# Chapter Seven by 71pSO0

VIEWS: 0 PAGES: 20

• pg 1
```									Chapter Seven
7.1   The probability distribution of the population data is called the population distribution. Tables 7.1 and
7.2 on page 309 of the text provide an example of such a distribution. The probability distribution of a
sample statistic is called its sampling distribution. Tables 7.3 to 7.5 on page 311 of the text provide an
example of the sampling distribution of the sample mean.

7.3   Nonsampling errors are errors that may occur during collection, recording, and tabulation of data. The
second part of Example 7–1 on pages 312 and 313 of the text exhibits nonsampling error.
Nonsampling errors occur both in sample surveys and censuses.

7.5   a.   (20  25  13  19  9  15  11  7  17  30) / 10  166 / 10  16.60

b. x  (20  25  13  9  15  11  7  17  30) / 9  147 / 9  16.33

Sampling error = x    16.33  16.60  .27

c. Rich’s incorrect x  (20  25  13  9  15  11  17  17  30) / 9  157 / 9  17.44

x    17.44  16.60  .84
Sampling error (from part b) = –.27
Nonsampling error  .84  (.27)  1.11
d.
Sample                            x               x
25, 13 19,9,15,11,7,17,30                      16.22             –.38
20, 13, 19, 9, 15, 11, 7, 17, 30               15.67             –.93
20, 25 19, 9, 15, 11, 7, 17, 30                17.00              .40
20, 25, 13, 9, 15, 11, 7, 17, 30               16.33             –.27
20, 25, 13, 19, 15, 11, 7, 17, 30              17.44              .84
20, 25, 13, 19, 9, 11, 7, 17, 30               16.78              .18
20, 25, 13, 19, 9, 15, 7, 17, 30               17.22              .62
20, 25, 13, 19, 9, 15, 11, 17, 30              17.67             1.07
20, 25, 13, 19, 9, 15, 11, 7, 30               16.56             –.04
20, 25, 13, 19, 9, 15, 11, 7, 17               15.11            –1.49

119
120                                                                                                 Chapter Seven

7.7    a.
x                 P(x )
55               1/6=.167
53               1/6=.167
28               1/6=.167
25               1/6=.167
21               1/6=.167
15               1/6=.167

b.
Sample                 x
55, 53, 28, 25, 21        36.4
55, 53, 28, 25, 15        35.2
55, 53, 28, 21, 15        34.4
55, 53, 25, 21, 15        33.8
55, 28, 25, 21, 15        28.8
53, 28, 25, 21, 15        28.4

x                  P(x )
36.4              1/6=.167
35.2              1/6=.167
34.4              1/6=.167
33.8              1/6=.167
28.8              1/6=.167
28.4              1/6=.167

c. The mean for the population data is:   55  53  28  25  21 15  197  32.83
6               6
Suppose the random sample of five family members includes the observations: 55, 28,25, 21, and
55  28  25  21 15 144
15. The mean for this sample is:     x                             28.80
5            5
Then the sampling error is: x    28.80  32.83  4.03

7.9    a. Mean of x   x  

b. Standard deviation of x   x   / n where   population standard deviation and n = sample size.

7.11   An estimator is consistent when its standard deviation decreases as the sample size is increased. The
sample mean x is a consistent estimator of μ because its standard deviation decreases as
the sample size increases. This is obvious from the formula of  x   / n .
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual                                 121

7.13       60 and   10

a.  x    60 and  x   / n  10 / 18  2.357

b.  x    60 and  x   / n  10 / 90  1.054

7.15    a. n / N = 300 / 5000 = .06 > .05

        N n     25    5000  300
x                                       1.400
n       N 1     300    5000  1

b. n / N  100 / 5000  .02  .05,  x   / n  25 / 100  2.500

7.17       125 and   36

a.  x         n  3.6 so n    x 2  36 3.62  100

b. n    x 2  36 2.252  256

7.19       \$961,000;   \$180,000 , and n = 80

 x    \$961,000 and  x           n  180,000 / 80  \$20,124.61

7.21       \$233 ,   \$72 , and n = 25,  x    \$233 and  x      n  72   25  \$14.40

7.23       \$405 and  x  \$45

 x   / n , so n    x 2  405 452  81 players

7.25    a.
x             P( x )                x P( x )                  x 2 P( x )
76.00              .20                  15.200                   1155.200
76.67              .10                   7.667                    587.829
79.33              .10                   7.933                    629.325
81.00              .10                   8.100                    656.100
81.67              .20                  16.334                   1333.998
84.33              .20                  16.866                   1422.310
85.00              .10                   8.500                    722.500
∑ x P( x )=80.600        ∑ x P( x ) = 6507.262
2
122                                                                                                  Chapter Seven

∑ x P( x )=80.600 = same value found in Exercise 7.6 for μ.

