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Math 2201
Chapter 5
Final Review
Multiple Choice
____
1. Determine the range of the following test scores. History Test 1 Scores (out of 100)
[Use these test scores for #1 -3.]
90
84
77
66
89
84
77
65
86
82
75
65
86
81
72
61
84
79
70
56
a.
b.
c.
d.
____
2. Determine the mean of the following test scores. History Test 1 Scores (out of 100)
a.
b.
c.
d.
____
79.2
78.5
74.25
76.45
3. Determine the median of the following test scores. History Test 1 Scores (out of 100)
a.
b.
c.
d.
____
34
56
90
78
78
79
56
77
4. Determine the range of the following test scores. History Test 2 Scores (out of 100)
95
92
89
88
85
a.
b.
c.
d.
____
85
84
80
78
73
72
72
70
68
67
62
59
52
40
32
72
95
32
63
5. Environment Canada compiled data on the number of lightning strikes per square kilometre in Alberta and
British Columbia towns from 1999 to 2008.
0.42
0.04
0.81
0.40
0.03
0.74
0.28
0.03
0.70
0.23
0.03
0.66
0.13
0.02
0.61
0.12
0.01
0.58
0.10
0.00
0.49
0.07
1.08
0.43
0.05
0.91
0.42
0.04
0.88
What value goes in the fourth row of this frequency table?
Lightning
Strikes (per
square
Frequency
kilometre)
0.00–0.19
13
0.20–0.39
2
0.40–0.59
6
0.60–0.79
0.80–0.99
3
1.00–1.19
1
a.
b.
c.
d.
5
4
6
3
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Math 2201
____
Chapter 5
Final Review
6. A pear orchard has 40 trees with these heights, given in inches.
110
105
83
84
104
92
95
98
88
92
80
81
115
88
106
92
97
103
100
93
98
93
93
102
92
87
117
92
75
102
83
107
122
92
115
86
89
98
105
125
What value goes in the second row of this frequency table?
Height (in.)
Frequency
70–79
1
80–89
90–99
14
100–109
9
110–119
4
120–129
2
a.
b.
c.
d.
____
7. The range of a set of data is 122 and the minimum value is 87.
To display this data in a histogram, Nat chose intervals of 20 starting with 80–99.
How many intervals will her histogram have?
a.
b.
c.
d.
____
13
11
12
10
7
10
8
9
8. At the end of a bowling tournament, three friends analyzed their scores.
Erinn’s mean bowling score is 92 with a standard deviation of 14.
Declan’s mean bowling score is 130 with a standard deviation of 18.
Matt’s mean bowling score is 116 with a standard deviation of 22.
Who is the more consistent bowler?
a.
b.
c.
d.
____
Declan
Erinn
Impossible to tell.
Matt
9. At the end of a bowling tournament, three friends analyzed their scores.
Lada’s mean bowling score is 125 with a standard deviation of 27.
Quinn’s mean bowling score is 182 with a standard deviation of 28.
Kamal’s mean bowling score is 170 with a standard deviation of 20.
Who is the more consistent bowler?
a.
b.
c.
d.
Lada
Quinn
Kamal
Impossible to tell.
____ 10. A pear orchard has 20 trees with these heights, given in inches.
110
83
104
95
88
80
115
106
97
100
98
93
92
117
75
83
122
115
89
105
Determine the standard deviation, to one decimal place.
a.
b.
c.
d.
15.0 in.
11.0 in.
9.0 in.
13.0 in.
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Math 2201
Chapter 5
Final Review
____ 11. A set of data is normally distributed. What percent of the data is within one standard deviation of the mean?
a.
b.
c.
d.
about 95%
about 68%
100%
about 50%
____ 12. A set of data is normally distributed. What percent of the data is within two standard deviation of the mean?
a.
b.
c.
d.
about 95%
about 50%
100%
about 68%
____ 13. A set of data is normally distributed. What percent of the data is greater than the mean?
a.
b.
c.
d.
