1.
12
Figure 6.11
Normal distribution with mean μ and
standard deviation σ.
Standard
deviation = σ
Mean = μ
x
13
Figure 6.12
Total area under a normal curve.
The shaded area is
1.0 or 100%
μ
x
14
Figure 6.13
A normal curve is symmetric about the
mean.
Each of the two shaded
areas is .5 or 50%
.5
.5
μ
x
15
Figure 6.14
Areas of the normal curve beyond
μ ± 3σ.
Each of the two shaded areas is very
close to zero
μ – 3σ
μ
μ + 3σ
x
16
Figure 6.15
Three normal distribution curves with the
same mean but different standard deviations.
σ=5
σ = 10
σ = 16
μ = 50
x
17
Figure 6.16
σ=5
µ = 20
Three normal distribution curves with
different means but the same standard
deviation.
σ=5
σ=5
µ = 30
µ = 40
x
18
THE STANDARD NORMAL
DISTRIBTUION
Definition
The normal distribution with μ = 0 and σ =
1 is called the standard normal
distribution.
19
Figure 6.17
The standard normal distribution curve.
σ=1
µ=0
-3
-2
-1
0
1
2
3
20
z
THE STANDARD NORMAL
DISTRIBTUION
Definition
The units marked on the horizontal axis of
the standard normal curve are denoted by z
and are called the z values or z scores. A
specific value of z gives the distance between
the mean and the point represented by z in
terms of the standard deviation.
21
Figure 6.18
Area under the standard normal curve.
Each of these two
areas is .5
.5
-3
-2
-1
.5
0
1
2
3
z
22
Example 6-1
Find the area under the standard normal
curve between z = 0 and z = 1.95.
23
Table 6.2
Area Under the Standard Normal Curve from
z = 0 to z = 1.95
z
0.0
0.1
0.2
.
1.9
.
.
3.0
.00
.0000
.0398
.0793
.
.4713
.
.
.4987
.01
.0040
.0438
.0832
.
.4719
.
.
.4987
…
…
…
…
…
…
…
…
…
.05
.0199
.0596
.0987
.
.4744
.
.
.4989
…
…
…
…
…
…
…
…
…
.09
.0359
.0753
.1141
.
.4767
.
.
.4990
Required area
24
Figure 6.19
Area between z = 0 to z = 1.95.
Shaded area
is .4744
0
1.95
z
25
Example 6-2
Find the area under the standard normal
curve from z = -2.17 to z = 0.
26
Solution 6-2