b.  x      x 2 P( x )  ( x ) 2  6507.262  (80.60) 2  3.302

c.       n  8.09     3  4.67 is not equal to  x  3.30 in this case because n / N  3 / 5  .60, which is

greater than .05.

      N  n 8.09 5  3
d.  x                            3.302
n   N 1    3 5 1

7.27   The central limit theorem states that for a large sample, the sampling distribution of the sample mean is
approximately normal, irrespective of the shape of the population distribution. Furthermore,  x  

and  x         n , where μ and σ are the population mean and standard deviation, respectively. A

sample size of 30 or more is considered large enough to apply the central limit theorem to x .

7.29   a. Slightly skewed to the right

b Approximately normal because n  30 and the central limit theorem applies

c. Close to normal with perhaps a slight skew to the right

7.31   In both cases the sampling distribution of x would be normal because the population distribution is
normal.

7.33     46 miles per hour,   3 miles per hour, and n = 20

 x    46 miles per hour and  x           n 3     20  .671 mile per hour

The sampling distribution of x is normal because the population is normally distributed.

7.35     3.02 ,   .29 , N = 5540 and n = 48
 x    3.02
Since n / N  48 / 5540  .009 which is less than .05,

 x      n  .29      48  .042
The sampling distribution of x is approximately normal because the population is approximately
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual                                    123

normally distributed.

7.37       \$96 ,   \$27 , and n  90

 x    \$96 and  x         n  27    90  \$2.846

The sampling distribution of x is approximately normal because the sample size is large (n > 30).

7.39       \$8367,   \$2400, and n = 625

 x    \$8367 and  x         n  2400     625  \$96.00

The sampling distribution of x is approximately normal because the sample size is large (n > 30).

7.41     P(  1.50 x  x    1.50 x )  P(1.50  z  1.50)  P( z  1.50)  P( z  1.50)
 .9332 .0668  .8664 or 86.64%.

7.43       66 ,   7 , and n = 49

 x     n 7     49  1

a. z = ( x   ) /  x  (68.44  66) / 1.00  2.44

b. z = ( x   ) /  x  (58.75  66) / 1.00  7.25

c. z = ( x   ) /  x  (62.35  66) / 1.00  3.65

d. z = ( x   ) /  x  (71.82  66) / 1.00  5.82

7.45       48 ,   8 , and n  16

 x    48 and  x   / n  8 / 16  2.0

a. For x  49.6 : z  (49.6  48) / 2.0  .80

For x  52.2 : z  (52.2  48) / 2.0  2.10

P(49.6  x  52.2)  P(.80  z  2.10)  P( z  2.10)  P( z  .80)  .9821 .7881  .1940

b. For x  45.7 : z  (45.7  48) / 2.0  2.3 / 2.0  1.15
P( x  45.7)  P( z  1.15)  P( z  1.15)  .8749

TI-83: Start by pressing the 2nd key followed by the VARS key which bring up the distribution menu.
Highlight DISTR, scroll down to 2: normalcdf( and press ENTER. Next enter the smaller of the two x
124                                                                                               Chapter Seven

values, followed by the larger one, the µ, the σ, the ) symbol, and press ENTER. Remember to separate
your numbers with commas! For the first part of this problem, after normalcdf( we would enter 49.6,

52.2, 48, 8 / (16 ) ) and then press ENTER. Notice to get the square root of a number you need to press

2nd followed by the x2 key, then the number your wish to know the square root of, and followed by the
) key. The results for this problem are shown below.

Normalcdf(49.6, 52.2,
48, 8/ √(16))
.1939909763
Normalcdf( 45.7, E99,
48, 8 / √(16))
.8749280114

Note: if this interval has only 1 end point, use –E99 (negative infinity) or E99 (positive infinity) to
represent the missing endpoint. To type in E99 (positive infinity) to represent the missing endpoint
press the 2nd key followed by the , key and then typing in 99. To type in - E99 (negative infinity) to
represent the missing endpoint press (−) then press the 2 nd key followed by the , key and then typing in
99.

MINITAB: Enter the two z values in a column in your worksheet. For this example we will use
column C1. Now select Calc, the Probability Distribution, and Normal which will cause a new menu to
pop up. In the new menu make sure box beside Cumulative Probability is filled in, the Mean says 0.0,
and the Standard deviation says 1.0. Replace the standard normal mean and standard deviation to
match the µ and σ which is now stated in terms of xs. Beside the words Input column type C1 or the
number of whatever column your z vales are in and then click on OK. The result appears in the Session
box. Once we have the probabilities actually have P(x < a) and P (x < b), but we desire P(a < x < b) so
we P (x < b) − P(x < a) to get the answer. For this example µ is 48,  x is 2, a is 49.6 and b is 52.2.

