1. CONTINUOUS RANDOM VARIABLES AND THE NORMAL DISTRIBUTION
Chapter 6:
CONTINUOUS RANDOM VARIABLES AND THE NORMAL DISTRIBUTION

2. CONTINUOUS PROBABILITY DISTRIBUTION
Table 6.1 Frequency and Relative Frequency Distributions of Heights of Female Students
Height of a Female Student (inches) x f
Relative Frequency
60 to less than 61
61 to less than 62
62 to less than 63
63 to less than 64
64 to less than 65
65 to less than 66
66 to less than 67
67 to less than 68
68 to less than 69
69 to less than 70
70 to less than 71 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

3. Figure 6.1 Histogram and polygon for Table 6.1.

4. Figure 6.2 Probability distribution curve for heights.

5. CONTINUOUS PROBABILITY DISTRIBUTION
Two characteristics
The probability that x assumes a value in any interval lies in the range 0 to 1
The total probability of all the (mutually exclusive) intervals within which x can assume a value of 1.0

6. Figure 6.3 Area under a curve between two points.
Shaded area is between 0 and 1
x = a
x = b x

7. Figure 6.4 Total area under a probability distribution curve.
Shaded area is 1.0 or 100% x

8. Figure 6.5 Area under the curve as probability.
Shaded area gives the probability P (a ≤ x ≤ b) a b x

9. Figure 6.6 Probability that x lies in the interval 65 to 68 inches.

10. Figure 6.7 The probability of a single value of x is zero.

11. Figure 6.8 Probability “from 65 to 68” and “between 65 and 68”.

12. 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.
The curve is symmetric about the mean.
The two tails of the curve extend indefinitely.

13. Figure 6.11 Normal distribution with mean μ and standard deviation σ.x

14. Figure 6.12 Total area under a normal curve.
The shaded area is 1.0 or 100% μ x

15. 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

16. 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

17. Figure 6. 15 Three normal distribution curves with the
Figure 6.15 Three normal distribution curves with the same mean but different standard deviations.
σ = 5
σ = 10
σ = 16 x
μ = 50

18. Figure 6. 16 Three normal distribution curves with
Figure 6.16 Three normal distribution curves with different means but the same standard deviation.
σ = σ = σ = 5
µ = µ = µ = x

19. THE STANDARD NORMAL DISTRIBTUION
Definition
The normal distribution with μ = 0 and σ = 1 is called the standard normal distribution.

20. Figure 6.17 The standard normal distribution curve.
σ = 1
µ = 0 z

21. 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.

22. Figure 6.18 Area under the standard normal curve.
Each of these two areas is .5
. 5
. 5 z

23. Example 6-1
Find the area under the standard normal curve between z = 0 and z = 1.95.

24. Table 6.2 Area Under the Standard Normal Curve from z = 0 to z = 1.95
.00
.01 …
.05
.09
0.0
0.1
0.2 .
1.9
3.0
.0000
.0398
.0793
.4713
.4987
.0040
.0438
.0832
.4719
.0199
.0596
.0987
.4744
.4989
.0359
.0753
.1141
.4767
.4990
Required area

25. Figure 6.19 Area between z = 0 to z = 1.95.
Shaded area is .4744
1.95 z

26. Example 6-2
Find the area under the standard normal curve from z = to z = 0.

27. Solution 6-2
Because the normal distribution is symmetric about the mean, the area from z = to z = 0 is the same as the area from z = 0 to z = 2.17.
Area from to 0 = P(-2.17≤ z ≤ 0) = .4850

28. Because the symmetry the shaded areas are equal
Figure 6.20 Area from z = 0 to z = 2.17 equals area from z = to z = 0.
Because the symmetry the shaded areas are equal z z

29. Table 6.3 Area Under the Standard Normal Curve from z = 0 to z = 2.17
.00
.01 …
.07
.09
0.0
0.1
0.2 .
2.1
3.0
.0000
.0398
.0793
.4821
.4987
.0040
.0438
.0832
.4826
.0279
.0675
.1064
.4850
.4989
.0359
.0753
.1141
.4857
.4990
Required area

30. Figure 6.21 Area from z = -2.17 to z = 0.
Shaded area is .4850 z

31. Example 6-3 Find the following areas under the standard normal curve.
Area to the right of z = 2.32
Area to the left of z = -1.54

32. Solution 6-3
Area to the right of = P (z > 2.32) = = as shown in Figure 6.22

33. Figure 6.22 Area to the right of z = 2.32.
This area is from Table VII
Shaded area is
.4898
2.32 z

34. Solution 6-3
Area to the left of = P (z < -1.54) = = as shown in Figure 6.23

35. Figure 6.23 Area to the left of z = -1.54.
This area is from Table VII
Shaded area is =
.4382 z
-1.54

36. Example 6-4
Find the following probabilities for the standard normal curve.
P (1.19 < z < 2.12)
P (-1.56 < z < 2.31)
P (z > -.75)

