Introduction to Probability and

Statistics 13th Edition Mendenhall

Solutions Manual
Visit to download the full and correct content document: https://testbankdeal.com/dow
nload/introduction-to-probability-and-statistics-13th-edition-mendenhall-solutions-man
ual/
 7: Sampling Distributions

7.1 You can select a simple random sample of size n = 20 using Table 10 in Appendix I. First choose a
starting point and consider the first three digits in each number. Since the experimental units have already
been numbered from 000 to 999, the first 20 can be used. The three digits OR the (three digits – 500) will
identify the proper experimental unit. For example, if the three digits are 742, you should select the
experimental unit numbered 742 − 500 = 242 . The probability that any three digit number is selected is
2 1000 = 1 500 . One possible selection for the sample size n = 20 is

242 134 173 128 399

056 412 188 255 388
469 244 332 439 101
399 156 028 238 231

7.2 Each student will obtain a different sample, using Table 10 in Appendix I.

7.3 Each student will obtain a different sample, using Table 10 in Appendix I.

7.4 The student should use the same procedure as in Exercise 7.1 and refer to Table 10, Appendix I.

7.5 If all of the town citizenry is likely to pass this corner, a sample obtained by selecting every tenth person is
probably a fairly random sample.

7.6 The questionnaires that were returned do not constitute a representative sample from the 1000
questionnaires that were randomly sent out. It may be that the voters who chose to return the questionnaire
were particularly adamant about the Parks and Recreation surcharge, while the others had no strong
feelings one way or the other. The nonresponse of half the voters in the sample will undoubtedly bias the
resulting statistics.

7.7 Voter registration lists and DMV records will systematically exclude a large segment of the general
population – nondrivers, people who do not own cars, nonvoters and so on.

7.8 The wording of the question is biased to suggest that a “yes” response is the correct one. A more unbiased
way to phrase the question might be: “Is there too much sex and violence during prime TV viewing
hours?”

7.9 Use a randomization scheme similar to that used in Exercise 7.1. Number each of the 50 rats from 01 to
50. To choose the 25 rats who will receive the dose of MX, select 25 two-digit random numbers from
Table 10. Each two-digit number OR the (two digits – 50) will identify the proper experimental unit.

7.10 a Since the question is particularly sensitive to people of different ethnic origins, you may find that the
answers may not always be truthful, depending on the ethnicity of the interviewer and the person being
interviewed.
b Notice that the percentage in favor of affirmative action increases as the ethnic origin of the interviewer
changes from Caucasian to Asian to African-American. The people being interviewed may be changing
their response to match what they perceive to be the response which the interviewer wants to hear.

7.11 a The sample was chosen from Native American youth attending an after-school program in Minneapolis,
MN who were willing to participate. This sample is not randomly selected; it is a convenience sample.
b Valid inferences can be made from this study only if the convenience sample chosen by the researcher
behaves like a random sample. That is, the in this particular after-school program must be representative of
the population of Native American youth as a whole.

167
 c In order to increase the chances of obtaining a sample that is representative of the population of Native
American youth as a whole, the researcher might try to obtain a larger base of students to choose from.
Perhaps there is a computerized database from which he or she might select a random sample.

7.12 a Since each patient must be randomly assigned to either aspirin or the experimental drug with equal
probability, assign the digits 0-4 to the aspirin treatment, and the digits 5-9 to the experimental drug
treatment. As each patient enters the study, choose a random digit using Table 10 and assign the
appropriate treatment.
b The randomization scheme in part a does not guarantee an equal number of patients in each group.

7.13 a The first question is more unbiased.

b Notice that the percentage favoring the new space program drops dramatically when the phrase
“spending billions of dollars” is added to the question.

7.14 Answers will vary. Many of the questions used in this policy survey were biased towards the political
views held by members of the Republican Party.

7.15 Follow the instructions in the My Personal Trainer section. The blanks have been filled in below.

The sampling distribution of x will be approximately normal with mean 53 and standard deviation (or
standard error) 3.

7.16 Follow the instructions in the My Personal Trainer section. The blanks have been filled in below.

To find the probability that the sample mean is greater than 55, write down the event of interest
__ P ( x > 55) __. When x = 55 ,
x −μ 55 − 53
z= = = .67
σ/ n 3
Find the probability:
P ( x > 55) = P ( z > .67) = 1 − .7486 = .2514

7.17 Follow the instructions in the My Personal Trainer section. The blanks have been filled in below.

The sampling distribution of x will be approximately normal with mean 100 and standard deviation (or
standard error) 3.16.
7.18 Follow the instructions in the My Personal Trainer section. The blanks have been filled in below.

To find the probability that the sample mean is between 105 and 110, write down the event of interest
__ P (105 < x < 110) __. When x = 105 and x = 110,
x −μ 105 − 100 x − μ 110 − 100
z= = = 1.58 and z = = = 3.16
σ/ n 3.16 σ/ n 3.16
Find the probability:
P (105 < x < 110) = P(1.58 < z < 3.16) = .9992 − .9429 = .0563

7.19 Regardless of the shape of the population from which we are sampling, the sampling distribution of the
sample mean will have a mean μ equal to the mean of the population from which we are sampling, and a
standard deviation equal to σ n.
a μ = 10; σ n =3 36 = .5
b μ = 5; σ n =2 100 = .2
c μ = 120; σ n =1 8 = .3536

7.20 a If the sampled populations are normal, the distribution of x is also normal for all values of n.

168
 b The Central Limit Theorem states that for sample sizes as small as n = 25 , the sampling distribution of
x will be approximately normal. Hence, we can be relatively certain that the sampling distribution of x
for parts a and b will be approximately normal. However, the sample size is part c, n = 8 , is too small to
assume that the distribution of x is approximately normal.