The results appear below. To complete the problem we need to do is a little subtraction. For part a:
P(49.6 < x < 52.2) =P (x < 52.2) − P(x < 49.6) = .982136 − .788145 which equals 0.193991.
For part b: P (x > 45.7) = 1 − P (x < 45.7) = 1 − .125072 =.874928.
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual                                            125

Excel: In the first empty cell type in the formula =NORMDIST(x, µ, σ, cumulative) and then press
ENTER. Where x is the z value such that we calculate P(z < x ), µ is mean in terms of xs, σ is the
standard deviation in terms of xs, and True mean this is a cumulative distribution. Since we want
P(49.6 < x < 52.2) we calculate P (x < 52.2) − P(x < 49.6). Note we can calculate this in a single step
or we can make the calculations in a few steps like in 6.29. For this example µ is 48 and  x is

8 / (16 ) . Then all we have to do is subtract the second probability from the first. For part b, P (x >

45.7) = 1 − P (x < 45.7) and like the proceeding calculation it can be done in steps or all at once. In this
instance it was done all at once by adding 1− between = Normdist( so that the cell looks as follows =
1−Normdist(. The results are shown below where labels were added for easier reading.

7.47       90 ,   18 , and n  64

 x    90 and  x   / n  18 / 64  2.25

a. For x  82.3 : z  (82.3  90) / 2.25  3.42

P( x  82.3)  P( z  3.42)  .0000 approximately

b. For x  86.7 : z  (86.7  90) / 2.25  1.47           P( x  86.7)  P( z  1.47)  P( z  1.47)  .9292

7.49       3.02 ,   .29 , and n  20

x       n  .29   20  .06484597

a. For x  3.10 : z  (3.10  3.02) / .06484597 1.23

P( x  3.10)  P( z  1.23)  P( z  1.23)  .1093

b. For x  2.90 : z  (2.90  3.02) / .06484597 1.85

P( x  2.90)  P( z  1.85)  .0322

c. For x  2.95 : z  (2.95  3.02) / .06484597 1.08

For x  3.11 : z  (3.11  3.02) / .06484597 1.39

P(2.95  x  3.11)  P(1.08  z  1.39)  P( z  1.39)  P( z  1.08)  .9177 .1401 .7776
126                                                                                              Chapter Seven

7.51       \$18.96 per hour,   \$3.60 per hour, and n  25

x       n  3.60    25  \$.72 per hour

a. For x  18 : z  (18  18.96) / .72  1.33

For x  20 : z  (20  18.96) / .72  1.44

P(18  x  20)  P(1.33  z  1.44)  P( z  1.44)  P( z  1.33)  .9251 .0918  .8333

b. P( x within \$1.00 per hour of  )  P(17.963  x  19.96)

For x  17.96 : z  (17.96  18.96) / .72  1.39

For x  19.96 : z  (19.96  18.96) / .72  1.39

P(17.96  x  19.96)  P(1.39  z  1.39)  P( z  1.39)  P( z  1.39)  .9177 .0823  .8354

c. P( x greater than μ by \$1.50 per hour or more)  P( x  20.46)

For x  20.46 : z  (20.46  18.96) / .72  2.08

P( x  20.46)  P( z  2.08)  P( z  2.08)  .0188

7.53     \$1840;   \$453; and n  36

x       n  453    36  \$75.5

a. For x  1750 : z  (1750  1840) / 75.5  1.19

For x  1950 : z  (1950  1840) / 75.5  1.46

P(1750  x  1950)  P(1.19  z  1.46)  P( z  1.46)  P( z  1.19) = .9279 − .1170 = .8109

b. For x  1700 : z  (1700  1840) / 75.5  1.85             P( x  1700)  P( z  1.85)  .0322

7.55     \$80 ,   \$25 , and n  75

 x   / n  25      75  2.88675135

a. For x  72 : z  (72  80) / 2.88675135 2.77

For x  77 : z  (77  80) / 2.88675135 1.04

P(72  x  77)  P(2.77  z  1.04)  P( z  1.04)  P( z  2.77)
= .1492 – .0028 = .1464
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual                                          127

b. P(  6  x    6)  P(74  x  86)

For x  74 : z  (74  80) / 2.88675135 2.08

For x  86 : z  (86  80) / 2.88675135 2.08

P(74  x  86)  P(2.08  z  2.08)  P( z  2.08)  P( z  2.08)  .9812  .0188  .9624

c. P( x    5)  P( x  85)

For x  85 : z  (85  80) / 2.88675135 1.73          P( x  85)  P( z  1.73)  P( z  1.73)  .0418

7.57       68 inches,   4 inches, and n  100

x      n 4     100  .4 inch

