CHAPTER
Normal Probability Distributions
173
5
5.1 INTRODUCTION TO NORMAL DISTRIBUTIONS AND
THE STANDARD NORMAL DISTRIBUTION
5.1 Try It Yourself Solutions
1a. A: 45x=, B: 60x=, C: 45x= (B has the greatest mean.)
b. 0.9834 b. 0.0154
5.1 EXERCISE SOLUTIONS
2. Neither. In a normal distribution, the mean and median are equal.
3. 1
4. Points at which the curve changes from curving upward to curving downward; µσ− and µσ+
174 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
7. 0µ=, 1σ=
8. Transform each data value x into a z-score by subtracting the mean from x and dividing the result
by the standard deviation. In symbols,
x
zµ
σ
−
=.
11. No, the graph crosses the x-axis.
12. No, the graph is not symmetric.
13. Yes, the graph fulfills the properties of the normal distribution.
17. The histogram represents data from a normal distribution because it is bell-shaped.
18. The histogram does not represent data from a normal distribution because it is skewed right.
22. (Area right of z = 2.3−) = 1 – (Area left of z = 2.3−)
= 1 – 0.0107
= 0.9893
23. (Area left of z = 0) – (Area left of z = 2.25−) = 0.5 – 0.0122 = 0.4878
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 175
31. 1 – 0.3613 = 0.6387 32. 0.9463 0.9474
10.0532
2
+
−≈
33. 0.9979 – 0.5 = 0.4979
39. (a)
It is reasonable to assume that the life spans are normally distributed because the histogram is
symmetric and bell-shaped.
(b) 37,234.7
6259.2
x
s
=
=
40. (a)
It is reasonable to assume that the weekly milk consumptions are normally distributed because
the histogram is nearly symmetric and bell-shaped.
41. (a) A = 105; B = 113; C = 121; D = 127
176 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
(b) 105 115
105 2.78
3.6
x
xz
µ
σ
−−
=⇒= = ≈−
42. (a) A = 11.92; B = 11.99; C = 12.01; D = 12.12
(b) 11.92 12
11.92 1.6
0.05
x
xz
µ
σ
−−
=⇒= = =−
11.99 12
11.99 0.2
0.05
x
xz
µ
σ
−−
=⇒= = =−
43. (a) A = 1241; B = 1392; C = 1924; D = 2202
(b) 1241 1509
1241 0.86
312
x
xz
µ
σ
−−
=⇒= = ≈−
1392 1509
1392 0.375
312
x
xz
µ
σ
−−
=⇒= = =−
44. (a) A = 9; B = 15; C = 22; D = 35
(b) 9 21.1
92.42
5.0
x
xz
µ
σ
−−
=⇒= = =−
15 21.1
15 1.22
5.0
x
xz
µ
σ
−−
=⇒= = =−
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 177
(c) x = 9 is unusual because its corresponding z-score (2.42)− lies more than 2 standard
deviations from the mean. x = 35 is unusual because its corresponding z–score (2.78) lies
more than 2 standard deviations from the mean.
45. 0.9750 46. 0.1894 0.1922 0.1908
2
+= 47. 1 – 0.0225 = 0.9775
48. 1 – 0.8997 = 0.1003 49. 0.9987 – 0.1587 = 0.8400 50. 0.9382 – 0.5 = 0.4382
55. P(0.89 0z−<<) = 0.5 – 0.1867 = 0.3133
56. (0 0.525)Pz<< = 0.7002 – 0.5 = 0.2002
57. ( 1.65 1.65)Pz−<< = 0.9505 – 0.0495 = 0.901
61.
The normal distribution curve is centered at its mean (60) and has 2 points of inflection (48 and
72) representing µσ±.
178 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
63. (a) Area under curve = area of square = (base)(height) = (1)(1) = 1
64. (a)
Area under curve = area of rectangle
5.2 NORMAL DISTRIBUTIONS: FINDING PROBABILITIES
5.2 Try It Yourself Solutions
1a.
b. 70 67 0.86
3.5
x
zµ
σ
−−
== ≈
2a.
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 179
b. 33 45 1
12
x
zµ
σ
−−
== =−
60 45 1.25
x
zµ
−−
== =
3a. Read user’s guide for the technology tool.
b. Enter the data.