37. Solution 6-4
P (1.19 < z < 2.12) = Area between 1.19 and = = as shown in Figure 6.24

38. Figure 6.24 Finding P (1.19 < z < 2.12).
Shaded area = = .1000 z
1.19
2.12

39. Solution 6-4
P (-1.56 < z < 2.31) = Area between and = = as shown in Figure 6.25

40. Figure 6.25 Finding P (-1.56 < z < 2.31).
Total shaded area = = .9302
.4406
.4896 z
-1.56
2.31

41. Solution 6-4
P (z > -.75) = Area to the right of = = as shown in Figure 6.26

42. Figure 6.26 Finding P (z > -.75).
Total shaded area = = .7734
.2734 .5
-.75 z

43. Figure 6.27 Area within one standard deviation of the mean.
Total shaded area is = or 68.26%
.3413
.3413
-1.0
1.0 z

44. Figure 6.28 Area within two standard deviations of the mean.
Total shaded area is = or 95.44%
.4772
.4772
-2.0
2.0 z

45. Figure 6.29 Area within three standard deviations of the mean.
Total shaded area is = or 99.74%
.4987
.4987 z
-3.0
3.0

46. Example 6-5
Find the following probabilities for the standard normal curve.
P (0 < z < 5.67)
P (z < -5.35)

47. Solution 6-5
P (0 < z < 5.67) = Area between 0 and = .5 approximately as shown in Figure 6.30

48. Figure 6.30 Area between z = 0 and z = 5.67.
Shaded area is approximately .5 z
5.67

49. Solution 6-5
P (z < -5.35) = Area to the left of = = .00 approximately as shown in Figure 6.31

50. Figure 6.31 Area to the left of z = -5.35.
This area is approximately .5
Shaded area is approximately .00
-5.35 z

51. 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
where μ and σ are the mean and standard deviation of the normal distribution of x, respectively.

52. 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.
x = 55
x = 35

53. Solution 6-6
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54. Figure 6.32 z value for x = 55. x μ = 50 x = 55 z z value for x = 55
Normal distribution with μ = 50 and σ = 10 x
μ = 50
x = 55
Standard normal distribution
.50 z
z value for x = 55

55. Solution 6-6

56. Figure 6.33 z value for x = 35.
σ = 10
μ = 50 x 35 z
-1.50

57. 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
between x = 25 and x = 32
between x = 18 and x = 34

58. Solution 6-7 The z value for x = 25 is 0 The z value for x = 32 is
P (25 < x < 32) = P(0 < z < 1.75) = .4599

59. Figure 6.34 Area between x = 25 and x = 32.
.4599 25 32 x
These areas are equal
.4599 z
1.75

60. Solution 6-7 For x = 18: For x = 34:
P (18 < x < 34) = P (-1.75 < z < 2.25 ) = = .9477

61. Figure 6.35 Area between x = 18 and x = 34.
Shaded area = = .9477
.4599
.4878 x z

62. 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
P (x > 55)
P (x < 49)

63. Solution 6-8
For x = 55:
P (x > 55) = P (z > 3.00) = = .0013

64. Figure 6.36 Finding P (x > 55).
Shaded area = = .0013
.4987 x z 40 55
3.00

65. Solution 6-8
For x = 49:
P (x < 49) = P (z < 1.80) = = .9641

66. Figure 6.37 Finding P (x > 49).
Shaded area = = .9641 .5
.4641 x z 49 40
1.80

67. 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).

68. Solution 6-9 For x = 30: For x = 39:
P (30 ≤ x ≤ 39) = P (-2.50 ≤ z ≤ -1.38) = = .0776

69. Figure 6.38 Finding P (30 ≤ x ≤ 39). x z
Shaded area = = .0776 x 30 39 50 z
-2.50
-1.38

70. 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
from x = 70 to x = 135
to the left of 27

71. Solution 6-10 For x = 70: For x = 135:
P (70 ≤ x ≤ 135) = P (-.83 ≤ z ≤ 4.58) = = approximately

72. Figure 6.39 Area between x = 70 and x = 135.
Shaded area is approximately = .7967 x z

73. Solution 6-10
For x = 27:
P (x < 27) = P (z < -4.42) = = .00 approximately

74. Figure 6.40 Area to the left of x = 27.
Shaded area is approximately = .00 x z

75. 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).

76. 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.

77. Solution 6-11 For x = $46,000: For x = $49,000:
P ($46,000 < x < $49,000) = P (-1.33 < z < .55) = = = 61.70%

78. Solution 6-11
Thus, the probability is that a randomly selected three-year-old Porsche 911 sports car will sell for a price between $46,000 and $49,000.

79. Figure 6.41 Area between x = $46,000 and x = $49,000.
Shaded area is = .6170
.2088
.4082
$46,000
$48,125
$49,000 x z
-1.33
.55

80. 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?

81. Solution 6-12 μ = 55 minutes σ = 4 minutes For x = 60:
P (x ≤ 60) = P (z ≤ 1.25) = = .8944

82. Solution 6-12
Thus, the probability is that this worker will finish assembling this racing car before the company closes for the day.

83. Figure 6.42 Area to the left of x = 60.
Shaded area is = .8944 .5
.3944 x z

84. 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.