Because the normal distribution is
symmetric about the mean, the area from
z = -2.17 to z = 0 is the same as the area from z
= 0 to z = 2.17.
Area from -2.17 to 0 = P(-2.17≤ z ≤ 0) = .4850
27
Figure 6.20
Area from z = 0 to z = 2.17 equals area
from z = -2.17 to z = 0.
Because the symmetry
the shaded areas are
equal
-2.17
0
z
0
2.17
z
28
Table 6.3
Area Under the Standard Normal Curve from
z = 0 to z = 2.17
z
0.0
0.1
0.2
.
.
2.1
.
3.0
.00
.0000
.0398
.0793
.
.
.4821
.
.4987
.01
.0040
.0438
.0832
.
.
.4826
.
.4987
…
…
…
…
…
…
…
…
…
.07
.0279
.0675
.1064
.
.
.4850
.
.4989
…
.09
…
.0359
…
.0753
…
.1141
…
.
…
.
…
.4857
…
.
…
.4990
Required area 29
Figure 6.21
Area from z = -2.17 to z = 0.
Shaded area is
.4850
-2.17
0
z
30
Example 6-3
Find the following areas under the
standard normal curve.
Area to the right of z = 2.32
b) Area to the left of z = -1.54
a)
31
Solution 6-3
a) Area to the right of 2.32
= P (z > 2.32) = .5 - .4898 = .0102
as shown in Figure 6.22
32
Figure 6.22
Area to the right of z = 2.32.
This area is .4898 from
Table VII
Shaded area is
.5 - .4898 .0102
.4898
0
2.32
z
33
Solution 6-3
b) Area to the left of -1.54
= P (z < -1.54) = .5 - .4382 = .0618
as shown in Figure 6.23
34
Figure 6.23
Area to the left of z = -1.54.
This area is .4382 from
Table VII
Shaded area is
.5 - .4382 =
.0618
.4382
-1.54
0
z
35
Example 6-4
Find the following probabilities for the
standard normal curve.
P (1.19 < z < 2.12)
b) P (-1.56 < z < 2.31)
c) P (z > -.75)
a)
36
Solution 6-4
a) P (1.19 < z < 2.12)
= Area between 1.19 and 2.12
= .4830 - .3830
= .1000
as shown in Figure 6.24
37
Figure 6.24
Finding P (1.19 < z < 2.12).
Shaded area =
.4830 - .3830 =
.1000
0
1.19
2.12
z
38
Solution 6-4
b) P (-1.56 < z < 2.31)
= Area between -1.56 and 2.31
= .4406 + .4896
= .9302
as shown in Figure 6.25
39
Figure 6.25
Finding
P (-1.56 < z < 2.31).
Total shaded area
= .4406 + .4896 =
.9302
.4406
-1.56
.4896
0
2.31
z
40
Solution 6-4
c)
P (z > -.75) = Area to the right of -.75
= .2734 + .5
= .7734
as shown in Figure 6.26
41
Figure 6.26
Finding P (z > -.75).
Total shaded area =
.2734 + .5 = .7734
.2734
.5
-.75 0
z
42
Figure 6.27
Area within one standard deviation of
the mean.
Total shaded area
is .3413 + .3413 =
.6826 or 68.26%
.3413 .3413
-1.0
0
1.0
z
43
Figure 6.28
Area within two standard deviations of
the mean.
Total shaded area is
.4772 + .4772 =
.9544 or 95.44%
.4772
-2.0
.4772
0
2.0
z
44
Figure 6.29
Area within three standard deviations of
the mean.
Total shaded area
is .4987 + .4987 =
.9974 or 99.74%
.4987
-3.0
.4987
0
3.0
z
45
Example 6-5
Find the following probabilities for the
standard normal curve.
a)
b)
P (0 < z < 5.67)
P (z < -5.35)
46
Solution 6-5
a) P (0 < z < 5.67)
= Area between 0 and 5.67
= .5 approximately
as shown in Figure 6.30
47
Figure 6.30
Area between z = 0 and z = 5.67.
Shaded area is
approximately .5
0
5.67
z
48
Solution 6-5
b) P (z < -5.35) = Area to the left of -5.35
= .5 - .5
= .00 approximately
as shown in Figure 6.31
49
Figure 6.31
Area to the left of z = -5.35.
This area is
approximately .5
Shaded area is
approximately .00
-5.35
0
z
50
STANDARDIZING A NORMAL
DISTRIBUTION
Converting an x Value to a z Value
For a normal random variable x, a particular value
of x can be converted to its corresponding z value
by using the formula
z
x

where μ and σ are the mean and standard
deviation of the normal distribution of x,
respectively.
51
Example 6-6
Let x be a continuous random variable that
has a normal distribution with a mean of
50 and a standard deviation of 10. convert
the following x values to z values.
a) x = 55
b) x = 35
52
Solution 6-6
a) ajdaj
z
x

55  50

 .50
10
53
Figure 6.32
z value for x = 55.
Normal
distribution with
μ = 50 and σ = 10
μ=
50
x = 55
x
Standard normal
distribution
0 .50
z
z value for x = 55
54
Solution 6-6
b)
z
x

35  50

 1.50
10
55
Figure 6.33
z value for x = 35.
σ = 10
35 μ = 50
-1.50
0
x
z
56
Example 6-7
Let x be a continuous random variable
that is normally distributed with a mean
of 25 and a standard deviation of 4.
Find the area
a) between x = 25 and x = 32
b) between x = 18 and x = 34
57
Solution 6-7
a)



The z value for x = 25 is 0
The z value for x = 32 is
x   32  25
z

 1.75

4
P (25 < x < 32) = P(0 < z < 1.75)
= .4599
58
Figure 6.34
Area between x = 25 and x = 32.
.4599
25
32
These areas
are equal
x
.4599
0
1.75
z
59
Solution 6-7
b)