7.21 a The probability histogram is shown below. It is far from mound-shaped.

0.250

0.225

p(x)
0.200

0.175

0.150

1 2 3 4 5 6 7
x

b Answers will vary from student to student.

c Using the data given in the text as an example, the relative frequency histogram shows a clearly
mound-shape.

18/50

14/50
Relative frequency

10/50

6/50

2/50

0
3.2 4.0 4.8 5.6
Sample mean

7.22 a For the data given in the text, the mean and standard deviation are calculated as
(∑ x )
2
224.32
∑x − 1025.23 −
i

∑x
2
224.3 n
i
50 = .623
x= = = 4.486 and s = =
i

n 50 n −1 49
b The values calculated in part a are close to the theoretical values of the mean and standard deviation for
σ 2.15
the sampling distribution of x , given as μ x = μ = 4.4 and σ x = = = .680 .
n 10

7.23-24 For a population with σ = 1 , the standard error of the mean is

σ n =1 n
The values of σ n for various values of n are tabulated below and plotted below. Notice that the
standard error decreases as the sample size increases.
n 1 2 4 9 16 25 100
SE ( x ) = σ n 1.00 .707 .500 .333 .250 .200 .100

169
 1.0

0.8

0.6
SE

0.4

0.2

0.0
0 20 40 60 80 100
n

7.25 a If the sample population is normal, the sampling distribution of x will also be normal (regardless of the
sample size) with mean μ = 106 and standard deviation (or standard error) given as
σ n = 12 25 = 2.4

x −μ 110 − 106
b Calculate z = = = 1.67 , so that
σ n 2.4
P ( x > 110 ) = P ( z > 1.67 ) = 1 − .9525 = .0475
c P (102 < x < 110 ) = P ( −1.67 < z < 1.67 ) = .9525 − .0475 = .9050

7.26 a Since the sample size is large, the sampling distribution of x will be approximately normal with mean
μ = 64,571 and standard deviation σ n = 4000 60 = 516.3978 .
b From the Empirical Rule (and the general properties of the normal distribution), approximately 95% of
the measurements will lie within 2 standard deviations of the mean:

μ ± 2 SE ⇒ 64,571 ± 2 ( 516.3978 )
64,571 ± 1032.80 or 63,538.20 to 65, 603.80
c Use the mean and standard deviation for the distribution of x given in part a.
⎛ 66, 000 − 64,571 ⎞
P ( x > 66, 000 ) = P ⎜ z > ⎟
⎝ 516.3978 ⎠
= P ( z > 2.77 ) = 1 − .9972 = .0028
d Refer to part c. You have observed a very unlikely occurrence, assuming that μ = 64,571 . Perhaps
your sample was not a random sample, or perhaps the average salary of $64,571 is no longer correct.

7.27 a Age of equipment, technician error, technician fatigue, equipment failure, difference in chemical purity,
contamination from outside sources, and so on.
b The variability in the average measurement is measured by the standard error, σ n . In order to
decrease this variability you should increase the sample size n.

7.28 The weight of a package of 12 tomatoes is the sum of the 12 individual tomato weights. Hence, since
weights and heights are in general normally distributed, so will be the sum of the 12 weights, according to
the Central Limit Theorem.

170
7.29 The number of bacteria in one cubic foot of water can be thought of as the sum of 1728 random variables,
each of which is the number of bacteria in a particular cubic inch of water. Hence, the Central Limit
Theorem insures the approximate normality of the sum.

7.30 a Since the original population is normally distributed, the sample mean x is also normally distributed
(for any sample size) with mean μ and standard deviation
σ n≈2 10

b If μ = 21 and n = 10 , the z-value corresponding to x = 20 is

x −μ 20 − 21
z= = = −1.58
σ n 2 10
and
P ( x < 20 ) = P ( z < −1.58 ) = .0571

c Refer to part b and assume that μ is unknown. It is necessary to find a value μ0 so that
P ( x < 20 ) = .001 . Recall that if μ = μ0 , the z-value corresponding to x = 20 is
x − μ0 20 − μ0 20 − μ0
z= = = .
σ n 2 10 .632

⎛ 20 − μ0 ⎞ ⎛ 20 − μ0 ⎞
It is necessary then to have P ( x < 20 ) = P ⎜ z < ⎟ = A ⎜ .632 ⎟ = .001 .
⎝ .632 ⎠ ⎝ ⎠
Refer to the figure below.

From Table 3, the value of z that cuts off .001 in the left hand tail of the normal distribution is z0 = −3.08 .
Hence, the two values, ( 20 − μ0 ) .632 and z0 = −3.08 , must be the same. Solving for μ0 ,
20 − μ0
= −3.08 or μ0 = 21.948 .
.632

7.31 a The random variable T = ∑ xi , were xi is normally distributed with mean μ = 630 and standard
deviation σ = 40 for i = 1, 2,3 . The Central Limit Theorem states that T is normally distributed with mean
nμ = 3(630) = 1890 and standard deviation σ n = 40 3 = 69.282 .
b Calculate
T − 1890 2000 − 1890
z= = = 1.59 .
69.282 69.282

171
 Then P (T > 2000 ) = P ( z < 1.59 ) = 1 − .9441 = .0559 .

7.32 a Since the total daily sales is the sum of the sales made by a fixed number of customers on a given day, it
is a sum of random variables, which, according to the Central Limit Theorem, will have an approximate
normal distribution.
b Let xi be the total sales for a single customer, with i = 1, 2, …., 30. Then xi has a probability
distribution with μ = 8.50 and σ = 2.5 . The total daily sales can now be written as x = ∑ xi . If n = 30 ,
the mean and standard deviation of the sampling distribution of x are given as
nμ = 30(8.5) = 255 and σ n = 2.5 30 = 13.693

7.33 a Since the original population is normally distributed, the sample mean x is also normally distributed
(for any sample size) with mean μ and standard deviation
σ n = 0.8 130 = .07016
The z-value corresponding to x = 98.25 is
x − μ 98.25 − 98.6
z= = = −4.99
σ n 0.8 130
and
P ( x < 98.25 ) = P ( z < −4.99 ) ≈ 0
b Since the probability is extremely small, the average temperature of 98.25 degrees is very unlikely.

