Download Section 7.2 Homework Answers

 Section 7.2 Homework Answers P ! 0.1226
25.5 30
Sample Mean
___
sum "__n!
25 " 20(1.7)
________
___
b. The two z-scores are z ! _______
" !n ! 1.0!20 "
sum
"
n!
30
"
20(1.7)
________
__
___
"2.012 and z ! _______
!
"
"0.894,
" !n
1.0!20
so the probability is approximately 0.1635 (0.1645
using Table A).
P14. a. The sampling distribution of the sample total
should be approximately normal because of the
very large sample size. It has mean and standard
error
!sum ! n! ! 1000(0.9) ! 900
__
_____
"sum ! " !n ! 1.1!1000 " 34.785
The z-score for 1000 families is then
sum "__n!
z ! _________
"!n
1000 " 900
! __________
34.785
" 2.875
P ! 0.002
900
1000
Sample Total (n ! 1000)
The probability of getting at least 1000 children is
about 0.002. Thus, there is almost no chance the
network will get 1000 children.
b. Yes, the probability goes from practically 0 to
almost certain. The z-score for 1200 families is
sum "__n!
z ! _________
"!n
1000 " 1080
! ___________
38.105
" "2.099
P ! 0.982
1000
1080
Sample Total (n ! 1200)
The probability of getting at least 1000 children in
a random sample of 1200 families is 0.982.
Statistics in Action Instructor’s Guide, Volume 2
© 2008 Key Curriculum Press
Note on P15: You may wish to have each student work
only one part of this question. Then have the students
compare results and notice that as n increases, the range
of reasonably likely outcomes for the sample mean
decreases.
P15. a. 0.9 # 1.96(1.1#!25 ) ! 0.9 # 0.4312, or
(0.469, 1.331)
____
b. 0.9 # 1.96(1.1#!100 ) ! 0.9 # 0.2156, or
(0.684, 1.116)
_____
c. 0.9 # 1.96(1.1#!1000 ) " 0.9 # 0.0682, or
(0.832, 0.968)
_____
d. 0.9 # 1.96(1.1#!4000 ) " 0.9 # 0.0341, or
(0.866, 0.934)
Exercises
E15. a.
I. Histogram B; n ! 25
II. Histogram A; n ! 4
III. Histogram C; n ! 2
__
b. The theoretical standard error is 2.402#!n ,
which turns out to be 1.698, 1.201, and 0.480 for the
respective sample sizes of 2, 4, and 25. All of these
are fairly close to the standard errors estimated
from the simulations.
c. For samples of sizes 2 and 4, the simulated
sampling distributions of the mean reflect the
skewness of the population distribution. For
samples of size 25, the skewness is essentially
eliminated and the simulated sampling
distribution looks like a normal distribution.
d. The rule that about 95% of the observations
lie within two standard errors of the population
mean works well for n ! 25, and slightly less well
for the skewed distributions for n ! 4 and n ! 2.
E16. a. I. Histogram B; n ! 4
II. Histogram C; n ! 25
III. Histogram A; n ! 2
b. The population standard deviation for this
distribution is 3.5. This standard deviation divided
by the square root of 2, 4, and 25, respectively,
yields 2.47, 1.75, and 0.70. These values are quite
close to the observed standard deviations of the
simulated sampling distributions.
c. Despite the new peak centered at the mean,
the simulated sampling distribution for n ! 2 still
reflects much of the pattern of the population,
showing the mounds at the extremes. For n ! 4,
a little of the population pattern remains, but
by n ! 25 it disappears and all that is seen is an
essentially normal distribution.
d. The rule works well for n ! 25, but not nearly
so well for the smaller sample sizes. As usual,
the rule works well for the sampling distribution
of the sample mean as long as the sample size is
reasonably large.
Section 7.2 Solutions
71
Section 7.2 Homework Answers E17. The population is approximately like that in this
table. Students’ estimates will vary. (Many other
assignments for random numbers are possible.)
Value
Percentage
Assignment of
Random Numbers
0
26
01–26
1
18
27–44
2
16
45–60
3
10
61–70
4
8
71–78
5
8
79–86
6
6
87–92
7
4
93–96
8
2
97–98
9
2
99–00
To get a sample of size 5, divide the random digits
into groups of two and use the assignments given
in the third column.
E18. The population is approximately like that in this
table. Students’ estimates will vary. (Many other
assignments for random numbers are possible.)
Value
Percentage
Assignment of
Random Numbers
0
18
01–18
1
14
19–32
2
10
33–42
3
6
43–48
4
2
49–50
5
2
51–52
6
6
53–58
7
10
59–68
8
14
69–82
9
18
83–00
To get a sample of size 5, divide the random digits
into groups of two and use the assignments given
in the third column.