100%
about 95%
about 50%
about 68%
____ 14. The ages of participants in a bonspiel are normally distributed, with a mean of 40 and a standard deviation of 10
years. What percent of the curlers are between 40 and 50?
a.
b.
c.
d.
68%
95%
16%
34%
____ 15. The ages of participants in a bonspiel are normally distributed, with a mean of 40 and a standard deviation of 10
years. What percent of the curlers are between 20 and 30?
a.
b.
c.
d.
27%
13.5%
17.5%
32%
____ 16. The ages of participants in a bonspiel are normally distributed, with a mean of 40 and a standard deviation of 10
years. What percent of the curlers are between 30 and 50?
a.
b.
c.
d.
20%
68%
50%
34%
____ 17. The ages of participants in a bonspiel are normally distributed, with a mean of 40 and a standard deviation of 10
years. What percent of the curlers are older than 60?
a.
b.
c.
d.
1.25%
5%
0%
2.5%
____ 18. A teacher is analyzing the class results for a physics test. The marks are normally distributed with a mean (µ) of
76 and a standard deviation (σ) of 4.
Determine Olivia’s mark if she scored µ – σ.
a.
b.
c.
d.
84
68
80
72
____ 19. A teacher is analyzing the class results for a physics test. The marks are normally distributed with a mean (µ) of
76 and a standard deviation (σ) of 4.
Determine Guy’s mark if he scored µ + 2σ.
a.
b.
c.
d.
80
68
72
84
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Math 2201
Chapter 5
Final Review
____ 20. Determine the z-score for the given value.
µ = 120, σ = 10, x = 125
a.
b.
c.
d.
–2
0.5
2
–0.5
____ 21. Determine the z-score for the given value.
µ = 52, σ = 6, x = 64
a.
b.
c.
d.
0.5
2
–0.5
–2
____ 22. Determine the z-score for the given value.
µ = 184, σ = 8.6, x = 174
a.
b.
c.
d.
1.16
–0.86
–1.16
0.86
____ 23. Determine the z-score for the given value.
µ = 510, σ = 93, x = 412
a.
b.
c.
d.
0.95
–0.95
1.05
–1.05
____ 24. Determine the percent of data to the left of the z-score: z = 1.44.
a.
b.
c.
d.
94.95%
92.51%
93.82%
95.91%
____ 25. Determine the percent of data to the left of the z-score: z = –1.50.
a.
b.
c.
d.
6.81%
7.35%
6.68%
8.08%
____ 26. Determine the percent of data to the left of the z-score: z = 0.87.
a.
b.
c.
d.
79.71%
80.78%
77.94%
78.23%
____ 27. Determine the percent of data to the right of the z-score: z = –1.96.
a.
b.
c.
d.
2.50%
97.50%
1.50%
98.50%
____ 28. Determine the percent of data to the right of the z-score: z = –0.08.
a.
b.
c.
d.
21.19%
53.19%
46.81%
78.81%
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Math 2201
Chapter 5
Final Review
____ 29. Determine the percent of data to the right of the z-score: z = 2.26.
a.
b.
c.
d.
1.19%
2.12%
1.58%
0.91%
____ 30. Determine the percent of data between the following z-scores:
z = –1.50 and z = 1.50.
a.
b.
c.
d.
100%
88.82%
94.41%
47.20%
____ 31. Determine the percent of data between the following z-scores:
z = –0.45 and z = –0.15.
a.
b.
c.
d.
76.68%
44.04%
32.64%
11.40%
____ 32. A poll was conducted about an upcoming election. The results are considered accurate within ±2.7 percent
points, 19 times out of 20.
State the confidence level.
a.
b.
c.
d.
99%
19%
90%
95%
____ 33. A poll was conducted about an upcoming election. The results are considered accurate within ±3.8 percent
points, 9 times out of 10.
State the confidence level.
a.
b.
c.
d.
90%
19%
95%
99%
____ 34. A poll was conducted about an upcoming election. The result that 44% of people intend to vote for one of the
candidates is considered accurate within ±2.7 percent points, 19 times out of 20.