P(100 < x < 150) = P(0.97− < z < 0.46) = 0.5105
5.2 EXERCISE SOLUTIONS
1. ( 170) ( 0.2) 0.4207Px Pz<=<−=
5. (160 170) ( 0.7 0.2) 0.4207 0.2420 0.1787Px P z<< = − <<− = − =
6. (172 192) ( 0.1 0.9) 0.8159 0.4602 0.3557Px P z<< = − << = − =
7. (200 450) ( 2.64 0.39) 0.3483 0.0041 0.3442Px P z<< = − <<− = − =
8. (670 800) (1.34 2.46) 0.9931 0.9099 0.0832Px P z<< = << = − =
12. (116 125) ( 1.94 0) 0.5 0.0262 0.4738Px P z<< = − << = − =
180 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
15. (a) ( 15) ( 0.89) 0.1867Px Pz<= <− =
(b) (18 25) ( 0.41 0.70) 0.7580 0.3409 0.4171Px P z<< = − << = − =
(c) ( 34) ( 2.13) 1 ( 2.13) 1 0.9834 0.0166Px Pz Pz>= > =− < =− =
(d) Yes, the event in part (c) is unusual because its probability is less than 0.05.
17. (a) ( 5) ( 2) 0.0228Px Pz<= <−=
(b) (5.5 9.5) ( 1.5 2.5) 0.9938 0.0668 0.927Px P z<< = − << = − =
(c) ( 10) ( 3) 1 ( 3) 1 0.9987 0.0013Px Pz Pz>= >=− <=− =
19. (a) ( 4) ( 2.44) 0.0073Px Pz<= <− =
(b) (5 7) ( 1.33 0.89) 0.8133 0.0918 0.7215Px P z<< = − << = − =
(c) ( 8) ( 2) 1 0.9722 0.0228Px Pz>= >=− =
21. (a) ( 600) ( 0.96) 0.8315 83.15%Px Pz<=< = ⇒
(b) ( 550) ( 0.51) 1 ( 0.51) 1 0.6950 0.3050Px Pz Pz>=> =−< =− =
(1000)(0.3050) 305 305=⇒ scores
22. (a) ( 500) ( 0.13) 0.4483 44.83%Px Pz<=<−= ⇒
(b) ( 600) ( 0.73) 1 ( 0.73) 1 0.7673 0.2327Px Pz Pz>=> =−< =− =
(1500)(0.2327) 349.05 349=⇒ scores
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 181
26. (a) ( 125) ( 2.08) 1 ( 2.08) 1 0.9812 0.0188 1.88%Px Pz Pz>=> =−< =− = ⇒
(b) ( 90) ( 0.83) 0.2033Px Pz<= <− =
(300)(0.2033) 60.99 61=⇒ bills
27. ( 2065) ( 2.17) 1 ( 2.17) 1 0.9850 0.0150 1.5%Px Pz Pz>=>=−<=− = ⇒
It is unusual for a battery to have a life span that is more than 2065 hours because the probability
is less than 0.05.
29. (a) 0.3085
(b) 0.1499
(c) 0.0668
No, because 0.0668 > 0.05, this event is not unusual.
31. Out of control, because the 10th observation plotted beyond 3 standard deviations.
32. Out of control, because two out of three consecutive points lie more than 2 standard deviations
from the mean. (8th and 10th observations)
182 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
5.3 NORMAL DISTRIBUTIONS: FINDING VALUES
5.3 Try It Yourself Solutions
1ab. (1) (2)
c. (1) 1.77z=− (2) 1.96z=±
3a. 52, 15µσ==
b.
()()
2.33 52 2.33 15 17.05zxzµσ=− ⇒ = + = + − =
()()
3.10 52 3.10 15 98.50zxzµσ=⇒=+=+ =
()()
0.58 52 0.58 15 60.70zxzµσ=⇒=+=+ =
c. 17.05 pounds is below the mean, 60.7 pounds and 98.5 pounds are above the mean.
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 183
5.3 EXERCISE SOLUTIONS
1. 0.81z=− 2. 0.16z=− 3. 2.39z=
13. 0.67z=− 14. 0.25z=− 15. 0.67z=
16. 0.84z= 17. 0.38z=− 18. 0.25z=
19. 0.58z=− 20. 1.99z= 21. 1.645z=±
184 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
25. 1.18z⇒=
28. 2.575z⇒=±
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 185
33. (a) 5th percentile ⇒ Area = 0.05 ⇒ 1.645z=−
()()
204 1.645 25.7 161.72xzµσ=+ = +− ≈ days
(b) 3rd quartile ⇒ Area = 0.75 ⇒ 0.67z=
()()
204 0.67 25.7 221.22xzµσ=+ = + ≈ days
34. (a) 80th percentile ⇒ Area = 0.80 ⇒ 0.84z=
()( )
1674 0.84 212.5 1852.5xzµσ=+ = + = days
(b) 1st quartile ⇒ Area = 0.25 ⇒ 0.67z=−
()()
1674 0.67 212.5 1531.63xzµσ=+ = +− ≈ days
37. Upper 4.5% ⇒ Area = 0.955 ⇒ 1.70z=
()()
32 1.70 0.36 32.61xzµσ=+ = + ≈ ounces
38. Top 1% ⇒ Area = 0.99 ⇒ 2.33z=
()()
8 2.33 0.03 7.93xzµσ µ µ=+ ⇒=+ ⇒≈ ounces
186 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
41. Bottom 10% ⇒ Area = 0.10 ⇒ 1.28z=−
42. A: Top 10% ⇒ Area = 0.90 ⇒ 1.28z=
()()
72 1.28 9 83.52xzµσ=+ = + =
5.4 SAMPLING DISTRIBUTIONS AND
THE CENTRAL LIMIT THEOREM
5.4 Try It Yourself Solutions
1a.
