CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 193
24. 100
30 : 800, 18.257
30
xx
n
n
σ
µσ== ==≈
25. 60,000 63,500 3500 3.39
6100 1031.09
35
x
z
n
µ
σ
−−−
== ≈ ≈−
( 60,000) ( 3.39) 0.0003Px Pz<=<−=
Only 0.03% of samples of 35 specialists will have a mean salary less than $60,000. This is an
extremely unusual event.
27. 2.695 2.714 0.019 2.39
0.045 0.00795
32
x
z
n
µ
σ
−−−
== ≈ ≈−
2.725 2.714 0.011 1.38
0.045 0.00795
x
z
µ
σ
−−
== ≈ ≈
194 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
29. 66 64.3 1.7 5.06
2.6 0.336
60
x
z
n
µ
σ
−−
== ≈≈
( 66) ( 5.06) 0Px Pz>= > ≈
There is almost no chance that a random sample of 60 women will have a mean height greater
than 66 inches. This event is almost impossible.
31. 70 64.3 2.19
2.6
x
zµ
σ
−−
== ≈
( 70) ( 2.19) 0.9857Px Pz<= < =
70 64.3 5.7 9.81
2.6 0.581
20
x
z
n
µ
σ
−−
== ≈≈
(70) (9.81)1Px Pz<= < ≈
It is more likely to select a sample of 20 women with a mean height less than 70 inches because
the sample of 20 has a higher probability.
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 195
33. 127.9 128 0.1 3.16
0.20 0.0316
40
x
z
n
µ
σ
−−−
== ≈ ≈−
( 127.9) ( 3.16) 0.0008Px Pz<=<−≈
Yes, it is very unlikely that you would have randomly sampled 40 cans with a mean equal to
127.9 ounces because it is more than 2 standard deviations from the mean of the sample means.
35. (a) 96
0.5
96.25 96 0.25 3.16
0.5 0.079
40
x
z
n
µ
σ
µ
σ
=
=
−−
== ≈≈
36. (a) 10
0.5
10.21 10 0.21 2.1
0.5 0.1
25
( 10.21) ( 2.1) 1 ( 2.1) 1 0.9821 0.0179
x
z
n
Px Pz Pz
µ
σ
µ
σ
=
=
−−
== ==
≥=≥=−≤=− =
196 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
37. (a) 50,000
800
49,721 50,000 279 3.49
x
µ
σ
µ
=
=
−−−
38. (a) 38,000
1000
37,650 38,000 350 2.47
1000 141.42
50
( 37,650) ( 2.47) 0.0068
x
z
n
Px Pz
µ
σ
µ
σ
=
=
−−−
== ≈ ≈−
≤=≤−=
39.
501
112
515 501 14 0.88
112 15.84
50
x
z
n
µ
σ
µ
σ
=
=
−−
== ≈≈
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 197
40.
()
4
0.5
4.2 4 0.2 4
0.5 0.05
100
(4.2) (4)1 4110
x
z
n
Px Pz Pz
µ
σ
µ
σ
=
=
−−
== ==
≥=≥=−≤≈−≈
It is very unlikely the machine is calibrated to produce a bolt with a mean of 4 inches.
42. Use the finite correction factor since 30 25 0.05 .nN=>=
2.5 3.32 0.82 4.25
1.09 500 30 (0.199) 0.9419
1 500 1
30
x
z
Nn
N
n
µ
σ
−− −
== ≈ ≈−
−−
−−
43.
Sample Number of boys
from 3 births
Proportion of boys
from 3 births
bbb 3 1
bbg 2 2
3
bgb 2 2
198 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
44.
Proportion of boys
from 3 births
Probability
0 1
8
1
3 3
8
45.
Sample Numerical
representation
Sample mean
bbb 111 1
bbg 110 2
3
bgb 101 2
3
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 199
46.
Sample Number
of boys
from 4
births
Proportion
of boys
from 4
births
Sample Number
of boys
from 4
births
Proportion
of boys
from 4
births
Proportion
of boys
from 4
births
Probability
bbbb 4 1 gbbb 3 3
4 0 1
16
bbbg 3 3
4 gbbg 2 1
2 1
4 1
4
47.
l
0.70 0.77 0.07 1.70
0.0411
0.77(0.23)
pp
zpq
−−−
== = =−
200 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
5.5 NORMAL APPROXIMATIONS TO BINOMIAL DISTRIBUTIONS
5.5 Try It Yourself Solutions
1a. n = 125, p = 0.05, q = 0.95
b. np = 6.25, nq = 118.75
c. Because 5np ≥ and 5nq ≥, the normal distribution can be used.
d.