85. Example 6-13
What is the probability that a randomly selected can of Orange Cola contains to ounces of soda?
What percentage of the Orange Cola cans contain to ounces of soda?

86. Solution 6-13 For x = 11.97: For x = 11.99:
P (11.97 ≤ x ≤ 11.99) = P (-2.00 ≤ z ≤ -.67) = = .2286

87. Figure 6.43 Area between x = 11.97 and x = 11.99.
Shaded area = = .2286 x z

88. Solution 6-13 For x = 12.02: For x = 12.07:
P (12.02 ≤ x ≤ 12.07) = P (1.33 ≤ z ≤ 4.67) = = .0918

89. Figure 6.44 Area from x = 12.02 to x = 12.07.
Shaded area is = .0918
μ = x z

90. 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?

91. Solution 6-14
For x = 36:
P (x < 36) = P (z < -2.25) = = .0122
Hence, 1.22% of the calculators are expected to be replaced.

92. Figure 6.45 Area to the left of x = 36.
Shaded area is = .0122
.4878
μ = x z

93. 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 and the value of z is positive.

94. Figure 6.46 Finding the z value.
Shaded area is given to be .4251 z z
To find this z

95. Table 6.4 Finding the z Value When Area Is Known.
Solution 6-15 z
.00
.01 …
.04
.09
0.0
0.1
0.2 .
1.4
3.0
.0000
.0398
.0793
.4987
.0040
.0438
.0832
.4251
.4988
.0359
.0753
.1141
.4990
We locate this value in Table VII of Appendix C
Therefore, z = 1.44

96. Example 6-16
Find the value of z such that the area under the standard normal curve in the right tail is

97. Solution 6-16 Area between 0 and z = .5 - .005 = .4950
Look for in the normal distribution table
Table VII does not contain .4950
Find the value closest to .4950
It is either or .4951
z = 2.57 or 2.58.

98. Figure 6.47 Finding the z value.
Shaded area is given to be .0050
.4950
z z
To find this z value

99. Example 6-17
Find the value of z such that the area under the standard normal curve in the left tail is .05.

100. Solution 6-17
Area between 0 and z = = .4500
z =

101. Figure 6.48 Finding the z value.
Shaded area is given to be .05
.4500
z z
To find this z value

102. 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

103. 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?

104. 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 – = 35.36

105. Solution 6-18
Thus, the company should replace all calculators that start to malfunction within 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.

106. Figure 6.49 Finding an x value.
Shaded area is given as.01
= .4900
To find this x value x
μ = 54 x z
-2.33
From Table VII, this value of z is approximately -2.33

107. 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 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?

108. Solution 6-19 Area between μ and the x value = .5 - .10 = .4000
x = μ + zσ = (153) = = ≈ 1216
Thus, if Debbie scores 1216 on the SAT, only about 10% of the examinees are expected to score higher than she does.

109. Figure 6.50 Finding an x value.
Shaded area is given to be .10
.4000
μ = 1020 x x z
1.28
From normal distribution table, this value of z is approximately 1.28
To find this x value

110. THE NORMAL APPROXIMATION OF THE BINOMIAL DISTRIBUTION
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.

111. THE NORMAL APPROXIMATION OF THE BINOMIAL DISTRIBUTION cont.
The binomial formula, which gives the probability of x successes in n trials, is

112. 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 > and nq > 5

113. Table 6.5 The Binomial Probability Distribution for n = 12 and p = .50x
P(x) 1 2 3 4 5 6 7 8 9 10 11 12
.0002
.0029
.0161
.0537
.1208
.1934
.2256

114. Figure 6.51 Histogram for the probability distribution of Table 6.5.

115. 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?

116. Solution 6-20
n = 30
p = .50, q = 1 – p = .50
x = 19, n – x = 30 – 19 = 11
From the binomial formula,

117. Solution 6-20

118. 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.

119. Figure 6.52
Correction for continuity x

120. Solution 6-20 For x = 18.5: For x = 19.5:
P (18.5 ≤ x ≤ 19.5) = P (1.28 ≤ z ≤ 1.64) = = .0498

121. 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
Using the binomial formula, we obtain the exact probability
The error due to using the normal approximation is =

122. Figure 6.53 Area between x = 18.5 and x = 19.5.
Shaded area is = .0498 x
18.5
19.5 z
1.28
1.64

123. 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?

124. Solution 6-21

125. 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.

126. Solution 6-21 For x = 71.5: For x = 76.5:
P (71.5 ≤ x ≤ 76.5) = P (-2.13 ≤ z ≤ -.88) = = .1728

127. 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

128. Figure 6.54 Area from x = 71.5 to x = 76.5.
Shaded area is = .1728 x
71.5
76.5 80 z
-2.13
-.88

129. 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 10th-graders in a random sample of 200 watch three or more hours of television on a typical school day?

130. Solution 6-22

131. Solution 6-22
For the continuity correction, we subtract .5 from 86, which gives 85.5.

132. Solution 6-22
For x = 85.5:
P (x ≥ 85.5) = P (z ≥ 1.09) = = .1379

133. 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

134. Figure 6.55 Area to the right of x = 85.5.
The required probability = = .1379 78
85.5 x z
1.09