18  25
 1.75
For x = 18: z 
4

For x = 34: z  34  25  2.25
4

P (18 < x < 34) = P (-1.75 < z < 2.25 )
= .4599 + .4878
= .9477
60
Figure 6.35
Area between x = 18 and x = 34.
Shaded area =
.4599 + .4878 =
.9477
.4599
18
-1.75
.4878
25
34
x
0
2.25
z
61
Example 6-8
Let x be a normal random variable with
its mean equal to 40 and standard
deviation equal to 5. Find the following
probabilities for this normal distribution
a)
b)
P (x > 55)
P (x < 49)
62
Solution 6-8
a)

55  40
For x = 55: z  5  3.00

P (x > 55) = P (z > 3.00) = .5 - .4987
= .0013
63
Figure 6.36
Finding P (x > 55).
Shaded area
= .5 - .4987
= .0013
.4987
40
55
x
0
3.00
z
64
Solution 6-8
b)
49  40
z
 1.80
5

For x = 49:

P (x < 49) = P (z < 1.80) = .5 + .4641
= .9641
65
Figure 6.37
Finding P (x > 49).
Shaded area =
.5 + .4641 =
.9641
.5
.4641
40
49
x
0
1.80
z
66
Example 6-9
Let x be a continuous random variable that
has a normal distribution with μ = 50 and
σ = 8. Find the probability P (30 ≤ x ≤ 39).
67
Solution 6-9
For x = 30:
30  50
z
 2.50
8

For x = 39:
39  50
z
 1.38
8

P (30 ≤ x ≤ 39) = P (-2.50 ≤ z ≤ -1.38)

= .4938 - .4162
= .0776
68
Figure 6.38
Finding P (30 ≤ x ≤ 39).
Shaded area =
.4938 - .4162 =
.0776
30
39
50
x
-2.50
-1.38
0
z
69
Example 6-10
Let x be a continuous random variable
that has a normal distribution with a
mean of 80 and a standard deviation of
12. Find the area under the normal
distribution curve
a) from x = 70 to x = 135
b) to the left of 27
70
Solution 6-10
a)



For x = 70:
70  80
z
 .83
12
135  80
For x = 135: z  12  4.58
P (70 ≤ x ≤ 135) = P (-.83 ≤ z ≤ 4.58)
= .2967 + .5
= .7967 approximately
71
Figure 6.39
Area between x = 70 and x = 135.
Shaded area is
approximately .2967 + .5 =
.7967
70
80
135
x
-.83
0
4.58
z
72
Solution 6-10
b)

27  80
 4.42
For x = 27: z 
12

P (x < 27) = P (z < -4.42)
= .5 - .5 = .00 approximately
73
Figure 6.40
Area to the left of x = 27.
Shaded area is
approximately
.5 - .5 = .00
27
-4.42
80
x
0
z
74
APPLICATIONS OF THE NORMAL
DISTRIBUTION
Example 6-11
According to Automotive Lease Guide, the
Porsche 911 sports car is among the
vehicles that hold their value best. A
Porsche 911 (with price of $87,500 for a
new car) is expected to command a price
of $48,125 after three years (The Wall
Street Journal, August 6, 2002).
75
Example 6-11
Suppose the prices of all three-year old
Porsche 911 sports cars have a normal
distribution with a mean price of $48,125
and a standard deviation of $1600. Find
the probability that a randomly selected
three-year-old Porsche 911 will sell for a
price between $46,000 and $49,000.
76
Solution 6-11