()( )
()()()
125 0.05 6.25
125 0.05 0.95 2.44
np
npq
µ
σ
== =
== ≈
b. 6.25
2.44
np
npq
µ
σ
==
=≈
c. x > 9.5
d. 9.5 6.25 1.33
2.44
x
zµ
σ
−−
== ≈
e. ( 1.33) 0.9082
( 9.5) ( 1.33) 1 ( 1.33) 0.0918
Pz
Px Pz Pz
<=
>=> =−< =
The probability that more than 9 respond yes is 0.0918.
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 201
b. 36
5.23
np
npq
µ
σ
==
=≈
c.
()
26.5 27.5Px<<
5.5 EXERCISE SOLUTIONS
1. Properties of a binomial experiment:
(1) The experiment is repeated for a fixed number of independent trials.
(2) There are two possible outcomes: success or failure.
(3) The probability of success is the same for each trial.
(4) The random variable x counts the number of successful trials.
2. 5np ≥ and 5nq ≥
202 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
7. 10, 0.85, 0.15
8.55, 1.55
np q
np nq
== =
=≥ =<
Cannot use normal distribution because nq < 5.
8. 20, 0.63, 0.37
12.6 5, 7.4 5
np q
np nq
== =
=≥ =≥
9. 50, 0.55, 0.45
27.5 5, 22.5 5
np q
np nq
== =
=≥ =≥
10. 30, 0.19, 0.81
5.7 5, 24.3 5
np q
np nq
== =
=≥ = ≥
11. 20, 0.76, 0.24
np q
== =
12. 15, 0.61, 0.39
9.15 5, 5.85 5
np q
np nq
== =
=≥ =≥
13. a 14. d 15. c 16. b
17. The probability of getting fewer than 25 successes; (24.5)Px<
18. The probability of getting at least 110 successes; ( 109.5)Px>
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 203
22. The probability of getting between 55 and 60 successes; (55.5 59.5)Px<<
23. 100, 0.93
93 5, 7 5
np
np nq
==
=≥ =≥
Can use normal distribution.
0.9147
=
c. ( 90) ( 89.5)Px Px<= <
(1.37)
0.0853
Pz=<−
=
204 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
24. 80, 0.70
56 5, 24 5
np
np nq
==
=≥ =≥
Can use normal distribution.
0.0005
=
b. 49.5 56 1.59
4.10
x
zµ
σ
−−
=≈ ≈−
50.5 56 1.34
x
zµ
−−
=≈ ≈−
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 205
25. 150, 0.35
52.5 5, 97.5 5
np
np nq
==
=≥ =≥
Can use normal distribution.
a. 75.5 52.5 3.94
5.84
x
zµ
σ
−−
=≈ ≈
b. 40.5 52.5 2.05
5.84
x
zµ
σ
−−
=≈ ≈−
( 40) ( 40.5)Px Px>≈ >
(2.05)
1 ( 2.05)
1 0.0202
0.9798
Pz
Pz
=>−
=− <−
=−
=
d. No, none of the probabilities are less than 0.05.
206 CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS
26. 50, 0.34, 0.66
17 5, 33 5
np q
np nq
== =
=≥ =≥
0.0396
=
b. (23) (23.5)Px Px>≈ ≥
( 1.94)
1(1.94)
1 0.9738
Pz
Pz
=≥
=− ≤
=−
CHAPTER 5 │ NORMAL PROBABILITY DISTRIBUTIONS 207
d. 125, 0.34, 0.66
42.5 5, 82.5 5
np q
np nq
== =
=≥ =≥
27. 250, 0.05, 0.95
12.5 5, 237.5 5
np q
np nq
== =
=≥ = ≥
Can use normal distribution.
b. ( 9) ( 8.5) ( 1.16) 1 ( 1.16) 1 0.1230 0.8770Px Px Pz Pz≥≈≥=≥−=−≤−=− =