E19. a. No. Exactly two accidents happened in about
15% of the days. Two or more accidents happened
in about 27% of the days.
b. In Display 7.35, the first plot is the one for
8 days, and the second plot is the one for 4 days.
c. No, if the days can be viewed as a random
sample of all days. An average of 1.75 occurred
about 16 times out of 200, so an average of 1.75
accidents is reasonably likely.
d. Yes, if the days can be viewed as a random
sample of all days. An average of 1.75 or more
occurred about 4 times out of 200, so is a rare
event.
72
Section 7.2 Solutions
e. The sampling distributions used in parts a to
c are based on random samples of 4 and 8 days.
The means for consecutive days may not look
like a random sample at all because of the high
dependency from day to day due to seasonal
weather or holiday traffic, for example.
E20. a.
0.25
Relative Frequency
0.20
0.15
0.10
0.05
0
1 2 3 4 5
Exam Score
b. A has n ! 25, B has n ! 1, and C has n !
5. Remind students that repeated sampling with
samples of size 1 should produce a distribution
that looks very much like the population.
c. If the class can be considered a random sample
of the students who took this exam, then an
average of 3.6 would be very unusual for a class
size of 25. A reasonable conclusion is that the
class size is 5. It is always possible, however, that
25 students in a well-taught class would do much
better than a random sample of 25 students.
Note on E21: Because the scores are normally
distributed, students can use the normal approximation
with all sample sizes. Point out that a larger sample size
makes the denominator smaller, which makes the
z-score larger and the probability smaller.
E21. a. The z-score is
510 " 500 ! 0.10
z ! _________
100
which gives a probability of 0.4602.
Alternatively, on a TI-83 Plus or TI-84 Plus
normalcdf 510,1E99,500,100 ! 0.4602.
P ! 0.4602
x ! 510
300
400
500
600
SAT Score (n ! 1)
700
b. The z-score is
510 " 500
__ ! 0.20
z ! _________
100"#4
which gives a probability of 0.4207.
Statistics in Action Instructor’s Guide, Volume 2
© 2008 Key Curriculum Press
Section 7.2 Homework Answers b. The z-score is
P ! 0.4207
x– ! 510
400
0.148 " 0.15
__ $ "1.3333
z ! ___________
0.003!"4
which gives a probability of 0.0912.
450
500
550
Mean Score (n ! 4)
600
c. The z-score is
P ! 0.0912
510 " 500
___ ! 0.50
z ! _________
100!"25
which gives a probability of 0.3085.
x– ! 0.148
0.147 0.1485 0.15 0.1515 0.1530
Mean Weight (n ! 4)
c. The z-score is
P ! 0.3085
x– ! 510
0.148 " 0.15
___ $ "2.1082
z ! ___________
0.003!"10
which gives a probability of 0.0175.
460
480
500
520
540
Mean Score (n ! 25)
d. The probability that one randomly selected
score is 510 or greater is about 0.46. Group the
random digits in pairs and assign the digits 01
through 46 to be a score of 510 or greater. The
other pairs of digits represent a score less than
that. Take four pairs of random digits and see
whether all four represent scores of 510 or greater.
If so, this run is a success. If any of the pairs
represents a score less than 510, this would be
a failure. Repeat this process many times. The
estimate of the probability is the proportion of
runs that are successes.
Note on E21d: The probability can be computed exactly
using the Multiplication Rule for Independent Events:
(0.4602)4 # 0.045.
E22. Because the weights are normally distributed
you can use the normal approximation with all
sample sizes.
a. The z-score is
0.148 " 0.15 $ "0.667
z ! ___________
0.003
which gives a probability of 0.2525.
Alternatively, on a TI-83 Plus or TI-84 Plus
normalcdf−1E99,.148,.15,.003 $ 0.2525.
P ! 0.2525
x ! 0.148
0.144 0.147 0.15 0.153 0.156
Weight (n ! 1)
Statistics in Action Instructor’s Guide, Volume 2
© 2008 Key Curriculum Press
P ! 0.0175
x– ! 0.148
0.1481
0.15
0.1519
Mean Weight (n ! 10)
E23. a. You want the probability that the mean score
will be between 490 and 510. For a mean score of
(490 " 500)
___ # "0.632 and the probability
490, z ! ________
(100!"40 )
that the mean is less than or equal to 490 is about
0.2635.