7.34 The sampled population has a mean of 5.97 with a standard deviation of 1.95.
x − μ 6.5 − 5.97
a With n = 31 , calculate z = = = 1.51 , so that
σ n 1.95 / 31
P ( x ≤ 6.5 ) = P ( z ≤ 1.51) = .9345
x −μ 9.80 − 5.97
b Calculate z = = = 10.94 , so that
σ n 1.95 / 31
P ( x ≥ 9.80 ) = P ( z ≥ 10.94 ) ≈ 1 − 1 = 0
c The probability of observing an average diameter of 9.80 or higher is extremely unlikely, if indeed the
average diameter in the population of affected tendons was no different from that of unaffected tendons
(5.97). We would conclude that the average diameter in the population of patients affected with AT is
higher than 5.97.

7.35 Follow the instructions in the My Personal Trainer section. The blanks have been filled in below.

The sampling distribution of p̂ will be approximately normal with mean p = .7 and standard deviation (or
pq .7(.3)
standard error) = = .0648.
n 50

7.36 Follow the instructions in the My Personal Trainer section. The blanks have been filled in below.

To find the probability that the sample proportion is less than .8, write down the event of interest
__ P( pˆ < .8) __. When pˆ = .8 ,
pˆ − p .8 − .7
z= = = 1.54
pq .0648
n
Find the probability: P ( pˆ < .8) = P( z < 1.54) = .9382

172
 pq .3(.7)
7.37 a p = .3; SE ( pˆ ) = = = .0458
n 100
pq .1(.9)
b p = .1; SE ( pˆ ) = = = .015
n 400
pq .6(.4)
c p = .6; SE ( pˆ ) = = = .0310
n 250

7.38 For each of the three binomial distributions, calculate np and nq:
a np = 2.5 and nq = 47.5 b np = 7.5 and nq = 67.5 c np = 247.5 and nq = 2.5
The normal approximation to the binomial distribution is only appropriate for part b, when n = 75 and p =
.1.

pq .4(.6)
7.39 a Since p̂ is approximately normal, with standard deviation SE ( pˆ ) = = = .0566 , the
n 75
probability of interest is
.43 − .4
P( pˆ ≤ .43) = P( z ≤ ) = P( z ≤ .53) = .7019
.0566
b The probability is approximated as
⎡ .35 − .4 .43 − .4 ⎤
P (.35 ≤ pˆ ≤ .43) = P ⎢ ≤z≤
⎣ .0566 .0566 ⎥⎦
= P ( −.88 ≤ z ≤ .53) = .7019 − .1894 = .5125

7.40 a For n = 500 and p = .1, np = 50 and nq = 450 are both greater than 5. Therefore, the normal
approximation will be appropriate.
⎛ .12 − .1 ⎞
b P ( pˆ > .12 ) = P ⎜ z > ⎟ = P ( z > 1.49 ) = 1 − .9319 = .0681
⎝ .0134 ⎠
⎛ .10 − .1 ⎞
c P ( pˆ < .10 ) = P ⎜ z < ⎟ = P ( z < 0 ) = .5
⎝ .0134 ⎠
d P ( −.02 ≤ ( pˆ − p ) ≤ .02 ) = P ( −1.49 ≤ z ≤ 1.49 ) = .9319 − .0681 = .8638

7.41 The values SE = pq n for n = 100 and various values of p are tabulated and graphed below. Notice that
SE is a maximum for p = .5 and becomes very small for p near zero and one.
p .01 .10 .30 .50 .70 .90 .99
SE ( pˆ ) .0099 .03 .0458 .05 .0458 .03 .0099

0.05

0.04

0.03
SE

0.02

0.01

0.0 0.2 0.4 0.6 0.8 1.0

p

173
7.42 a For n = 400 and p = .8, np = 320 and nq = 80 are both greater than 5. Therefore, the normal
pq .8(.2)
approximation will be appropriate with SE = = = .02 .
n 400
⎛ .83 − .8 ⎞
b P ( pˆ > .83) = P ⎜ z > ⎟ = P ( z > 1.5 ) = 1 − .9332 = .0668
⎝ .02 ⎠
⎛ .76 − .8 .84 − .8 ⎞
c P (.76 < pˆ < .84 ) = P ⎜ <z< ⎟ = P ( −2 < z < 2 ) = .9772 − .0228 = .9544
⎝ .02 .02 ⎠

7.43 a For n = 100 and p = .19, np = 19 and nq = 81 are both greater than 5. Therefore, the normal
pq .19(.81)
approximation will be appropriate, with mean p = .19 and SE = = = .03923 .
n 100
⎛ .25 − .19 ⎞
b P ( pˆ > .25 ) = P ⎜ z > ⎟ = P ( z > 1.53) = 1 − .9370 = .0360
⎝ .03923 ⎠
⎛ .25 − .19 .30 − .19 ⎞
c P (.25 < pˆ < .30 ) = P ⎜ <z< ⎟ = P (1.53 < z < 2.80 ) = .9974 − .9370 = .0604
⎝ .03923 .03923 ⎠
d The value pˆ = .30 lies
pˆ − p .30 − .19
z= = = 2.80
pq .03923
n
standard deviations from the mean. Also, P ( pˆ ≥ .30) = P( z ≥ 2.80) = 1 − .9974 = .0026 . This is an unlikely
occurrence, assuming that p = .19 . Perhaps the sampling was not random, or the 19% figure is not correct.