State the confidence interval.
a.
b.
c.
d.
41.3%–46.7%
42.65%–45.35%
44%–46.7%
41.3%–44%
____ 35. A poll was conducted about an upcoming election. The result that 71% of people intend to vote for one of the
candidates is considered accurate within ±3.0 percent points, 9 times out of 10.
State the confidence interval.
a.
b.
c.
d.
71%–74%
71%–77%
74%–77%
69.5%–72.5%
____ 36. Which sample size will have the greatest margin of error?
a.
b.
c.
d.
200
100
It is impossible to tell.
50
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Math 2201
Chapter 5
Final Review
____ 37. Which sample size will have the least margin of error?
a.
b.
c.
d.
It is impossible to tell.
1000
3000
2000
____ 38. In a recent survey of high school students, 42% of those surveyed said that the food in the cafeteria was
overpriced. The survey is considered accurate to within 6 percent points, 19 times out of 20.
If a high school has 1000 students, state the range of the number of students who would agree with the survey.
a.
b.
c.
d.
520–640
360–420
420–480
360–480
____ 39. In a recent survey of high school students, 72% of those surveyed agreed that school should start half an hour
later. The survey is considered accurate to within 3.5 percent points, 19 times out of 20.
If a high school has 1200 students, state the range of the number of students who would agree with the survey.
a.
b.
c.
d.
822–864
864–906
822–906
864–948
____ 40. In a recent survey of high school students, one third of those surveyed said they would vote for Melissa as
student council treasurer. The survey is considered accurate to within 5 percent points, 19 times out of 20.
If a high school has 1200 students, state the range of the number of votes Melissa should expect.
a.
b.
c.
d.
300–500
370–430
200–600
340–460
Short Answer
41. Environment Canada compiled data on the number of lightning strikes per square kilometre in Ontario towns
from 1999 to 2008.
3.60
2.11
0.96
0.65
3.47
2.04
0.90
0.65
2.53
1.90
0.85
0.65
2.38
1.90
0.72
0.63
2.25
1.33
0.70
0.38
2.25
1.13
0.66
0.19
Complete the frequency table.
Lightning Strikes
(per square
kilometre)
Frequency
0.00–0.99
1.00–1.99
2.00–2.99
3.00–3.99
42. An apple orchard has 32 trees with these heights, given in inches.
116
90
91
99
114
110
124
102
82
89
104
102
95
105
118
118
110
97
92
93
91
116
101
101
116
86
101
83
117
93
132
104
If the interval width is 5 and starts at 80, what is the last interval?
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Math 2201
Chapter 5
43. Four groups of students recorded their pulse rates after a 2 km run.
Group 1 126 168 158 192 146 166 104 164
Group 2 158 132 156 160 108 150 178 136
Group 3 136 174 156 176 150 166 142 156
Group 4 144 150 142 152 174 176 118 152
Final Review
116
172
130
178
138
140
182
164
172
126
180
128
136
154
166
158
152
130
148
158
128
160
172
166
Determine the standard deviation of Group 2, to one decimal place.
44. Determine the mean of Group 3, to one decimal place. [Use data from #43.]
45. A company measured the lifespan of a random sample of 20 batteries in their MP3 players. Times are in
hours.
7.8
11.0
10.5
8.8
9.0
9.4
10.2
9.4
8.6
9.0
9.3
8.5
7.9
9.1
7.9
9.3
7.4
9.7
10.6
8.5
Determine the standard deviation, to two decimal places.
46. The ages of members in a hiking club are normally distributed, with a mean of 32 and a standard deviation of 6
years. What percent of the members are between 26 and 32?
47. The ages of members in a hiking club are normally distributed, with a mean of 32 years and a standard deviation
of 6 years. What percent of the members are between 32 and 44?
48. The ages of members in a hiking club are normally distributed, with a mean of 32 and a standard deviation of 6
years. What percent of the members are younger than 20?