(510 " 500)
___ # 0.632
For a mean score of 510, z ! ________
(100!"40 )
and the probability that the mean is less than or
equal to 510 is about 0.7365. So the probability
that the mean score is between 490 and 510 is
0.7365 " 0.2635 ! 0.4729.
P ! 0.473
490 500 510
Sample Mean Score
____
Or, normalcdf490,510,500,100!"40 ) $ 0.473.
Note on E23b: This exercise contains material that
will be covered in Chapter 9 of the student book and is
optional for now. You can handle this exercise two ways:
Students can work on finding the method themselves,
or they can be given the formula below. Whichever way
you choose, be sure students understand that the smaller
the interval, the larger the sample size must be. They
should notice that to cut the interval in half, they must
quadruple the sample size.
Section 7.2 Solutions
73
Section 7.2 Homework Answers b. You know that if the sampling distribution is
approximately normal, about 95% of all sample
__
means are in the interval !x_ ! 1.96 "!"n . Thus,
to be 95% sure that the sample mean is within a
value E of the population mean, you must have
__
E ! 1.96 "!"n . Solving for the square root
__
gives "n ! 1.96 "!E. Squaring both sides and
simplifying gives
2
2
2
You need a sample size of about 384.
Note on E24: See the note about E23b.
7 # 6.5
___ " 0.1235 and the
E24. a. The z-score is z " ______
12.8$"10
probability that the mean increase is greater than 7
is about 0.451. There is about a 45.1% chance that
the mean increase in her stock prices
will exceed
___
7%. Or, normalcdf7,1E99,6.5,12$ "10  # 0.451.
P " 0.451
6.5
Sample Mean Increase
__
X # 6.5__
5 # 6.5__
b. Jenny wants P(X # 5) " P ! ______
% ______
"
12.8!"n
12.8!"n "
0.95. Because the sample mean is approximately
normally distributed and the z-score cutting off an
area of 0.95 to the right is #1.645, it must be that
5 # 6.5__ " #1.645
________
12.8!"n
Solving this equation gives n " 197.05. Jenny
should choose about 197 randomly selected stocks.
E25. a. You would expect to see 2 ! 67.4 " 134.8
children.
b. Since the numbers of children seen on a
summer weekend were approximately normally
distributed, the sampling distribution of the mean,
even with n " 2, will be approximately normally
73 # 134.8
__ # #8.40
distributed. The z-score is z " _______
10.4$"2
and the probability is almost 0.
c. Conclude that the low number of children
seen in the emergency room is not due to chance.
Note on E25c: The article goes on to say: “We observed
a significant fall in the numbers of attendees to the
emergency department on the weekends that the two
most recent Harry Potter books were released. Both these
weekends were in mid-summer with good weather.”
E26. a. 16,597,552 ! $103.20 " $1,712,867,366
b. With a sample of 100 people, the sampling
distribution of the mean will be approximately
74
Section 7.2 Solutions
P " 0.248
2
1.96 " " __________
1.96 ! 100 # 384.16
n ! _______
E2
102
__
" 100 ! $103.20 " $10,320 and
normal with !sum____
"sum " $100 ! "100 " $1000.
11,000 # 10,320
The z-score is z " __________
" 0.68 and the
1,000
probability is about 0.248.
10,320
11,000
Sample Mean Increase
0 # 10,320
c. z " _______
" #10.32. P(sum $ 0) is very
1,000
close to zero. There is virtually no chance that the
casino would lose money on a randomly selected
group of 100 customers.
Note on E27: This exercise is difficult and contains
optional material.
E27. a. Solve this problem using the sampling
distribution of the sample sum, which has a mean
__
of 4.3n and a standard error of 1.4 ! "n .
The point that cuts off the lower 0.02 (so 98%
is above it) of the standard normal distribution is
z " #2.054. If the total number of people is above
100, then choose the sample size, n, so that the
point 100 lies 2.054 standard errors below 4.3n, or
__
100 " 4.3n # 2.054 ! 1.4 "n
or
__
4.3n # 2.8756"n # 100 " 0
This equation is quadratic in form and can be
__
solved for "n by using the quadratic formula
and then squaring the positive solution to get n.
__
The negative solution for "n can be ignored.
Alternatively, the solution can be estimated by
plotting the equation on a graphing calculator.
Because you need only the nearest integer solution,
either method works well. The solution is about
n " 27, so the director should select about 27 names.
b. By drawing 27 names, the director expects to
get 4.3 ! 27 " 116.1 people. The 16.1 extra people
will cost him 16.1 ! $250 " $4025. This was a
pretty costly oversight!
Note on E28: This exercise is difficult and contains
optional material.
E28. From Display 7.38, ! " 2.236, " " 1.115.