7.44 a The random variable p̂ , the sample proportion of adults who say that prescription drugs are
unreasonably high, has a binomial distribution with n = 1000 and p = .66 . Since np = 660 and
nq = 340 are both greater than 5, this binomial distribution can be approximated by a normal distribution
.66(.34)
with mean p = .66 and SE = = .01498 .
1000
⎛ .68 − .66 ⎞
b P ( pˆ > .68 ) = P ⎜ z > ⎟ = P ( z > 1.34 ) = 1 − .9099 = .0901
⎝ .01498 ⎠
c P (.64 < pˆ < .68 ) = P ( −1.34 < z < 1.34 ) = .9099 − .0901 = .8198
d The value pˆ = .70 lies
pˆ − p
.70 − .66
z= = = 2.67
pq .01498
n
standard deviations from the mean. Also, P ( pˆ ≥ .70) = P( z ≥ 2.67) = 1 − .9962 = .0038 . This is an unlikely
occurrence, assuming that p = .66 , and would tend to contradict the reported figure.

7.45 a The random variable p̂ , the sample proportion of brown M&Ms in a package of n = 55 , has a binomial
distribution with n = 55 and p = .13 . Since np = 7.15 and nq = 47.85 are both greater than 5, this
binomial distribution can be approximated by a normal distribution with mean p = .13 and
.13(.87)
SE = = .04535.
55
⎛ .2 − .13 ⎞
b P ( pˆ < .2 ) = P ⎜ z < ⎟ = P ( z < 1.54 ) = .9382
⎝ .04535 ⎠

174
 ⎛ .35 − .13 ⎞
c P ( pˆ > .35 ) = P ⎜ z > ⎟ = P ( z > 4.85 ) ≈ 1 − 1 = 0
⎝ .04535 ⎠
d From the Empirical Rule (and the general properties of the normal distribution), approximately 95% of
the measurements will lie within 2 (or 1.96) standard deviations of the mean:
p ± 2SE ⇒ .13 ± 2(.04535)
.13 ± .09 or .04 to .22

7.46 The random variable p̂ , the sample proportion of overweight children in a random sample of n = 100 , has
a binomial distribution with n = 100 and p = .15 . Since np = 15 and nq = 85 are both greater than 5, this
binomial distribution can be approximated by a normal distribution with mean p = .15 and
.15(.85)
SE = = .03571
100
⎛ .25 − .15 ⎞
a P ( pˆ > .25 ) = P ⎜ z > ⎟ = P ( z > 2.80 ) = 1 − .9974 = .0026
⎝ .03571 ⎠
⎛ .12 − .15 ⎞
b P ( pˆ < .12 ) = P ⎜ z < ⎟ = P ( z < −.84 ) = .2005
⎝ .03571 ⎠
⎛ .30 − .15 ⎞
c P ( pˆ > .30 ) = P ⎜ z > ⎟ = P ( z > 4.20 ) ≈ 1 − 1 = 0 . Since the z-score is so high and the probability
⎝ .03571 ⎠
is so low, it would be highly unlikely to find 30% of the children to be overweight.

7.47 a The random variable p̂ , the sample proportion of consumers who like nuts or caramel in their
chocolate, has a binomial distribution with n = 200 and p = .75 . Since np = 150 and nq = 50 are both
greater than 5, this binomial distribution can be approximated by a normal distribution with
.75(.25)
mean p = .75 and SE = = .03062
200
⎛ .80 − .75 ⎞
b P ( pˆ > .80 ) = P ⎜ z > ⎟ = P ( z > 1.63) = 1 − .9484 = .0516
⎝ .03062 ⎠
c From the Empirical Rule (and the general properties of the normal distribution), approximately 95% of
the measurements will lie within 2 (or 1.96) standard deviations of the mean:
p ± 2SE ⇒ .75 ± 2(.03062)
.75 ± .06 or .69 to .81

7.48 a The upper and lower control limits are

s .87
UCL = x + 3 = 20.74 + 3 = 20.74 + .83 = 21.57
n 10
s .87
LCL = x − 3 = 20.74 − 3 = 20.74 − .83 = 19.91
n 10
b Control charts are used to monitor the process variable, detecting shifts that might indicate control
problems.
c The control chart is constructed by plotting two horizontal lines, one the upper control limit and one the
lower control limit (see Figure 7.15 in the text). Values of x are plotted, and should remain within the
control limits. If not, the process should be checked.

7.49 Similar to Exercise 7.48.

a The upper and lower control limits are

175
 s 4.3
UCL = x + 3 = 155.9 + 3 = 155.9 + 5.77 = 161.67
n 5
s 4.3
LCL = x − 3 = 155.9 − 3 = 155.9 − 5.77 = 150.13
n 5
b The control chart is constructed by plotting two horizontal lines, one the upper control limit and one the
lower control limit (see Figure 7.15 in the text). Values of x are plotted, and should remain within the
control limits. If not, the process should be checked.

7.50 The x chart is used to monitor the average value of a sample of quantitative data, while the p chart is used
to monitor qualitative data by counting the number of defective items and tracking the percentage
defective.