a. For x  67.8 : z  (67.8  68) / .4  .50                      P( x  67.8)  P( z  .50)  .3085

b. For x  67.5 : z  (67.5  68) / .4  1.25

For x  68.7 : z  (68.7  68) / .4  1.75

P(67.5  x  68.7)  P(1.25  z  1.75)  P( z  1.75)  P( z  1.25) = .9599 − .1056 = .8543

c. P( x within .6 inches of  )  P(67.4  x  68.6)

For x  67.4 : z  (67.4  68) / .4  1.50

For x  68.6 : z  (68.6  68) / .4  1.50

P(67.4  x  68.6)  P(1.50  z  1.50)  P( z  1.50)  P( z  1.50)  .9322  .0668  .8664

d. P( x is lower than  by .5 inches or more)  P( x  67.5)

For x  67.5 : z  (67.5  68) / .4  1.25                    P( x  67.5)  P( z  1.25)  .1056

7.59     x  3 inches,   .1 inch, and n  25

 x  .1 .25  .02 inch

For x  2.95 : z  (2.95  3) / .02  2.50

For x  3.05 : z  (3.05  3) / .02  2.50

P( x  2.95)  P( x  3.05)  P( z  2.50)  P( z  2.50)  P( z  2.50)  P( z  2.50)
= .0062 + .0062 = .0124
128                                                                                           Chapter Seven

7.61   p  600 / 5000  .12 and p  18 / 120  .15
ˆ

7.63   Number in population with characteristic  9500 .75  7125
Number in sample with characteristic  400.78  312

7.65   Sampling error = p  p  .66  .71  .05
ˆ

7.67                                               ˆ
The estimator of p is the sample proportion p .

ˆ                                                  ˆ
The sample proportion p is an unbiased estimator of p, since the mean of p is equal to p.

7.69       p  pq / n , hence  p decreases as n increases.
ˆ                    ˆ

7.71   p  .21, q  1  p  1  .21  .79

a. n  400,  p  p  .21 , and  p 
ˆ                   ˆ      pq / n  .21(.79) / 400  .020

b. n  750,  p  p  .21 , and  p 
ˆ                   ˆ      pq / n  .21(.79) / 750  .015

7.73   N  1400, p  .47 , and q  1  p  1  .47  .53

pq   N n   .47(.53) 1400  90
a. n / N  90 / 1400  .064  .05             p 
ˆ                                       .051
n    N 1      90     1400  1

b. n / N  50 / 1400  .036  .05 ,  p 
ˆ     pq / n  .47(.53) / 50  .071

7.75   a. np  400(.28)  112 and nq  400(.72)  288

Since np  5 and nq  5 , the central limit theorem applies.

b. np  80(.05)  4 ; since np  5 , the central limit theorem does not apply.

c. np  60(.12)  7.2 and nq  60(.88)  52.8

Since np  5 and nq  5 , the central limit theorem applies.

d. np  100(.035)  3.5 ; since np  5 , the central limit theorem does not apply.

7.77   a. The proportion of these TV sets that are good is 4 6  .667

b. Total number of samples of size 5 is: ( 6 C5 )  6
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual                                    129

c & d. Let: G = good TV set and D = defective TV set

Let the six TV sets be denoted as: 1 = G, 2 = G, 3 = D, 4 = D, 5 = G, and 6 = G. The six possible
samples, their sample proportions, and the sampling errors are given in the table below.

Sample                    TV sets              ˆ
p               Sampling error
1, 2, 3, 4, 5            G, G, D, D, G        3/5=.60          .60 – .667 = –.067
1, 2, 3, 4, 6            G, G, D, D, G        3/5=.60           .60 – .667 = .067
1, 2, 3, 5, 6            G, G, D, G, G        4/5=.80           .80 – .667 = .133
1, 2, 4, 5, 6            G, G, D, G, G        4/5=.80           .80 – .667 = .133
1, 3, 4, 5, 6            G, D, D, G, G        3/5=.60          .60 – .667= –.067
2, 3, 4, 5, 6            G, D, D, G, G        3/5=.60          .60 – .667= –.067

pˆ                 f        Relative Frequency
.60                  4            4/6=.667
.80                  2            2/6=.333
   f 6

pˆ                ˆ
P( p )
.60             .667
.80             .333

7.79     n  400 , p  .45 , q  1  p  1  .45  .55

 p  p  .45 and  p  pq / n  .45(.55) / 400  .025
ˆ                 ˆ

np  400(.45)  180 and nq  400(.55)  220
ˆ
Since np and nq are both greater than 5, the sampling distribution of p is approximately normal.