Solve this problem using the sampling
distribution of the sample sum, which has a
mean of 2.236n and a standard error of
__
1.115 ! "n .
Statistics in Action Instructor’s Guide, Volume 2
© 2008 Key Curriculum Press
Section 7.2 Homework Answers The point that cuts off the lower 0.05 (so 95%
is above it) of the standard normal distribution is
z ! "1.645. If the total number of color television
sets is above 1000, then choose the sample size, n,
so that the point 1000 lies 1.645 standard errors
below 2.235n, or
__
1000 ! 2.236n " 1.645 ! 1.115 !n
c.
or
__
2.236n " 1.834!n " 1000 ! 0
This equation is quadratic in form and can be
__
solved for !n by using the quadratic formula
and then squaring the positive solution to get n.
__
The negative solution for !n can be ignored.
Alternatively, the solution can be estimated by
plotting the equation on a graphing calculator.
Because you need only the nearest integer
solution, either method works well. The solution
is about n ! 465 households.
E29. a. The mean of the sampling distribution of the
sample mean is equal to the population mean !
for all sample sizes. However, as the sample size
increases, the standard error of the sampling
distribution of the sample mean decreases by
a factor of 1 divided by the square root of n.
__
Specifically, !x_ ! ! and "_x ! ""!n .
b. The mean of the sampling distribution
of the sample total increases by a factor of
n. The standard error of the sampling distribution
of the sample mean increases by a factor of the
square root of n. Specifically, !sum ! n! and
__
"sum ! !n ! ".
E30. a. As long as n is greater than 1, the numerator
will be smaller than
the denominator, so
____
N"n
multiplying by !____
N " 1 will decrease the standard
error. This makes sense when sampling because
extreme values of the mean are more difficult
to get when you sample without replacement.
For example, consider selecting two numbers
from the set {1, 2, 3, 4, 5}. With replacement it is
possible to draw 1 twice, for a mean of 1. Without
replacement the smallest mean you can get is 1.5
from drawing 1 and 2. The distribution becomes
less spread out.
_
b. ! ! 76.6, " ! 13.74
Remaining Test Scores
Mean
55, 65, 75, 80
68.75
55, 65, 75, 90
71.25
55, 65, 75, 95
72.5
55, 65, 80, 90
72.5
55, 65, 80, 95
73.75
55, 65, 90, 95
76.25
55, 75, 80, 90
75
55, 75, 80, 95
76.25
55, 75, 90, 95
78.75
55, 80, 90, 95
80
65, 75, 80, 90
77.5
65, 75, 80, 95
78.75
65, 75, 90, 95
81.25
65, 80, 90, 95
82.5
75, 80, 90, 95
85
The sampling distribution is shown here.
68
70
72
74
76
78
Mean Score
80
82
84
86
d. The mean of the sampling
distribution of the
_
sample mean, !x_ , is 76.6, the same as !.
e. Mean Score x ! x " ux_ "2
68.75
62.6736
71.25
29.3403
72.50
17.3611
72.50
17.3611
73.75
8.5069
76.25
0.1736
75.00
2.7778
76.25
0.1736
78.75
4.3403
80.00
11.1111
77.50
0.6944
78.75
4.3403
81.25
21.0069
82.50
34.0278
85.00
69.4444
Sum
283.3
________
!283.3"15 # 4.346
13.74
"__
__ !
! ____
f. The first formula gives "_x ! ___
!n
!4
6.87.
____
N"n
"__
____
The____
second formula gives "_x ! ___
!
!n ! N " 1
13.74
6"4
____
____
__
#
4.345.
!4 ! 6 " 1
Statistics in Action Instructor’s Guide, Volume 2
© 2008 Key Curriculum Press
Section 7.2 Solutions
75
Section 7.2 Homework Answers g. The second formula gives the correct standard
error for sampling without replacement. (The
small difference is purely due to rounding error.)
E31. a. !v!m " !v ! !m " 503 ! 518 " 1021
_______
__________
"v!m " !"v2 ! "m2 " !1132 ! 1152
" 161.23
b. The sampling distribution of the sum will be
approximately normal because both distributions
are normal. The z-score is
800 # 1021 " #1.371
z " __________
161.23
e. The mean of the distribution of the sum will
still be 1021. The standard error is unpredictable
because the two scores are certainly not
independent. Students who score high on the
critical reading section also tend to score high on
the math section. The shape is also unpredictable.
E32. a. !m!f " !m ! !f " 69.3 ! 64.1 " 133.4 inches
"m!f "
_______
2
2
___________
!"m ! "f " !2.92
2
!2.752
" 4.01 inches
b. The distributions of heights of both males
and females are approximately normal, so the
distribution of the sum will be as well.