49. The ages of members in a hiking club are normally distributed, with a mean of 32 and a standard deviation of 6
years. What percent of the members are older than 38?
50. A teacher is analyzing the class results for a computer science test. The marks are normally distributed with a
mean (µ) of 77.4 and a standard deviation (σ) of 4.2.
Determine Kath’s mark if she scored µ – 2σ.
51. A teacher is analyzing the class results for a physics test. The marks are normally distributed with a mean (µ) of
76 and a standard deviation (σ) of 4.
Sketch the normal curve for the test.
52. Determine the z-score for the given value.
µ = 360, σ = 20, x = 315
53. Determine the z-score for the given value.
µ = 9.3, σ = 0.4, x = 8.8
54. Determine the z-score for the given value.
µ = 55, σ = 1, x = 54.6
55. Determine the z-score for the given value.
µ = 52.9, σ = 1.7, x = 58.2
56. Determine the z-score for the given value.
µ = 5.89, σ = 0.53, x = 5.23
57. Determine the percent of data to the left of the z-score: z = 1.05.
58. Determine the percent of data to the left of the z-score: z = –2.20.
59. Determine the percent of data to the left of the z-score: z = 0.71.
60. Determine the percent of data to the right of the z-score: z = –0.19.
61. Determine the percent of data to the right of the z-score: z = –0.68.
62. Determine the percent of data to the right of the z-score: z = 1.60.
June 2014
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Math 2201
Chapter 5
Final Review
63. Determine the percent of data between the following z-scores:
z = –0.68 and z = 1.74.
64. Determine the percent of data between the following z-scores:
z = –0.14 and z = 0.09.
65. Determine the percent of data between the following z-scores:
z = 0.34 and z = 1.70.
66. Determine the percent of data between the following z-scores:
z = –2.99 and z = 1.50.
67. A poll was conducted about an upcoming election. The results are considered accurate within ±4.0 percent
points, 9 times out of 10.
State the confidence level.
68. A poll was conducted about an upcoming election. The results are considered accurate within ±2.8 percent
points, 19 times out of 20.
State the confidence level.
69. The results of a survey have a confidence interval of 29% to 37%, 9 times out of 10.
Determine the margin of error.
70. The results of a survey have a confidence interval of 42.8% to 51.6%, 19 times out of 20.
Determine the margin of error.
71. The results of a survey have a confidence interval of 77.2% to 91.6%, 99 times out of 100.
Determine the margin of error.
72. A poll was conducted about an upcoming election. The result that 18% of people intend to vote for one of the
candidates is considered accurate within ±4.5 percent points, 9 times out of 10.
State the confidence interval.
73. A poll was conducted about an upcoming election. The result that 52% of people intend to vote for one of the
candidates is considered accurate within ±8.0 percent points, 19 times out of 20.
State the confidence interval.
74. A poll was conducted about an upcoming election. The result that 65% of people intend to vote for one of the
candidates is considered accurate within ±4.2 percent points, 9 times out of 10.
State the confidence interval.
75. A poll was conducted about an upcoming election. The result that 46% of people intend to vote for one of the
candidates is considered accurate within ±6.7 percent points, 19 times out of 20.
State the confidence interval.
76. The results of a survey has a confidence interval of 74.9% to 79.1%, 9 times out of 10.
Determine the margin of error.
77. The results of a survey have a confidence interval of 44.8% to 51.6%, 19 times out of 20.
Determine the margin of error.
78. The results of a survey have a confidence interval of 33.4% to 46.0%, 9 times out of 10.
Determine the margin of error.
79. In a recent survey of high school students, 43% of those surveyed agreed that the lunch period should be
extended. The survey is considered accurate to within 4.6 percent points, 19 times out of 20.
If a high school has 1500 students, state the range of the number of students who would agree with the survey.
80. In a recent survey of high school students, 65% of those surveyed said they would vote for Chuck as student
council vice president. The survey is considered accurate to within 8.2 percent points, 19 times out of 20.