7.51 a The upper and lower control limits for a p chart are
p (1 − p ) .035(.965)
UCL = p + 3 = .035 + 3 = .035 + .055 = .090
n 100
p (1 − p ) .035(.965)
LCL = p − 3 = .035 − 3 = .035 − .055 = −.020
n 100
or LCL = 0 (since p cannot be negative).
b The control chart is constructed by plotting two horizontal lines, one the upper control limit and one the
lower control limit (see Figure 7.16 in the text). Values of p̂ are plotted, and should remain within the
control limits. If not, the process should be checked.

7.52 The upper and lower control limits for a p chart are
p (1 − p ) .041(.959)
UCL = p + 3 = .041 + 3 = .041 + .042 = .083
n 200
p (1 − p ) .041(.959)
LCL = p − 3 = .041 − 3 = .041 − .042 = −.001
n 200
or LCL = 0 (since p cannot be negative).
b The control chart is constructed by plotting two horizontal lines, one the upper control limit and one the
lower control limit (see Figure 7.16 in the text). Values of p̂ are plotted, and should remain within the
control limits. If not, the process should be checked.

7.53 a The upper and lower control limits are

s 1605
UCL = x + 3 = 10, 752 + 3 = 10, 752 + 2153.3 = 12,905.3
n 5
s 1605
LCL = x − 3 = 10, 752 − 3 = 10, 752 − 2153.3 = 8598.7
n 5
b The x chart will allow the manager to monitor daily gains or losses to see whether there is a problem
with any particular table.

7.54 The upper and lower control limits for a p chart are
p (1 − p ) .021(.979)
UCL = p + 3 = .021 + 3 = .021 + .022 = .043
n 400
p (1 − p ) .021(.979)
LCL = p − 3 = .021 − 3 = .021 − .022 = −.001
n 400
or LCL = 0 (since p cannot be negative). The manager can use the control chart to detect changes in the
production process which might produce an unusually large number of defectives.

176
 ∑ pˆ i .14 + .21 + " + .26
7.55 Calculate p = = = .197 . The upper and lower control limits for the p chart are then
k 30
p (1 − p ) .197(.803)
UCL = p + 3 = .197 + 3 = .197 + .119 = .316
n 100
p (1 − p ) .197(.803)
LCL = p − 3 = .197 − 3 = .197 − .119 = .078
n 100

7.56 The upper and lower control limits are

s .07
UCL = x + 3 = 7.24 + 3 = 7.24 + .12 = 7.36
n 3
s .07
LCL = x − 3 = 7.24 − 3 = 7.24 − .12 = 7.12
n 3

7.57 Using all 104 measurements, the value of s is calculated to be s = .006717688 and x = .0256 . Then the
upper and lower control limits are
s .06717688
UCL = x + 3 = .0256 + 3 = .0357
n 4
s .06717688
LCL = x − 3 = .0256 − 3 = .0155
n 4

7.58 The upper and lower control limits are

s .2064
UCL = x + 3 = 31.7 + 3 = 31.7 + .277 = 31.977
n 5
s .2064
LCL = x − 3 = 31.7 − 3 = 31.7 − .277 = 31.423
n 5
The control chart is constructed by plotting two horizontal lines, one the upper control limit and one the
lower control limit (see Figure 7.15 in the text). Values of x are plotted, and should remain within the
control limits. If not, the process should be checked.

7.59 Refer to Exercise 7.58. The sample mean is outside the control limits in hours 2, 3 and 4. The process
should be checked.
4!
7.60 a C24 = = 6 samples are possible.
2!2!
b-c The 6 samples along with the sample means for each are shown below.

Sample Observations x
1 6, 1 3.5
2 6, 3 4.5
3 6, 2 4.0
4 1, 3 2.0
5 1, 2 1.5
6 3, 2 2.5

d Since each of the 6 distinct values of x are equally likely (due to random sampling), the sampling
distribution of x is given as
1
p(x) = for x = 1.5, 2, 2.5,3.5, 4, 4.5
6
The sampling distribution is shown on the next page.

177
 p(x)

.167

1.5 2.0 2.5 3.0 3.5 4.0 4.5

x

e The population mean is μ = ( 6 + 1 + 3 + 2 ) 4 = 3 . Notice that none of the samples of size n = 2 produce
a value of x exactly equal to the population mean.

7.61 Refer to Exercise 7.60. If samples of size n = 3 are drawn without replacement, there are 4 possible
samples with sample means shown below.
Sample Observations x
1 6, 1, 3 3.333
2 6, 1, 2 3
3 6, 3, 2 3.667
4 1, 3, 2 2

The sampling distribution of x is then

1
p(x) = for x = 2,3,3.333,3.667
4
The sampling distribution is shown below.
p(x)

.250

2 3 3.333 3.667 4
x

7.62 a The distribution of lead content readings for individual water specimens is probably skewed to the right,
with a few specimens containing a very large amount of lead. This conclusion is confirmed by looking at
the mean and standard deviation, .033 and .10, respectively. The large standard deviation does not allow
the measurements to spread over the range μ ± 2σ without ranging into negative levels (which are not
possible).

178
 b Since the sample of 23 daily lead levels is drawn from a population of means based on n = 40
observations, the sample is being drawn from an approximately normal population, according to the Central
Limit Theorem.
c For the sample means in part b, the sampling distribution has mean μ = .033 and standard deviation
σ n = .10 40 = .0158 .