(125 # 133.4)
z " ____________ " #2.09
4.01
P ! 0.0852
sum ! 800
699
860 1021 1182 1343
Sum of Scores (n ! 2)
The probability is about 0.0852 (0.0853 using
Table A).
c. 1021 $ 1.96(161.23), or between
approximately 705 and 1337
d. You need c # m " 100. The sampling
distribution of the difference will be
approximately normal because both distributions
are normal. The mean and standard error of the
sampling distribution of the difference are
!c#m " !c # !m " 503 # 518 " #15
_______
__________
"c#m " !" c2 ! "m2 " !1132 ! 1152
P ! 0.018
sum ! 125
125.4
129.4 133.4 137.4 141.4
Sum of Heights (n ! 2)
P(sum # 125) " 0.018
c. The middle 95% of sums will be between
133.4 $ 1.96(4.01), or approximately between
125.54 and 141.26.
d. You want P(male height # female height " 2).
The distribution of the difference has mean !m#f "
!m # !f " 69.3 # 64.1 " 5.2 inches. "m#f " 4.01.
(2 # 5.2)
z " ________ " #0.798
4.01
" 161.23
The z-score for a difference of 100 is
100 # (#15)
z " ___________ " 0.713
161.23
P ! 0.2378
difference ! 100
–337 –176 –15 146
307
Difference of Scores (n ! 2)
The probability is 0.2378 (0.2389 using
Table A).
Note on E31d: This does not imply that 23.78% of
students have an SAT critical reading score of at least
100 points higher than their SAT math score. See part e.
76
Section 7.2 Solutions
difference ! 2
P ! 0.788
–2.8
1.2
5.2
9.2
13.2
Difference of Heights (n ! 2)
P(difference " 2) " 0.788
e. The mean of the distribution of the sum
will still be 133.4 inches. The standard error is
unpredictable because the two scores are certainly
not independent. Relatives of taller than average
people tend to also be taller than average. The
shape is also unpredictable.
E33. a. !sum " 3(3.5) " 10.5; "2sum " 3(2.917) " 8.75
b. !sum " 7(3.5) " 24.5; "2sum " 7(2.917) "
20.419
c. The sampling distribution of the sum is
approximately normal because E31 says that
Statistics in Action Instructor’s Guide, Volume 2
© 2008 Key Curriculum Press
Section 7.2 Homework Answers the distribution of critical reading scores is
approximately normal. It has mean and variance
!sum ! 20 ! 503 ! 10,060
"2sum ! n"2 ! 20 ! 1132 ! 255,380
The z-score for a total of 10,000 is
10,000_______
" 10,060
z ! ______________
" "0.1187
!255,380
sum ! 10,000
P ! 0.4527
9,050
10,060
11,070
Sum of Scores (n ! 20)
The probability is about 0.4527 (0.4522 using
Table A).
E34. a. Because the distribution of the weights is
approximately normal, the sampling distribution
of the sum of two (or any number of) weights will
be approximately normal as well. The mean and
standard error are
The z-score for a total of 750 is
750 " 754 " "0.1714
z ! _________
23.338
The probability that the total weight is more than
750 kg is 0.5680.
707.3 730.7 754 777.3 800.7
Sum of Weights (n ! 10)
d. The first ram is more than 5 kg heavier than
the second ram if ram1 " ram2 $ 5. The sampling
distribution of the difference ram1 " ram2 has
mean !1"2 ! !1 " !2 ! 75.4 " 75.4 ! 0.
The standard error is the same as that in part a,
10.437.
The z-score for a difference of 5 is
5 " 0 " 0.4791
z ! ______
10.437
!sum ! 75.4 # 75.4 ! 150.8
P ! 0.3159
___________
"sum ! !7.382 # 7.382 " 10.437
b. Using the results from part a, the z-score is
"0.556 and the probability is 0.2892.
P ! 0.5680
sum ! 750
difference ! 5
–20.9 –10.4
0
10.4
Difference (n ! 2)
20.9
The probability that the difference is greater than
5 is about 0.3159.
P ! 0.2892
sum ! 145
129.9 140.4 150.8 161.2 171.7
Sum of Weights (n ! 2)
c. Using the formulas in E33, the mean and
standard error of the sampling distribution are
!sum ! n! ! 10(75.4) ! 754
____
________
"sum ! !n"2 ! !10(7.38)2 " 23.338
Statistics in Action Instructor’s Guide, Volume 2
© 2008 Key Curriculum Press
Section 7.2 Solutions
77