7.81     n  50 , p  .28 , q  1  p  1  .28  .72

 p  p  .28 and  p  pq / n  .28(.72) / 50  .063
ˆ                 ˆ

np  50(.28)  14 and nq  50(.72)  36
ˆ
Since np and nq are both greater than 5, the sampling distribution of p is approximately normal.

7.83     P( p  2.0 p  p  p  2.0 p )  P(2.00  z  2.00)  P( z  2.00)  P( z  2.00)
ˆ   ˆ            ˆ

 .9772 .0228  .9544
Thus, 95.44% of the sample proportions will be within 2 standard deviations of the population
proportion.
130                                                                                                Chapter Seven

7.85   n  100; p  .59; q  1  .59  .41

 p  pq / n  .59(.41) / 100  .04918333
ˆ

a. z  ( p  p) /  p  (.56  .59) / .04918333 .61
ˆ          ˆ

b. z  ( p  p) /  p  (.68  .59) / .04918333 1.83
ˆ          ˆ

c. z  ( p  p) /  p  (.53  .59) / .04918333 1.22
ˆ          ˆ

d. z  ( p  p) /  p  (.65  .59) / .04918333 1.22
ˆ          ˆ

7.87   p  .616 , q  1  p  1  .616  .384 , and n  200

 p  pq / n  .616(.384) / 200  .03439070
ˆ

a. For p  .60 : z  (.60  .616) / .03439070 .47
ˆ

For p  .66 : z  (.66  .616) / .03439070 1.28
ˆ

P(.60  p  .66)  P(.47  z  1.28)  P( z  1.28)  P( z  .47) =.8997 − .3192 = .5805
ˆ

b. For p  .64 : z  (.64  .616) / .03439070 .70
ˆ                                                 P( p  .64)  P( z  .70)  P( z  .70)  .2420
ˆ

TI-83: Start by pressing the 2nd key followed by the VARS key which bring up the distribution menu.
Highlight DISTR, scroll down to 2: normalcdf( and press ENTER. Next enter the smaller of the two x
values, followed by the larger one, the µ, the σ, the ) symbol, and press ENTER. Remember to separate
your numbers with commas! For the first part of this problem, after normalcdf( we would enter .60,

.66, .616,   (.616)(.384) / 200 ) and then press ENTER. Notice to get the square root of a number you

need to press 2nd followed by the x2 key, then the number your wish to know the square root of, and
followed by the ) key. The results for this problem are shown below.

Normalcdf(.60, .66,
.616, √(.616*.384/200))
.5787455319
Normalcdf( .64, E99,
.616, √(.616*.384/200))
.2426313144.

Note: if this interval has only 1 end point, use –E99 (negative infinity) or E99 (positive infinity) to
represent the missing endpoint. To type in E99 (positive infinity) to represent the missing endpoint
press the 2nd key followed by the , key and then typing in 99. To type in - E99 (negative infinity) to
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual                                        131

represent the missing endpoint press (−) then press the 2 nd key followed by the , key and then typing in
99.

MINITAB: Enter the two z values in a column in your worksheet. For this example we will use
column C1. Now select Calc, the Probability Distribution, and Normal which will cause a new menu to
pop up. In the new menu make sure box beside Cumulative Probability is filled in, the Mean says 0.0,
and the Standard deviation says 1.0. Replace the standard normal mean and standard deviation to
match the µ and σ which is now stated in terms of xs. Beside the words Input column type C1 or the
number of whatever column your z vales are in and then click on OK. The result appears in the Session
box. Once we have the probabilities actually have P(z < a) and P (z < b), but we desire P(a < z < b) so

we P (z < b) − P(z < a) to get the answer. For this example µ is .616,  p is
ˆ       (.616)(.384) / 200 =

.03439070, a is .60 and b is .66. The results appear below. To complete the problem we need to do is a
little subtraction. For part a:
P(.60 < z < .66) =P (z < .66) − P(z < .60) = .899625 − .320879 which equals 0.578746.
For part b: P (x > 45.7) = 1 − P (x < 45.7) = 1 − .757369 =.242631.

Excel: In the first empty cell type in the formula =NORMDIST(x, µ, σ, cumulative) and then press
ENTER. Where x is the z value such that we calculate P(z < x ), µ is mean in terms of xs, σ is the
standard deviation in terms of xs, and True mean this is a cumulative distribution. Since we want
P(.60 < z < .66) we calculate P (z < .66) − P(z < .60). Note we can calculate this in a single step like in
7.45 or we can make the calculations in a few steps like in 6.29. For this example µ is .616 and  p
ˆ

is (.616)(.384) / 200 . Then all we have to do is subtract the second probability from the first. For part

b, P (z > .64) = 1 − P (x < .64) and like the proceeding calculation it can be done in steps or all at once.