46,000  48,125
 1.33
For x = $46,000: z 
1600

For x = $49,000: z  49,000  48,125  .55
1600

P ($46,000 < x < $49,000)
= P (-1.33 < z < .55)
= .4082 + .2088
= .6170 = 61.70%
77
Solution 6-11
Thus, the probability is .6170 that a
randomly selected three-year-old Porsche
911 sports car will sell for a price between
$46,000 and $49,000.
78
Figure 6.41
Area between x = $46,000 and
x = $49,000.
.4082
.2088
Shaded area is
.4082 + .2088 =
.6170
$46,000 $48,125 $49,000
-1.33
0 .55
x
z
79
Example 6-12
A racing car is one of the many toys
manufactured by Mark Corporation. The
assembly times for this toy follow a normal
distribution with a mean of 55 minutes and a
standard deviation of 4 minutes. The company
closes at 5 P.M. everyday. If one worker starts to
assemble a racing car at 4 P.M., what is the
probability that she will finish this job before the
company closes for the day?
80
Solution 6-12


μ = 55 minutes
σ = 4 minutes
60  55
z
 1.25
4

For x = 60:

P (x ≤ 60) = P (z ≤ 1.25) = .5 + .3944
= .8944
81
Solution 6-12
Thus, the probability is .8944 that this
worker will finish assembling this racing car
before the company closes for the day.
82
Figure 6.42
Area to the left of x = 60.
Shaded area is
.5 + .3944 = .8944
.5
.3944
55
60
x
0
1.25
z
83
Example 6-13
Hupper Corporation produces many types of soft
drinks, including Orange Cola. The filling
machines are adjusted to pour 12 ounces of soda
into each 12-ounce can of Orange Cola. However,
the actual amount of soda poured into each can is
not exactly 12 ounces; it varies from can to can.
It has been observed that the net amount of soda
in such a can has a normal distribution with a
mean of 12 ounces and a standard deviation of
.015 ounce.
84
Example 6-13
a)
b)
What is the probability that a randomly
selected can of Orange Cola contains 11.97
to 11.99 ounces of soda?
What percentage of the Orange Cola cans
contain
12.02 to 12.07 ounces of soda?
85
Solution 6-13
a)



For x = 11.97:
11.97  12
z
 2.00
.015
For x = 11.99:
11.99  12
z
 .67
.015
P (11.97 ≤ x ≤ 11.99)
= P (-2.00 ≤ z ≤ -.67) = .4772 - .2486
= .2286
86
Figure 6.43
Area between x = 11.97 and x = 11.99.
Shaded area
= .4772 - .2486
= .2286
11.97 11.99 12
-2.00
-.67
0
x
z
87
Solution 6-13
b)



For x = 12.02:
12.02  12
z
 1.33
.015
For x = 12.07:
12.07  12
z
 4.67
.015
P (12.02 ≤ x ≤ 12.07)
= P (1.33 ≤ z ≤ 4.67) = .5 - .4082
= .0918
88
Figure 6.44
Area from x = 12.02 to x = 12.07.
Shaded area is
.5 - .4082 = .0918
μ = 12
0
12.02
1.33
12.07
4.67
x
z
89
Example 6-14
The life span of a calculator manufactured by
Texas Instruments has a normal distribution with
a mean of 54 months and a standard deviation of
8 months. The company guarantees that any
calculator that starts malfunctioning within 36
months of the purchase will be replaced by a new
one.
About what percentage of calculators made by
this company are expected to be replaced?
90
Solution 6-14
36  54
z
 2.25
8

For x = 36:

P (x < 36) = P (z < -2.25)

= .5 - .4878
= .0122
Hence, 1.22% of the calculators are
expected to be replaced.
91
Figure 6.45
Area to the left of x = 36.
Shaded area
is .5 - .4878
= .0122
.4878
36
μ = 54
x
-2.25
0
z
92
DETERMINING THE z AND x VALUES WHEN
AN AREA UNDER THE NORMAL
DISTRIBUTION CURVE IS KNOWN
Example 6-15
Find a point z such that the area under the
standard normal curve between 0 and z is
.4251 and the value of z is positive.
93
Figure 6.46
Finding the z value.
Shaded area is given
to be .4251
0
z
z
To find this z
94
Table 6.4
Finding the z Value When Area Is
Known.
Solution 6-15
z
0.0
0.1
0.2
.
1.4
.
3.0
.00
.0000
.0398
.0793
.
.
.4987
.01
.0040
.0438
.0832
.
.
.4987
Therefore, z = 1.44
…
…
…
…
…
…
…
.04
.4251
.
.4988
…
…
…
…
…
…
…
…
.09
.0359
.0753
.1141
.
.
…
.4990
We locate this value in
Table VII of Appendix C
95
Example 6-16
Find the value of z such that the area
under the standard normal curve in the
right tail is .0050.
96
Solution 6-16


