36. a. E (
p
) = .76
(1 ) .76(1 .76) .0214
pp
−−
b.
.79 .76 1.40
.0214
z−
==
P(z ≤1.40) = .9192
P(z < -1.40) = .0808
p
c.
(1 ) .76(1 .76) .0156
750
p
pp
n
−−
= = =
.79 .76 1.92
.0156
z−
==
P(z ≤ 1.92) = .9726
p
37. a. E(
p
) = .17
(1 ) (.17)(1 .17) .0133
800
p
pp
n
−−
= = =
Distribution is approximately normal because np = 800(.17) = 136 > 5
b.
.19 .17 1.51
.0133
z−
==
P(z ≤ 1.51) = .9345
P(z < -1.51) = .0655
P(.15
p
.19) = P(-1.51 z 1.51) = .9345 – .0655 = .8690
c.
(1 ) (.17)(1 .17) .0094
1600
p
pp
n
−−
= = =
.19 .17 2.13
.0094
z−
==
P(z ≤ 2.13) = .9834
38. a. Sorting the list of companies and random numbers to identify the five companies associated with the
five smallest random numbers provides the following sample.
Random
Number
.008012
.055369
.059279
.144127
.227759
39. a. With n = 100, we can approximate the sampling distribution with a normal distribution having
E(
x
) = 8086
2500 250
b.
200 .80
/ 2500 / 100
x
zn
−
= = =
P(z ≤ .80) = .7881
P(z < -.80) = .2119
P(7886
x
8286) = P(-.80 z .80) = .7881 – .2119 = .5762
c. At 9000,
9000 8086 3.66
2500 / 100
z−
==
P(
x
≥ 9000) = P(z ≥ 3.66)
0
40. a. Normal distribution with
b.
15 1.50
/ 80 / 64
x
zn
−
= = =
P(z ≤ 1.50) = .9332
P(z < -1.50) = .0668
P(391
x
421) = P(-1.50 z 1.50) = .9332 – .0668 = .8664
c. At
x
= 380,
380 406 2.60
/ 80 / 64
x
zn
−−
= = = −
P(
x
≤ 380) = P(z ≤ -2.60) = .0047
approximately normal.
a. With a mean base rate for the fare of $89 and a mean of $39 for additional charges, the population
mean total cost per flight is µ = $128.
b. This means 118
x
138
At
x
= 138,
138 128 1.94
40 / 60
z−
==
P(z ≤ 1.94) = .9738
P(z < -1.94) = .0262
x
c. This means 123
x
133
At
x
= 133,
133 128 .97
40 / 60
z−
==
P(z ≤ .97) = .8340
P(z < -.97) = .1660
x
a.
7400 / 60 955
x
==
b.
00
955
x
x
z
−
= = =
P(
x
> 27,175) = P(z > 0) = .50
Note: This could have been answered easily without any calculations; 27,175 is the expected value
1000 1.05
x
z
−
= = =
955
x
P(z < -1.05) = .1469
P(26,175
x
28,175) = P(-1.05 z 1.05) = .8531 – .1469 = .7062
Using Excel: NORM.DIST(28175,27175,955,TRUE)-
1000 1.35
740
x
x
z
−
= = =
P(z ≤ 1.35) = .9115
P(z < -1.35) = .0885
P(26,175
x
28,175) = P(-1.35 z 1.35) = .9115 – .0885 = .8230
43. a.
x
N n
Nn
=−
−1
N = 2000
x=−
−=
2000 50
2000 1
144
50 2011.
N = 5000
x=−
−=
5000 50
5000 1
144
50 2026.
=
1.9
+
2.1
2
=
2
The area below
x
= 2.1 must be 1 – .05 = .95. An area of .95 in the standard normal table shows
z = 1.645.
Thus,
2.1 2.0 1.645
/ 30
z
−
==
Solve for 
(.1) 30 .33
(1 ) .15(1 .15) .0230
240
p
pp
n
−−
= = =
b. Within ± .04 means .11 ≤
p
≤ .19
.19 .15 1.74
.0230
p
pp
z
−−
= = =
.11 .15 1.74
.0230
z−
= = −
P(.13
p
.17) = P(–.87 ≤ z ≤ .87) = .8078 – .1922 = .6156
Using Excel: NORM.DIST(.17,.15,.0230,TRUE)-NORM.DIST(.13,.15,.0230,TRUE) = .6155
47.
(1 ) (.40)(.60) .0245
400
p
pp
n
−
= = =
p
.375 .40 1.02
.0245
z−
= = −
P(z < -1.02) = .1539
P (
p
.375) = 1 – .1539 = .8461
Using Excel: 1-NORM.DIST(.375,.40,0245,TRUE) = .8462
p
.44 .40 1.59
.0251
z−
==
.36 .40 1.59
.0251
z−
= = −
P(.36
p
.44) = P(–1.59 ≤ z ≤ 1.59) = .9441 – .0559 = .8882
Using Excel: NORM.DIST(.44,.40,.0251,TRUE)-NORM.DIST(.36,.40,.0251,TRUE) = .8900
p
P(
p
.45) = P(z
1.99) = 1 – .9767 = .0233
Using Excel: 1-NORM.DIST(.45,.28,.0251,TRUE) = .0232
49. a. Normal distribution with E (
p
) = .15 and
(1 ) (.15)(.85) .0292
150
p
pp
n
−
= = =
.18 .15 1.03
.0292
z−
==
P(z ≤ 1.03) = .8485
P(.12
p
.18) = P(-1.03 z 1.03) = .8485 – .1515 = .6970
Using Excel: NORM.DIST(.18,.15,0292,TRUE)-NORM.DIST(.12,.15,.0292,TRUE) = .6958
50. a.
p
p p
n n
=−= =
( ) . (. ).
125 75 0625
(. )
0625 48
2
b. Normal distribution with E(
p
) = .25 and
p
= .0625
(Note: (48)(.25) = 12 > 5, and (48)(.75) = 36 > 5)
c. P (
p
.30) = ?
.30 .25 .80
.0625
z−
==
P(z ≤ .80) = .7881