In this instance it was done all at once by adding 1− between = Normdist(…, so that the cell looks as
follows = 1−Normdist(…, The results are shown below where labels were added for easier reading.
132                                                                                                  Chapter Seven

7.89   p  .39; , q  1  .39  .61; and n  300

 p  pq / n  .39(.61) / 300  .02816026
ˆ

a. For p  .35 : z  (.35  .39) / .02816026 1.42
ˆ

For p  .45 : z  (.45  .39) / .02816026 2.13
ˆ

P(.35  p  .45)  P(1.42  z  2.13)  P( z  2.13)  P( z  1.42) = .9834 − .0778 = .9056
ˆ

b. For p  .36 : z  (.36  .39) / .02816026 1.07
ˆ                                                   P( p  .36)  P( z  1.07)  P( z  1.07)  .8577
ˆ

7.91   p  .06 , q  1  p  1  .06  .94 , and n  100

 p  pq / n  .06(.94) / 100  .02374868
ˆ

For p  .08 : z  (.08  .06) / .02374868 .84
ˆ                                                      P( p  .08)  P( z  .84)  P( z  .84)  .2005
ˆ

7.93     750 hours,   55 hours, and n  25

 x    750 hours and  x          n  55       25  11 hours

The sampling distribution of x is normal because the population is normally distributed.

7.95     750 hours,   55 hours, and n  25

x       n  55    25  11 hours

a. For x  735 : z  (735  750) / 11  1.36              P( x  735)  P( z  1.36)  P( z  1.36)  .9131

b. For x  725 : z  (725  750) / 11  2.27

For x  740 : z  (740  750) / 11  .91

P(725  x  740)  P(2.27  z  .91)  P( z  .91)  P( z  2.27) = .1814 – .0116 = .1698

c. P( x within 15 hours of  )  P(735  x  765)

For x  735 : z  (735  750) / 11  1.36

For x  765 : z  (765  750) / 11  1.36

P(735  x  765)  P(1.36  z  1.36)  P( z  1.36)  P( z  1.36)  .9131 .0869  .8262

d. P( x is lower than  by 20 hours or more)  P( x  730)
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual                                            133

For x  730 : z  (730  750) / 11  1.82                     P( x  730)  P( z  1.82)  .0344

7.97       16.87 hours,   5 hours, and n  100

x       n 5    100  .5 hours

a. For x  17 : z  (17  16.87) / .5  .26               P( x  17)  P( z  .26)  P( z  .26)  .3974

b. For x  16.5 : z  (16.5  16.87) / .5  .74

For x  17.5 : z  (17.5  16.87) / .5  1.26

P(16.5  x  17.5)  P(.74  z  1.26)  P( z  1.26)  P( z  .74) = .8962 − .2296 = .6666

c. P(  .75  x    .75)  P(16.12  z  17.62)

For x  16.12 : z  (16.12  16.87) / .5  1.5

For x  17.62 : z  (17.62  16.87) / .5  1.5

P(16.12  x  17.62)  P(1.50  z  1.50)  P( z  1.50)  P( z  1.50) = .9332 + .0668 = .8664

d. P( x    .75)  P( x  16.12)

For x  16.12 : z  (16.12  16.87) / .5  1.50              P( x  16.12)  P( z  1.50)  .0668

7.99     p  .88, q  1  p  1  .88  .12 , and n  80

 p  p  .88 , and  p  pq / n  .88(.12) / 80  .036
ˆ                   ˆ

np  80(.88)  70.4 and nq  80(.12)  9.6
ˆ
Since np and nq are both greater than 5, the sampling distribution of p is approximately normal.