If a high school has 2000 students, state the range of the number of votes Chuck should expect.
81. In a recent survey of high school students, 52.4% of those surveyed said they would vote for Sammie as student
council president. The survey is considered accurate to within 1.9 percent points, 19 times out of 20.
If a high school has 800 students, state the range of the number of votes Sammie should expect.
June 2014
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Math 2201
Chapter 5
Final Review
Problem
82. Four groups of students recorded their pulse rates after a 2 km run.
Group 1 126 168 158 192 146 166 104 164
Group 2 158 132 156 160 108 150 178 136
Group 3 136 174 156 176 150 166 142 156
Group 4 144 150 142 152 174 176 118 152
116
172
130
178
138
140
182
164
a) Make a frequency table with five intervals to organize the pulse rates. Show what you did.
b) Which range of pulse rates occurs most frequently? Which occurs least frequently?
83. Billy has 15 min to get to his after-school job. Despite his best efforts, he is frequently late. His employer says
that unless he arrives to work on time consistently, he will lose his job. He has recorded his travel times (in
minutes) for the last two weeks: 11, 13, 19, 18, 13, 15, 17, 17, 20, 16. Over the next two weeks, he continues
to record his travel times: 16, 15, 13, 18, 13, 26, 14, 16, 14, 16.
Do you think Billy will lose his job? Use statistics to justify your answer.
84. The manager of a customer support line currently has 200 unionized employees. Their contract states that the
mean number of calls that an employee should handle per day is 40, with a maximum standard deviation of 8
calls. The manager tracked the number of calls that each employee handles. Does the manager need to hire
more employees if the calls continue in this pattern?
Daily Calls (min)
Frequency
21–25
6
26–30
15
31–35
35
36–40
62
41–45
54
46–50
21
51–55
6
56–60
1
85. Leon keeps track of the amount he spends, in dollars, on weekly lunches during one semester:
25
19
36
19
17
10
24
33
24
28
25
31
28
26
29
26
18
32
a) Determine the range, mean, and standard deviation, correct to two decimal places.
b) Remove the greatest and the least weekly amounts. Then determine the range, mean, and standard deviation
for the remaining amounts.
c) What effect does removing the greatest and the least amounts have on the three values?
86. Yumi always waits until her gas tank is nearly empty before refuelling. She keeps track of the distance she
drives on each tank of gas. The distance varies depending on the weather and the amount she drives on the
highway. The distance has a mean of 520 km and a standard deviation of 14 km.
a) Sketch a normal curve to show the distribution of the driving distances for a tank of gas. Mark the
kilometres driven for values that are 1, 2, and 3 standard deviations from the mean.
b) What percent of the time does Yumi drive between 492 km and 520 km on a tank of gas?
87. Yumi always waits until her gas tank is nearly empty before refuelling. She keeps track of the distance she
drives on each tank of gas. The distance varies depending on the weather and the amount she drives on the
highway. The distance has a mean of 520 km and a standard deviation of 14 km.
a) What percent of the time does she drive between 534 km and 562 km on a tank of gas?
b) Between what two values will she drive 95% of the time?
88. Jackson raises Siberian husky sled dogs at his kennel. He knows, from the data he has collected over the
years, that the masses of adult male dogs are normally distributed, with a mean of 23.6 kg and a standard
deviation of 1.8 kg.
How many of his 87 adult male dogs would you expect to be in the range 21.8 kg to 25.4 kg?
89. Jackson raises Siberian husky sled dogs at his kennel. He knows, from the data he has collected over the
years, that the masses of adult male dogs are normally distributed, with a mean of 23.6 kg and a standard
deviation of 1.8 kg.
Jackson has 48 puppies this year. How many of them could he expect to have a mass greater than 20 kg when
they grow up?