Area between 0 and z = .5 - .005
= .4950
Look for .4950 in the normal distribution
table


Find the value closest to .4950


Table VII does not contain .4950
It is either .4949 or .4951
z = 2.57 or 2.58.
97
Figure 6.47
Finding the z value.
Shaded area is
given to be .0050
.4950
0
z
z
To find this z value
98
Example 6-17
Find the value of z such that the area under
the standard normal curve in the left tail is
.05.
99
Solution 6-17


Area between 0 and z = .5 - .05
= .4500
z = -1.65.
100
Figure 6.48
Finding the z value.
Shaded area is given
to be .05
.4500
z
0
z
To find this z value
101
Finding an x Value for a Normal
Distribution
For a normal curve, with known values of
μ and σ and for a given area under the
curve between the mean and x, the x
value is calculated as
x = μ + zσ
102
Example 6-18
It is known that the life of a calculator
manufactured by Texas Instruments has a
normal distribution with a mean of 54
months and a standard deviation of 8
months. What should the warranty period
be to replace a malfunctioning calculator if
the company does not want to replace
more than 1% of all the calculators sold?
103
Solution 6-18



Area between the mean and the x value
= .5 - .01 = .4900
z = -2.33
x = μ + zσ = 54 + (-2.33)(8)
= 54 – 18.64 = 35.36
104
Solution 6-18
Thus, the company should replace all
calculators that start to malfunction within
35.36 months (which can be rounded to
35 months) of the date of purchase so that
they will not have to replace more than 1%
of the calculators.
105
Figure 6.49
Finding an x value.
Shaded area is
given as.01
.5 - .01
= .4900
To find this x value
x
μ = 54
x
-2.33
0
z
From Table VII, this value
of z is approximately -2.33
106
Example 6-19
Almost all high school students who intend to go
to college take the SAT test. In 2002, the mean
SAT score (in verbal and mathematics) of all
students was 1020. Debbie is planning to take
this test soon. Suppose the SAT scores of all
students who take this test with Debbie will have
a normal distribution with a mean of 1020 and a
standard deviation of 153. What should her score
be on this test so that only 10% of all examinees
score higher than she does?
107
Solution 6-19



Area between μ and the x value
= .5 - .10 = .4000
x = μ + zσ = 1020 + 1.28(153)
= 1020 + 195.84
= 1215.84 ≈ 1216
Thus, if Debbie scores 1216 on the SAT,
only about 10% of the examinees are
expected to score higher than she does.
108
Figure 6.50
Finding an x value.
Shaded area is
given to be .10
.4000
μ = 1020
0
From normal distribution table, this
value of z is approximately 1.28
x
x
1.28
z
To find this
x value
109
THE NORMAL APPROXIMATION OF
THE BINOMIAL DISTRIBUTION
1.
2.
3.
4.
The binomial distribution is applied to a discrete
random variable.
Each repetition, called a trial, of a binomial
experiment results in one of two possible
outcomes, either a success or a failure.
The probabilities of the two outcomes remain
the same for each repetition of the experiment.
The trials are independent.
110
THE NORMAL APPROXIMATION OF
THE BINOMIAL DISTRIBUTION cont.

The binomial formula, which gives the
probability of x successes in n trials, is
P( x)  n C x p q
x
n x
111
THE NORMAL APPROXIMATION OF
THE BINOMIAL DISTRIBUTION cont.
Normal Distribution as an Approximation to
Binomial Distribution
Usually, the normal distribution is used as
an approximation to the binomial
distribution when np and nq are both
greater than 5 - that is , when
np > 5
and nq > 5
112
Table 6.5
x
0
1
2
3
4
5
6
7
8
9
10
11
12
The Binomial Probability Distribution for
n = 12 and p = .50
P(x)
.0002
.0029
.0161
.0537
.1208
.1934
.2256
.1934
.1208
.0537
.0161
.0029
.0002
113
Figure 6.51
Histogram for the probability distribution
of Table 6.5.
114
Example 6-20
According to an estimate, 50% of the
people in America have at least one credit
card. If a random sample of 30 persons is
selected, what is the probability that 19 of
them will have at least one credit card?
115
Solution 6-20