7.101    p  .70, q  1  p  1  .70  .30 , and n  400

 p  pq / n  .70(.30) / 400  .02291288
ˆ

a. i. For p  .65 : z  (.65  .70) / .02291288 2.18
ˆ                                                        P( p  .65)  P( z  2.18)  .0146
ˆ

ii. For p  .73 : z  (.73  .70) / .02291288 1.31
ˆ

For p  .76 : z  (.76  .70) / .02291288 2.62
ˆ
134                                                                                                  Chapter Seven

P(.73  p  .76)  P(1.31  z  2.62)  P( z  2.62)  Pz  1.31) =.9956–.9049=.0907
ˆ

b. P( p  .06  p  p  .06)  P(.70  .06  p  .70  .06)  P(.64  p  .76)
ˆ                            ˆ                        ˆ

For p  .64 : z  (.64  .70) / .02291288 2.62
ˆ

For p  .76 : z  (.76  .70) / .02291288 2.62
ˆ

P(.64  p  .76)  P(2.62  z  2.62)  P( z  2.62)  P( z  2.62)  .9956  .0044  .9912
ˆ

c. P( p  p  .05)  P( p  .70  .05)  P( p  .75)
ˆ                 ˆ                   ˆ

For p  .75 : z  (.75  .70) / .02291288 2.18
ˆ                                               P( p  .75)  P( z  2.18)  P( z  2.18)  .0146
ˆ

d. P( p is less than p by .06 or more) = P( p  .34)
ˆ                                     ˆ

For p  .34 : z  (.34  .40) / .044721359 1.34
ˆ                                                        P( p  .34)  P( z  1.34)  .0901
ˆ

7.103     \$105,000 and n  100

 x   / n  105,000 / 100  \$10,500

The required probability is: P( 10,000  x    10,000)

For x   10,000 : z  ( 10,000)    / 10,500  .95

For x    10,000 : z  (  10,000)    / 10,500  .95

P( 10,000  x    10,000)  P(.95  z  .95)  P( z  .95)  P( z  .95)  .8289  .1711 .6578

7.105     c,   .8 ppm
We want P(  .5  x    .5)  .95

1.96 x  .5 or  x  .255

2        (.8) 2
 x   / n , so n                       9.84         Ten measurements are necessary.
( x ) 2 (.255) 2

7.107   a.   p  .53, n  200, and we assume that n / N  .05

 p  pq / n  (.53)(.47) / 200  .03529164
ˆ

The shape of the sampling distribution is approximately normal.
.50  .53
z  ( p  p) /  p 
ˆ          ˆ                 .85   P( p  .50)  P( z  .85)  .5  .3032  .8023
ˆ
.03529164

b. P( z  1.65)  .9505
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual                                       135

.5  .53
z  ( p  p) /  p , so  p  ( p  p) / z 
ˆ          ˆ        ˆ     ˆ                          .01818182
 1.65
(.53)(.47)
 p  pq / n , hence n  pq /( p ) 2 
ˆ                             ˆ                                 753.53
(.01818182 2
)
The sample should include at least 754 voters.

7.109        160 pounds,   25 pounds, and n  35

 x   / n  25       35  4.22577127

Since n  30, x is approximately normally distributed.

P (sum of 35 weights exceeds 6000 pounds) = P (mean weight exceeds 6000/35) = P( x  171.43)

For x  171.43 : z  (171.43 160) / 4.22577127 2.70

P( x  171.43)  P( z  2.70)  P( z  2.70)  .0035

Self – Review Test for Chapter Seven

1.    b       2.   b         3.    a             4.    a                 5.    b          6.    b

7.    c       8.   a         9.    a             10.       a             11.   c          12.   a

13. According to the central limit theorem, for a large sample size, the sampling distribution of the sample
mean is approximately normal irrespective of the shape of the population distribution. The mean and

standard deviation of the sampling distribution of the sample mean are:  x   and  x     n.

The sample size is usually considered to be large if n  30.
ˆ
From the same theorem, the sampling distribution of p is approximately normal for large samples. In the

case of proportion, the sample is large if np  5 and nq  5.

14.       145 pounds and   18 pounds

a.  x    145 pounds and  x            n  18        25  3.60 pounds

b.  x    145 pounds and  x            n  18        100  1.80 pounds
136                                                                                                  Chapter Seven

In both cases the sampling distribution of x is approximately normal because the population has an
approximate normal distribution.

15.     11 minutes and   2.7 minutes

a.  x    11minutes and  x        n  2.7     25  .54 minute

Since the population has an unknown distribution and n  30 , we can draw no conclusion about the
shape of the sampling distribution of x .

b.  x    11minutes and  x        n  2.7 / 75  .312 minutes

Since n  30 , the sampling distribution of x is approximately normal.