June 2014
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Math 2201
Chapter 5
Final Review
90. A manufacturer offers a warranty on its vacuum cleaners. The vacuum cleaners have a mean lifespan of 3.4
years, with a standard deviation of 0.4 years. For how long should the vacuum cleaners be covered by the
warranty, if the manufacturer wants to repair no more than 2.5% of the vacuum cleaners sold?
91. The result of all the rolls in a game using four dice are recorded.
Sum
Frequency
4–6
5
7–9
13
10–12
31
13–15
35
16–18
29
19–21
11
22–24
2
a) Graph the data. Does the data appear to have a normal distribution?
b) Determine the mean and standard deviation of the data. Do these values validate your answer to part a)?
92. A tile company produces floor tiles that has an average thickness of 55 mm, with a standard deviation of
0.6 mm. For premium-quality floors, the tiles must have a thickness between 54 mm and 55 mm. What
percent, to the nearest whole number, of the total production can be sold for premium-quality floors?
93. A tile company produces glass kitchen tiles that has an average thickness of 71 mm, with a standard deviation
of 0.4 mm. For premium-quality tiles, the tiles must have a thickness between 70 mm and 71.5 mm. What
percent, to the nearest whole number, of the total production can be sold as premium-quality tiles?
94. In a population, 50% of the adults are taller than 172 cm and 10% are taller than 190 cm. Determine the mean
height and standard deviation for this population.
95. In a population, 50% of the adults are taller than 175 cm and 25% are taller than 185 cm. Determine the mean
height and standard deviation for this population.
96. In a population, 80% of the adults are taller than 165 cm and 20% are taller than 187 cm. Determine the mean
height and standard deviation for this population.
97. A manufacturer of plasma televisions has determined that the televisions require servicing after a mean of 62
months, with a standard deviation of 4.5 months. What length of warranty should be offered, if the
manufacturer wants to repair less than 1% of the televisions under the warranty?
98. A manufacturer of computer screens has determined that the screens require servicing after a mean of 70
months, with a standard deviation of 8.8 months. What length of warranty should be offered, if the
manufacturer wants to repair less than 0.5% of the screens under the warranty?
99. A manufacturer of mobile phones has determined that the phones require servicing after a mean of 33 months,
with a standard deviation of 2.4 months. What length of warranty should be offered, if the manufacturer wants
to repair less than 5% of the phones under the warranty?
100. a) Use confidence intervals to interpret the following statement and apply the result to a graduating class of
1400 students.
In a recent survey, 72% of post-secondary graduates indicated that they expected to earn at least $6000/month
by the time they were ready to retire. The survey is considered accurate within ±5.2%, 19 times in 20.
b) How could they increase the confidence level of their result?
101. In a pre-election survey in Calgary, 18% of those surveyed said they were undecided about whom to vote for
in the mayoral election. The survey is considered accurate to within 4.3 percent points, 99 times out of 100.
a) Determine the confidence level and the confidence interval.
b) If there are 680 000 eligible voters in Calgary, state the range of the number of people who are undecided.
June 2014
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Math 2201
Chapter 5
Final Review
Answer - MULTIPLE CHOICE
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A
C
A
D
B
D
A
B
C
D
B
A
C
D
B
B
D
D
D
B
B
C
D
B
C
B
B
B
A
B
D
D
A
A
C
D
C
D
C
B
SHORT ANSWER
41.
ANS:
Lightning Strikes (per
square kilometre)
0.00–0.99
Frequency
12
1.00–1.99
2.00–2.99
3.00–3.99
4
6
2
42.
ANS:
120–124
43.
ANS:
18.6
44.
ANS:
159.6
45.
ANS:
0.95 h
46.
ANS:
34%
47.
ANS:
47.5%
48.
ANS:
2.5%
June 2014
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Math 2201
49.
ANS:
16%
50.
ANS:
69.0
51.
ANS:
52.
ANS:
–2.25
53.
ANS:
–1.25
54.
ANS:
–0.4
55.
ANS:
3.12
56.
ANS:
–1.25
57.
ANS:
85.31%
58.
ANS:
1.39%
59.
ANS:
76.11%
60.