7.63 a Using the range approximation, the standard deviation σ can be approximated as
R 55 − 5
σ≈ = = 12.5
4 4
b The sampling distribution of x is approximately normal with mean μ and standard error
σ n ≈ 12.5 400 = .625
⎛ −2 2 ⎞
Then P ( x − μ ≤ 2 ) = P ⎜ ≤z≤ ⎟ = P ( −3.2 ≤ z ≤ 3.2 ) = .9993 − .0007 = .9986 .
⎝ .625 .625 ⎠
c If the scientists are worried that the estimate of μ = 35 is too high, and if your estimate is x = 31.75 ,
then your estimate lies
x − μ 31.75 − 35
z= = = −5.2
σ n .625
standard deviations below the mean. This is a very unlikely event if in fact μ = 35 . It is more likely that
the scientists are correct in assuming that the mean is an overestimate of the mean biomass for tropical
woodlands.

7.64 It is given that n = 3, μ = 900 and σ = 100 . Then

P [sample mean satisfies ANSI ] = P ( x < 850 )
⎛ x − μ 850 − 900 ⎞
= P ⎜⎜ < ⎟ = P ( z < −.87 )
⎝σ n 100 3 ⎟⎠
= .1922

7.65 a To divide a group of 20 people into two groups of 10, use Table 10 in Appendix I. Assign an
identification number from 01 to 20 to each person. Then select ten two digit numbers from the random
number table to identify the ten people in the first group. (If the number is greater than 20, subtract
multiples of 20 from the random number until you obtain a number between 01 and 20.)
b Although it is not possible to select an actual random sample from this hypothetical population, the
researcher must obtain a sample that behaves like a random sample. A large database of some sort should
be used to ensure a fairly representative sample.
c The researcher has actually selected a convenience sample; however, it will probably behave like a
simple random sample, since a person’s enthusiasm for a paid job should not affect his response to this
psychological experiment.

7.66 a Since the data already existed before the researcher decided to study it, this is an observational study.
b Since the subject of the study is a sensitive one, there will be problems of nonresponse and/or
inaccurate responses to the questions.

7.67 a From the 50 lettuce seeds, the researcher must choose a group of 26 and a group of 13 for the
experiment. Identify each seed with a number from 01 to 50 and then select random numbers from Table
10. The first 26 numbers chosen will identify the seeds in the first Petri dish, and the next 13 will identify
the seeds in the third Petri dish. If the same number is picked twice, simply ignore it and go on to the next
number. Use a similar procedure to choose the groups of 26 and 13 radish seeds.
b The seeds in these two packages must be representative of all seeds in the general population of lettuce
and radish seeds.

7.68 a This is an observational study, since the data existed before you decided to observe or describe it.

179
 b The researcher should be concerned about nonresponse and untruthful responses, due to the sensitive
nature of the question.

7.69 Referring to Table 10 in Appendix I, we will select 20 numbers. First choose a starting point and consider
the first four digits in each number. If the four digits are a number greater than 7000, discard it. Continue
until 20 numbers have been chosen. The customers have already been numbered from 0001 to 7000. One
possible selection for the sample size n = 20 is

1048 2891 5108 4866

2236 6355 0236 5416
2413 0942 0101 3263
4216 1036 5216 2933
3757 0711 0705 0248

7.70 a For this binomial random variable with n = 500 and p = .85 , the mean and standard deviation of p̂ are
.85(.15)
μ = p = .85 and SE = = .01597
500
b Since np = 500(.85) = 425 and nq = 75 are both greater than 5, the normal approximation is
appropriate.
⎛ .82 − .85 ⎞
c P ( pˆ > .82 ) = P ⎜ z > ⎟ = P ( z > −1.88 ) = 1 − .0301 = .9699
⎝ .01597 ⎠
⎛ .83 − .85 .88 − .85 ⎞
d P (.83 < pˆ < .88 ) = P ⎜ <z< ⎟ = P ( −1.25 < z < 1.88 ) = .9699 − .1056 = .8643
⎝ .01597 .01597 ⎠
e For a normal (or approximately normal) random variable, the interval μ ± 2.58σ will contain 99% of
the measurements. For this binomial random variable x, this interval is
p ± 2.58SE ⇒ .85 ± 2.58(.01597)
.85 ± .04 or .81 to .89

7.71 a Since each cluster (a city block) is censused, this is an example of cluster sampling.
b This is a 1-in-10 systematic sample.
c The wards are the strata, and the sample is a stratified sample.
d This is a 1-in-10 systematic sample.
e This is a simple random sample from the population of all tax returns filed in the city of San
Bernardino, California.

7.72 Define xi to be the weight of a particular man or woman using the elevator. It is given that xi is
approximately normally distributed with μ = 150 and σ = 35 . Then according to the Central Limit
Theorem, the sum or total weight, ∑ xi , will be normally distributed with mean nμ = 150n and standard
deviation nσ = 35 n . It is necessary to find a value of n such that
P ( ∑ xi > 2000 ) = .01
The z-value corresponding to ∑ xi = 2000 is
∑ xi − nμ 2000 − 150n
z= =
σ n 35 n
so that we need
⎛ 2000 − 150n ⎞
P⎜ z > ⎟ = .01
⎝ 35 n ⎠

2000 − 150n
From Table 3, we conclude that = 2.33 . Manipulating the above equation, we obtain a
35 n
quadratic equation in n.

180
 ( 2000 − 150n ) = 81.55 n
4, 000, 000 − 606, 650.4025n + 22,500n = 02

Using the quadratic formula, the necessary value of n is found.