16.     42 seconds,   10 seconds, and n  50

x      n  10    50  1.41421356seconds

a. For x  38 : z  (38  42) / 1.4142136 2.83

For x  41 : z  (41  42) / 1.4142136  .71

P(38  x  41)  P(2.83  z  .71)  P( z  .71)  P( z  2.83) = .2389 – .0023 = .2366

b. For x  42  2  40 : z  (40  42) / 1.4142136 1.41

For x  42  2  44 : z  (44  42) / 1.4142136 1.41

P(40  x  44)  P(1.41  z  1.41)  P( z  1.44)  P( z  1.41)  .9207  .0793  .8414

c. P( x greater than  by 1 second or more)

For x  43 : z  (43  42) / 1.4142136 .71             P( x  43)  P( z  .71)  P( z  .71)  .2389

d. For x  39 : z  (39  42) / 1.4142136  2.12

For x  44 : z  (44  42) / 1.4142136  1.41

P(39  x  44)  P(2.12  z  1.41)  P( z  1.41)  P( z  2.12) = .9207 − .0170 = .9037

e. x  43 : z  (43  42) / 1.4142136 .71                 P( x  43)  P( z  .71)  .7611
Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual                                       137

17.     16 ounces,   .18 ounce, and n  16

x       n  .18    16  .045

a. i. For x  15.90 : z  (15.90  16) / .045  2.22

For x  15.95 : z  (15.95  16) / .045  1.11

P(15.90  x  15.95)  P(2.22  z  1.11)  P( z  1.11)  P( z  2.22) = .1335 – .0132 = .1203

ii. For x  15.95 : z  (15.95  16) / .045  1.11     P( x  15.95)  P( z  1.11)  .1335

iii. For x  15.97 : z  (15.97  16) / .045  .67     P( x  15.97)  P( z  .67)  P( z  .67)  .7486

b. P(16  .10  x  16  .10)  P(15.90  x  16.10)

For x  15.90 : z  (15.90  16) / .045  2.22

For x  16.10 : z  (16.10  16) / .045  2.22

P(15.90  x  16.10)  P(2.22  z  2.22)  P( z  2.22)  P( z  2.22)  .9868 .0132  .9736

c. P( x  16  .135)  P( x  15.865)

For x  15.865 : z  (15.865  16) / .045  3.00                P( x  15.865)  P( z  3.00)  .0013

18.   p  .07, q  1  .07  .93

a. n  80 ,  p  p  .07, and  p 
ˆ                  ˆ       pq / n  .07(.93) / 80  .029

np  80(.07)  5.6 and nq = 80(.93) = 74.4
ˆ
Since np and nq are both greater than 5, the sampling distribution of p is approximately normal.

b. n  200 ,  p  p  .07, and  p 
ˆ                  ˆ       pq / n  .07(.93) / 200  .018

np  200(.07)  14 and nq  200(.93)  186
ˆ
Since np and nq are both greater than 5, the sampling distribution of p is approximately normal.

c. n  1000,  p  p  .07, and  p 
ˆ                  ˆ        pq / n  .07(.93) / 1000  .008

np  1000(.07)  70 and nq  1000(.93)  930
138                                                                                                Chapter Seven

ˆ
Since np and nq are both greater than 5, the sampling distribution of p is approximately normal.

19.   p  .41, q  1  .41  .59 and n  300

 p  pq / n  .41(.59) / 300  .02839601
ˆ

a. i. For p  .45 : z  (.45  .41) / .02839601 1.41
ˆ                                               P( p  .45)  P( z  1.41)  P( z  1.41)  .0793
ˆ

ii. For p  .40 : z  (.40  .41) / .02839601 .35
ˆ

For p  .46 : z  (.46  .41) / .02839601 1.76
ˆ

P(.40  p  .46)  P(.35  z  1.76)  P( z  1.76)  P( z  .35) = .9608 − .3632 = .5976
ˆ

iii. For p  .43 : z  (.43  .41) / .02839601 .70
ˆ                                                      P( p  .43)  P( z  .70)  .7580
ˆ

iv. For p  .36 : z  (.36  .41) / .02839601 1.76
ˆ                                              For p  .39 : z  (.39  .41) / .02839601 .70
ˆ

P(.36  p  .39)  P(1.76  z  .70)  P( z  .70)  P( z  1.76) = .2420 – .0392 = .2028
ˆ

b. P(.41  .04  p  .41  .04)  P(.37  p  .45)
ˆ                        ˆ

For p  .37 : z  (.37  .41) / .02839601 1.41
ˆ                                                  For p  .45 : z  (.45  .41) / .02839601 1.41
ˆ

P(.37  p  .45)  P(1.41  z  1.41)  P( z  1.41)  P( z  1.41) = .9207 −.0793 = .8414
ˆ

c. P( p  .41  .05)  P( p  .36)
ˆ                   ˆ

For p  .36 : z  (.36  .41) / .02839601 1.76
ˆ                                                           P( p  .36)  P( z  1.76)  .0392
ˆ

d. P(.41 .03  p  .41 .03)  P(.38  p  .44)
ˆ                       ˆ

For p  .38 : z  (.38  .41) / .02839601 1.06
ˆ                                                  For p  .44 : z  (.44  .41) / .02839601 1.06
ˆ

P(.36  p  .44)  P(1.06  z  1.06)  P( z  1.06)  P( z  1.06) = .8554 – .1446 = .7108
ˆ

e.    P( p  .41  .03)  P( p  .44)
ˆ                   ˆ

For p  .44, z  (.44  .41) / .02893601 1.06
ˆ                                                     P( p  .44)  P( z  1.06)  P( z  1.06)  .1446
ˆ

```
To top