n = 30
p = .50,
q = 1 – p = .50
x = 19,
n – x = 30 – 19 = 11
From the binomial formula,
P(19)  30 C19 (.5) (.5)  .0509
19
11
116
Solution 6-20
  np  30(.50)  15
  npq  30(.50)(.50)  2.73861279
117
Continuity Correction Factor
Definition
The addition of .5 and/or subtraction of .5
from the value(s) of x when the normal
distribution is used as an approximation to
the binomial distribution, where x is the
number of successes in n trials, is called
the continuity correction factor.
118
Figure 6.52
Correction for continuity
x
119
Solution 6-20

For x = 18.5:
18.5  15
z
 1.28
2.73861279
19.5  15
z
 1.64
2.73861279

For x = 19.5:

P (18.5 ≤ x ≤ 19.5) = P (1.28 ≤ z ≤
1.64) = .4495 - .3997
= .0498
120
Solution 6-20



Thus, based on the normal approximation,
the probability that 19 persons in a sample
of 30 will have at lease on credit card is
approximately .0498.
Using the binomial formula, we obtain the
exact probability .0509.
The error due to using the normal
approximation is .0509 - .0498 = .0011.
121
Figure 6.53
Area between x = 18.5 and x = 19.5.
Shaded area is
.4495 - .3997 = .0498
15
18.5
19.5
x
z
0
1.28
1.64
122
Example 6-21
In a recent survey conducted for Money
magazine, 80% of the women (married or single)
surveyed said that they are more knowledgeable
about investing now than they were just five years
ago (Money, June 2002). Suppose this result is
true for the current population of all women. What
is the probability that in a random sample of 100
women, 72 to 76 will say that they know more
about investing now than just five years ago?
123
Solution 6-21
  np  100(.80)  80
  npq  100(.80)(.20)  4.0
124
Solution 6-21
To make the continuity correction, we
subtract .5 from 72 and add .5 to 76 to
obtain the interval 71.5 to 76.5.
125
Solution 6-21
For x = 71.5:
71.5  80
z
 2.13
4.0

For x = 76.5:
76.5  80
z
 .88
4.0

P (71.5 ≤ x ≤ 76.5) = P (-2.13 ≤ z ≤ -.88)

= .4834 - .3106
= .1728
126
Solution 6-21
Thus, the probability that 72 to 76 women
in a random sample of 100 say that they
are more knowledgeable about investing
now than they were just five years ago is
approximately .1728.
127
Figure 6.54
Area from x = 71.5 to x = 76.5.
Shaded area is
.4834 - .3106 = .1728
71.5
76.5
80
x
-2.13
-.88
0
z
128
Example 6-22
According to the 2001 Youth Risk Behavior
Surveillance by the Centers for Disease Control
and Prevention, 39% of the 10th-graders
surveyed said that they watch three or more
hours of television on a typical school day.
Assume that this percentage is true for the
current population of all 10th-graders. What is
the probability that 86 or more of the 10thgraders in a random sample of 200 watch three
or more hours of television on a typical school
day?
129
Solution 6-22
  np  200(.39)  78
  npq  200(.39)(.61)  6.89782574
130
Solution 6-22
For the continuity correction, we subtract .5
from 86, which gives 85.5.
131
Solution 6-22
85.5  78
z
 1.09
6.89782574

For x = 85.5:

P (x ≥ 85.5) = P (z ≥ 1.09)
= .5 - .3621
= .1379
132
Solution 6-22
Thus, the probability that 86 or more of the
10th-graders in a random sample of 200
watch three or more hours of television on
a typical school say is approximately .1379.
133
Figure 6.55
Area to the right of x = 85.5.
The required
probability = .5 .3621 = .1379
78
0
85.5
1.09
x
z
134