−b ± b 2 − 4ac 606, 650.4025 ± 89,580.734
n= =
2a 45, 000

Choosing the smaller root of the equation,

517, 069.67
n= = 11.49 .
45, 000

Hence, n = 12 will provide a sample size with P ( ∑ xi > 2000 ) ≈ .01 . In fact, for n = 11 ,
P ( ∑ xi > 2000 ) = P ( z > 3.02 ) = 1 − .9987 = .0013
while for n = 12 ,
P ( ∑ xi > 2000 ) = P ( z > 1.65 ) = 1 − .9505 = .0495

7.73 a The number of packages which can be assembled in 8 hours is the sum of 8 observations on the random
variable described here. Hence, its mean is nμ = 8(16.4) = 131.2 and its standard deviation is
σ n = 1.3 8 = 3.677 .
b If the original population is approximately normal, the sampling distribution of a sum of 8 normal
random variables will also be approximately normal. Since the original population is exactly normal, so
will be the sampling distribution of the sum.
⎛ 135 − 131.2 ⎞
c P ( x > 135 ) = P ⎜ z > ⎟ = P ( z > 1.03) = 1 − .8485 = .1515
⎝ 3.677 ⎠

7.74 a The total production is the sum of 10 observations on the random variable x described in Exercise 7.73.
Hence, its mean is 10(131.2) = 1312 and its standard deviation is 3.677 10 = 11.628 .
⎛ 1280 − 1312 ⎞
b P [ production < 1280] = P ⎜ z < ⎟ = P ( z < −2.75 ) = .0030
⎝ 11.628 ⎠

7.75 a The average proportion of defectives is

.04 + .02 + " + .03
p= = .032
25
and the control limits are
p (1 − p ) .032(.968)
UCL = p + 3 = .032 + 3 = .0848
n 100
p (1 − p ) .032(.968)
and LCL = p − 3 = .032 − 3 = −.0208
n 100
If subsequent samples do not stay within the limits, UCL = .0848 and LCL = 0 , the process should be
checked.
b From part a, we must have pˆ > .0848 .
c An erroneous conclusion will have occurred if in fact p < .0848 and the sample has produced pˆ = .15
by chance. One can obtain an upper bound on the probability of this particular type of error by calculating
P ( pˆ ≥ .15 when p = .0848 ) .

181
7.76 The probability that a shipment of 100 bulbs will have no more than 4% defective is P ( pˆ ≤ .04 ) . Since p̂
p (1 − p ) .032(.968)
is approximately normal with mean p ≈ .032 and SE ≈ = = .0176 ,
n 100
⎛ .04 − .032 ⎞
P ( pˆ ≤ .04 ) = P ⎜ z ≤ ⎟ = P ( z ≤ .45 ) = .6736
⎝ .0176 ⎠

7.77 Refer to Exercise 7.75, in which UCL = .0848 and LCL = 0 . For the next 5 samples, the values of p̂ are
.02, .04, .09, .07, .11. Hence, samples 3 and 5 are producing excess defectives. The process should be
checked.

23.1 + 21.3 + " + 21.3 636.1

7.78 a If the process is in control, σ = 1.20 . Calculate x = = = 21.2033 .
30 30
With n = 5 , the upper and lower control limits are
s 1.2
x ±3 = 21.2033 ± 3 = 21.2033 ± 1.6010
n 5
or LCL = 19.6023 and UCL = 22.8043 .
b The centerline is x = 21.2033 and the graph is omitted. Only one sample exceeds the UCL; the process
is probably in control.

7.79 Answers will vary from student to student. Paying cash for opinions will not necessarily produce a random
sample of opinions of all Pepsi and Coke drinkers.

7.80 Answers will vary. One solution is to number the containers from 01 to 20. Refer to Table 10 in Appendix
I, and select 20 two-digit numbers, choosing a random starting point. If the two digit number is greater
than 20, subtract 20 from the number consecutively, until you have a number between 01 and 20. Continue
until 10 numbers have been chosen. These ten containers will be stored at one temperature; the other ten
will be stored at the second temperature.

7.81 a Since the fill per can is normal with mean 12 and standard deviation 0.2, the total fill for a pack of six
cans will also have a normal distribution with mean nμ = 24(12) = 288 and standard deviation
σ n = 0.2 24 = .9798
b Let T be the total fill for the case of soda. Then
286 − 288
P (T < 286) = P ( z < ) = P ( z < −2.04) = .0207
.9798
11.8 − 12
c P ( x < 11.8) = P ( z < ) = P ( z < −2.45) = .0071
0.2 / 6

7.82 Similar to Exercise 7.81. Since the package weights are normal with mean 16 and standard deviation 0.6,
the total weight for a box of 24 packages will also have a normal distribution with mean nμ = 24(16) = 384
and standard deviation σ n = 0.6 24 = 2.939 . Let T be the total weight for the box. Then
392 − 384
P (T > 392) = P ( z > )
2.939
= P ( z > 2.72) = 1 − .9967 = .0033

7.83 a The average proportion of inoperable components is

6 + 7 + " + 5 75
p= = = .10
50(15) 750
and the control limits are

182
 p (1 − p ) .10(.90)
UCL = p + 3 = .10 + 3 = .2273
n 50
p (1 − p ) .1(.9)
and LCL = p − 3 = .10 − 3 = −.0272
n 50
If subsequent samples do not stay within the limits, UCL = .2273 and LCL = 0 , the process should be
checked.

1
7.84 a The random variable x has a discrete probability distribution given as p ( x ) = for x = 1, 2,3, 4,5, 6 .
6
Using the formulas given in Chapter 4,

⎛1⎞ 21
μ = ∑ xp( x) = (1 + 2 + " + 6 ) ⎜ ⎟ = = 3.5
⎝6⎠ 6
⎛1⎞
σ 2 = ∑ x 2 p ( x) − μ 2 = (12 + 22 + " + 62 ) ⎜ ⎟ − ( 3.5 )
2

6 ⎝ ⎠
91
= − 12.25 = 2.9167
6
and σ = 2.9167 = 1.708
b-c Answers will vary from student to student. The distribution should be relatively uniform with mean
and standard deviation close to those given in part a.