ANS:
57.53%
61.
ANS:
75.17%
62.
ANS:
5.48%
63.
ANS:
71.08%
64.
ANS:
9.16%
65.
ANS:
32.23%
66.
ANS:
93.18%
67.
ANS:
90%
68.
ANS:
95%
69.
ANS:
±4%
70.
ANS:
±4.4%
71.
ANS:
±7.2%
72.
ANS:
13.5% to 22.5%
73.
ANS:
44% to 60%
74.
ANS:
60.8% to 69.2%
75.
ANS:
39.3% to 52.7%
76.
ANS:
2.1%
77.
ANS:
3.4%
June 2014
Chapter 5
Final Review
12
Math 2201
78.
ANS:
6.3%
79.
ANS:
576–714
80.
ANS:
1136–1464
81.
ANS:
404–434
PROBLEM
82.
Chapter 5
Final Review
ANS:
a) The maximum value is 192 and the minimum value is 104, so the range is 88.
If the interval width is 20 and starts at 100, then there will be five intervals and all values are included:
Interval
Frequency
100–119
4
120–139
6
140–159
160–179
180–199
15
13
2
b) The range 140–159 occurs most frequently and the range 180–199 occurs least frequently.
83.
ANS:
Using technology, for the first set of times, the mean is 15.9 min and the standard deviation is about 2.7 min.
For the second set of times, the mean is 16.1 min and the standard deviation is about 3.6 min.
Billy was late the same number of times for both periods. His mean time of arrival and standard deviation got worse. However, If Billy had been on
time the day it took him 15 min to arrive, his mean and standard deviation times would have improved. I think he will lose his job, unless there was a
very good reason for the one day he was 11 min late.
84.
ANS:
Determine the midpoint of each interval.
Midpoint of
Interval (min)
23
Frequency
6
28
15
33
35
38
62
43
54
48
21
53
6
58
1
Using technology, the mean is 38.875 calls and the standard deviation is about 6.6 calls.
The manager does not need to hire more employees if the calls continue in this pattern.
85.
ANS:
a) The maximum value is 33 and the minimum value is 10 so the range is 23.
Using technology, the mean is 25 and the standard deviation is about 6.3.
b) The maximum value is now 32 and the minimum value is now 17 so the range is 15.
Using technology, the mean is about 25.4 and the standard deviation is about 5.2.
c) Removing the greatest and the least amounts decreases the range and the standard deviation. The mean stayed about the same.
86.
ANS:
a)
b) 492 = 520 – 2(14)
Since 520 km is the mean and 492 is two standard deviations below the mean, the percent of the time Judy drives between these two distance is half of
95%, or 47.5%.
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Math 2201
Chapter 5
Final Review
87.
ANS:
a) 534 = 520 + 1(14)
562 = 520 + 3(14)
Since 534 km is one standard deviation above the mean and 562 is three standard deviations above the mean, the percent of the time Judy drives
between these two distance is half of 99.7% – 68%, or 15.85%.
b) She will drive 95% of the time between the two values two standard deviations from the mean.
520 – 2(14) = 492
520 + 2(14) = 548
The two distances are 492 km and 548 km.
88.
ANS:
21.8 = 23.6 – 1(1.8)
25.4 = 23.6 + 1(1.8)
Since 21.8 kg is one standard deviation below the mean and 25.4 kg is one standard deviation above the mean, I would expect 68% of the adult male
dogs to be in this range.
0.68(87) = 59.16
I would expect 59 adult male dogs to be in the range of 21.8 kg to 25.4 kg.
89.
ANS:
20 = 23.6 – 2(1.8)
Since 20 kg is two standard deviations below the mean, I would expect 95% plus half of the remaining 5% of the puppies to be in this range.
95% + 0.5(5%) = 97.5%
0.975(48) = 47
I would expect 47 puppies to have a mass greater than 20 kg when they grow up.
90.
ANS:
Since 2.5% of the vacuum cleaners should have a lifespan less than two standard deviations below the mean, the manufacturer should set that time as
the warranty.