Sample Mean Sample Mean Sample Mean Sample Mean
1, 1, 1 1 3, 1, 1 1.67 5, 1, 1 2.33 7, 1, 1 3
1, 1, 3 1.67 3, 1, 3 2.33 5, 1, 3 3 7, 1, 3 3.67
1, 1, 5 2.33 3, 1, 5 3 5, 1, 5 3.67 7, 1, 5 4.33
1, 1, 7 3 3, 1, 7 3.67 5, 1, 7 4.33 7, 1, 7 5
1, 3, 1 1.67 3, 3, 1 2.33 5, 3, 1 3 7, 3, 1 3.67
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 187
b.
x
f Probability
1 1 0.0156
1.67 3 0.0469
2
4, 1.667, 1.291
xx x
µσ σ=≈ ≈
c.
2
2
4,
51.667,
3
x
x
n
µµ
σ
σ
==
==≈
2a. 11
63, 1.4
64
xx
n
σ
µµ σ== = = ≈
b. 64n=
c. With a smaller sample size, the mean stays the same but the standard deviation increases.
188 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
4a. 1.5
25, 0.15
100
xx
n
σ
µµ σ== = = =
c. ( 2) 0.0228
( 3.33) 0.9996
(24.7 25.5) ( 2 3.33) 0.9996 0.0228 0.9768
Pz
Pz
Px Pz
<− =
<=
<< = −<< = − =
d. Of the samples of 100 drivers ages 15 to 19, 97.68% will have a mean driving time that is
between 24.7 and 25.5 minutes.
c. ( 265,000) ( 2.46) 1 ( 2.46) 1 0.0069 0.9931Px Pz Pz>=>−=−<−=−=
d. 99.31% of samples of 12 single-family houses will have a mean sales price greater than $265,000.
6a. 200 190
200 : 0.21
48
x
xz
µ
σ
−−
=== ≈
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 189
5.4 EXERCISE SOLUTIONS
1. 150
25 3.536
50
x
xn
µµ
σ
σ
==
== ≈
2. 150
25 2.5
100
x
xn
µµ
σ
σ
==
== =
5. False. As the size of a sample increases, the mean of the distribution of sample means does not
change.
6. False. As the size of a sample increases, the standard deviation of the distribution of sample
means decreases.
7. False. A sampling distribution is normal if either 30n≥ or the population is normal.
190 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
11.
Sample Mean Sample Mean Sample Mean
2, 2, 2 2 4, 4, 8 5.33 8, 16, 2 8.67
2, 2, 4 2.67 4, 4, 16 8 8, 16, 4 9.33
2, 2, 8 4 4, 8, 2 4.67 8, 16, 8 10.67
5.36
7.5, 3.09
3
xx
µσ
=≈≈
The means are equal but the standard deviation of the sampling distribution is smaller.
12.
Sample Mean
130, 130 130
130, 200 165
130, 230 180
130, 270 200
200, 130 165
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 191
207.5, 51.17
51.17
207.5, 36.18
2
xx
µσ
µσ
=≈
=≈≈
The means are equal but the standard deviation of the sampling distribution is smaller.
14. 24.3 24 0.3 2.40
1.25 0.125
100
x
z
n
µ
σ
−−
== ==
( 24.3) ( 2.40) 1 ( 2.40) 1 0.9918 0.0082Px Pz Pz>=>=−<=− =
The probability is unusual because it is less than 0.05.
16. 12,753 12,750 3 10.60
1.7 0.283
36
x
z
n
µ
σ
−−
== ≈=
( 12,750 or 12,753) ( 0 or 10.60) 0.5 0.000 0.5Px x Pz z<>=<>=+=
The probability is not unusual because it is greater than 0.05.
192 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
21. 188.4
54.5 10.9
25
x
xn
µ
σ
σ
=
== =
22. 24.2
8.1 1.479
30
x
xn
µ
σ
σ
=
== ≈