7.85 a The theoretical mean and standard deviation of the sampling distribution of x when n = 2 are
μ = 3.5 and SE = σ n = 1.708 2 = 1.208
b-c Answers will vary from student to student. The distribution should be relatively uniform with mean
and standard deviation close to those given in part a.

7.86 a The theoretical mean and standard deviation of the sampling distribution of x when n = 3 are
μ = 3.5 and SE = σ n = 1.708 3 = .986
b-c Answers will vary from student to student. The distribution should be relatively uniform with mean
and standard deviation close to those given in part a.

7.87 a The theoretical mean and standard deviation of the sampling distribution of x when n = 4 are
μ = 3.5 and σ n = 1.708 4 = .854
b-c Answers will vary from student to student. The distribution should be relatively uniform with mean
and standard deviation close to those given in part a.

7.88 a If the sample population is normal, the sampling distribution of x will also be normal (regardless of the
sample size) with mean μ = 1 and standard deviation (or standard error) given as
SE = σ n = .36 5 = .161

b Use the Normal Probabilities for Means applet. Enter the values of μ , σ , n (the warning reminds you
that x must be normal) and x = 1.3 , choosing Area to Right from the dropdown list. The applet shows
P ( x > 1.3) = .0312 . If you choose to use hand calculations, calculate
x −μ 1.3 − 1
z= = = 1.86 , so that
σ n .161
P ( x > 1.3) = P ( z > 1.86 ) = 1 − .9686 = .0314 (the applet uses full decimal accuracy)

183
 c Change the applet to x = .5 and Area to Left from the dropdown list. The applet shows
P ( x < 0.5 ) = 9.0 (10−4 ) = .0009 . If you choose to use hand calculations, calculate
x −μ .5 − 1
z= = = −3.11 , so that
σ n .161
P ( x < 0.5 ) = P ( z < −3.11) = .0009

d Change the applet to x = 1.4 and Two-tails from the dropdown list. The probability that x deviates
from μ = 1 by more than .4 is the probability that x exceeds 1.4 or is less than .6. The applet shows
P ( x > 1.4 ) + P ( x < .6 ) = .013 . If you choose to use hand calculations, calculate
x1 − μ 1.4 − 1 x2 − μ 0.6 − 1
z1 = = = 2.48 and z2 = = = −2.48
σ n .161 σ n .161
Then P ( x > 1.4 ) + P ( x < .6 ) = P ( z > 2.48 ) + P ( z < −2.48 ) = 2 (.0066 ) = .0132

7.89 Use the Normal Probabilities for Means applet. Enter the values of μ , σ , n .
a Enter x = 1100 , choosing Area to Left from the dropdown list. The applet shows P ( x < 1100 ) = .0062.
Since the entire area to the left of 1110 is 0.5, the area between 1100 and 1110 is .5 − .0062 = .4938 .
b Enter x = 1120 , choosing Area to Right from the dropdown list. The applet
shows P ( x > 1120 ) = .0062.
c Enter x = 900 , choosing Area to Left from the dropdown list. The applet shows P ( x < 900 ) = .0000.

Case Study: Sampling the Roulette at Monte Carlo

1 Each bet results in a gain of (– $5) if he loses and $175 if he wins. Thus, the probability distribution of the
gain x on a single $5 bet is

x p(x)
–5 37/38
175 1/38

2 Then
E ( x ) = ∑ xp ( x) = ( −5 )( 37 38 ) + 175 (1 38 ) = −.2632
σ = ∑ x p( x) − μ 2 = ( −5) ( 37 38 ) + (175 ) (1 38 ) − ( −.2632 ) = 830.1939
2 2 2 2 2
x

3 The gain for the evening is the sum S = ∑ xi of the gains or losses for the 200 bets of $5 each. When the
sample size is large, the Central Limit Theorem assures that this sum will be approximately normal with mean
μ S = nμ = 200 ( −.2632 ) = −$52.64
and variance
σ S2 = nσ x2 = 200(830.1939) = 166, 038.78
σ S = 166, 038.78 = 407.48

The total winnings will vary from –$1000 (if the gambler loses all 200 bets) to $35,000 (if the gambler wins all 200
bets), a range of $36,000. However, most of the winnings (95%) will fall in the interval

μ S ± 2σ S = −52.64 ± 2(407.43)

184
or –$867.50 to $762.22. The large gains are highly improbable.

4 The loss of $1000 on any one night will occur only if there are no wins in 200 bets of $5. The probability
200
⎛ 37 ⎞
of this event is ⎜ ⎟ = .005 . Define y to be the number of evenings on which a loss of $1000 occurs. Then y has
⎝ 38 ⎠
a binomial distribution with p = .005 and n = 365 . Using the Poisson approximation to the binomial with
μ = np = 1.825 , the probability of interest is approximately
(1.825) e −1.825
7

p(7) ≈ = .002
7!
which is highly improbable.

5 The largest evening’s winnings, $1160, is not surprising. It lies

1160 − ( −52.64 )
z= = 2.98
407.43
standard deviations above the mean, so that
P ( winnings ≥ 1160 ) = P ( z ≥ 2.98 ) = 1 − .9986 = .0014
for any one evening. The probability of observing winnings of $1160 or greater on one evening out of 365 is then
approximated using the Poisson approximation with μ = 365(.0014) = .511 or
(.511) e−.511
1

p(1) ≈ = .3065
1!

185