3.4 – 2(0.4) = 2.6
The warranty period should be less than 2.6 years.
91.
ANS:
a) Use the midpoint:
Midpoint of
Interval
5
8
11
14
17
20
23
Frequency
5
13
31
35
29
11
2
The data appears to have a normal distribution.
b) Using technology, the mean is about 13.6 and standard deviation is about 3.9.
Of the 126 rolls, 95 are within one standard deviation of the mean. This is about 75% of the data. Also, 119 of the rolls are within two standard
deviations of the mean. This is about 94% of the data.
Since these percents are so close to the expected values for a normal distribution (68% within one standard deviation and 95% within two standard
deviations), the measures of central tendency validate my answer to part a).
92.
ANS:
Determine the two z-scores:
The z-scores are –1.67 and 0.
Using the z-score table, 50.00% – 4.75% = 45.25% of the data is between these two z-scores.
About 45% of the total production can be sold for premium-quality floors.
93.
ANS:
June 2014
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Math 2201
Chapter 5
Final Review
Determine the two z-scores:
The z-scores are –2.5 and 1.25.
Using the z-score table, 89.44% – 1.62% = 87.82% of the data is between these two z-scores.
About 88% of the total production can be sold as premium-quality tiles.
94.
ANS:
Using the z-score table, 50% or 0.50 corresponds to a z-score of 0.0. This is the mean height of the population. So the mean height is 172 cm.
Then 10% of the population to the right of 190 cm is 90% or 0.90 of the population to the left of 190 cm. Using the z-score table, 0.90 corresponds to a
z-score of 1.28.
The standard deviation is 14.0 cm.
95.
ANS:
Using the z-score table, 50% or 0.50 corresponds to a z-score of 0.0. This is the mean height of the population. So the mean height is 175 cm.
Then 25% of the population to the right of 185 cm is 75% or 0.75 of the population to the left of 185 cm. Using the z-score table, 0.75 corresponds to a
z-score of 0.67.
The standard deviation is 15.0 cm.
96.
ANS:
If 20% of the population is to the left of 165 cm and 20% of the population is to the right of 187 cm, then the mean of these heights is the mean height
of the population:
The mean height is 176 cm.
Then 20% of the population to the right of 187 cm is 80% or 0.80 of the population to the left of the 187 cm. Using the z-score table, 0.80 corresponds
to a z-score of 0.84.
The standard deviation is 13.1 cm.
97.
ANS:
µ = 62 months
σ = 4.5 months
Repair rate, r = 1%
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Math 2201
Chapter 5
Final Review
Using the z-score table, 1% or 0.01 corresponds to a z-score of –2.33.
I will round down so that fewer than 1% of the televisions require repairs under warranty.
The warranty period should be 51 months.
98.
ANS:
µ = 70 months
σ = 8.8 months
Repair rate, r = 0.5%
Using the z-score table, 0.5% or 0.005 corresponds to a z-score of –2.58.
The warranty period should be 47 months.
99.
ANS:
µ = 33 months
σ = 2.4 months
Repair rate, r = 5%
Using the z-score table, 5% or 0.05 corresponds to a z-score of –1.65.
The warranty period should be 29 months.
100.
ANS:
a) 72% – 5.2% = 66.8%
0.668(1400) = 935.2
72% + 5.2% = 77.2%
0.772(1400) =1080.8
With 95% confidence, it can be said that 935 to 1081 of the 1400 graduates expect to earn at least $6000/month by the time they were ready to retire.
b) To increase the confidence level, either increase the sample size or decrease the margin of error.
101.
ANS:
a) The confidence level is 99 times out of 100 or 99%.
18% – 4.3% = 13.7%
18% + 4.3% = 22.3%
The confidence interval is 13.7% to 22.3%.
b) 0.137(680 000) = 93 160
0.223(680 000) = 151 640
The number of people who are undecided should be in the range of 93 160 to 151 640.
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