Math 2201 – Review Unit 5
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. Determine the range of the following test scores.
History Test 1 Scores (out of 100)
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)
90
84
77
66
89
84
77
65
86
82
75
65
86
81
72
61
84
79
70
56
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)
90
84
77
66
89
84
77
65
86
82
75
65
86
81
72
61
84
79
70
56
a.
b.
c.
d.
____
34
56
90
78
78
79
56
77
4. 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
kilometre)
Frequency
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. The range of a set of data is 7.5 and the minimum value is 2.8.
To display this data in a histogram, Yul chose intervals of 1.5 starting with 2.5–3.9.
How many intervals will his histogram have?
a.
b.
c.
d.
____
5
4
6
3
7
4
6
5
6. 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
Which range of data occurs least frequently?
a.
b.
c.
d.
____
0.60–0.69
0.70–0.79
0.50–0.59
0.40–0.49
7. 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 had the highest scoring game during the tournament?
a. Declan
b. Matt
c. Impossible to tell.
d. Erinn
____
8. 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.
9. Chinedu recorded the time it takes him to get to school using three different routes.
Hour
Route 1 (min)
Route 2 (min)
Route 3 (min)
1
13
20
16
2
15
18
17
3
12
20
15
4
12
12
17
5
16
17
22
On which route does Chinedu have a more consistent travel time?
a. Route 3
b. Route 1
c. Route 2
____ 10. 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%
____ 11. Which description does not describe the normal curve?
a.
b.
c.
d.
starts off increasing
symmetrical
shaped like a bell
always increasing
____ 12. 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%
____ 13. Which set is normally distributed?
0–9
10–19
Interval
20–29
30–39
40–49
50–59
Set A.
Set B.
Set C.
Set D.
a.
b.
c.
d.
100
800
950
400
500
750
420
620
850
700
180
760
820
650
220
820
450
600
460
900
150
550
990
850
Set C.
Set B.
Set D.
Set A.
____ 14. Determine the z-score for the given value.
µ = 52,  = 6, x = 64
a.
b.
c.
d.
0.5
2
–0.5
–2
____ 15. 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%
____ 16. 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%
____ 17. A poll was conducted about an upcoming election. The result that 54% of people intend to vote for one of the
candidates is considered accurate within ±7.1 percent points, 19 times out of 20.
State the confidence interval.
a.
b.
c.
d.
54%–61.1%
47.1%–60.9%
46.9%–61.1%
46.9%–54%
____ 18. The results of a survey have a confidence interval of 56.0% to 64.6%, 9 times out of 10.
Determine the margin of error.
a.
b.
c.
d.
±64.6%
±16.6%
±8.3%
±56.0%
____ 19. 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
____ 20. 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 () of 4.
Determine Guy’s mark if he scored µ + 2.
a.
b.
c.
d.
80
68
72
84
Short Answer
21. A teacher is analyzing the class results for a computer science test. The marks are normally distributed with a mean
(µ) of 79.5 and a standard deviation () of 3.5.
Determine Daryl’s mark if he scored µ + .
22. 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.
23. Khamid and Gerbrand are laying interlocking bricks. Their supervisor records how many bricks they lay each
hour.
1
2
3
4
5
6
Hour
212
193
204
195
182
216
Khamid
230
195
214
207
218
191
Gerbrand
Which worker is more consistent?
Problem
24. 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
a) Make a frequency table with five intervals to organize the pulse rates.
b) Construct a histogram of the data.
138
140
182
164
25. 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?
26. In a pre-election survey in Vancouver, 16.5% of those surveyed said they were undecided about whom to vote
for in the mayoral election. The survey is considered accurate to within 5.2 percent points, 19 times out of 20.
a) Determine the confidence level and the confidence interval.
b) If there are 425 000 eligible voters in Vancouver, state the range of the number of people who are undecided.
27. 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?
28. 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?
29. 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.
Math 2201 - ICA ch. 5
Answer Section
MULTIPLE CHOICE
1. ANS: A
PTS: 1
DIF: Grade 11
REF: Lesson 5.1
TOP: Exploring data
KEY: range
2. ANS: C
PTS: 1
DIF: Grade 11
REF: Lesson 5.1
TOP: Exploring data
KEY: mean
3. ANS: A
PTS: 1
DIF: Grade 11
REF: Lesson 5.1
TOP: Exploring data
KEY: median
4. ANS: B
PTS: 1
DIF: Grade 11
REF: Lesson 5.2
TOP: Frequency tables, histograms, and frequency polygons
KEY: frequency distribution | histogram | frequency polygon
5. ANS: C
PTS: 1
DIF: Grade 11
REF: Lesson 5.2
TOP: Frequency tables, histograms, and frequency polygons
KEY: frequency distribution | histogram
6. ANS: C
PTS: 1
DIF: Grade 11
REF: Lesson 5.2
TOP: Frequency tables, histograms, and frequency polygons
KEY: frequency distribution
7. ANS: C
PTS: 1
DIF: Grade 11
REF: Lesson 5.3
OBJ: 1.3 Explain, using examples, the properties of a normal curve, including the mean, median, mode,
standard deviation, symmetry and area under the curve.
TOP: Standard deviation
KEY: mean | standard deviation
8. ANS: D
PTS: 1
DIF: Grade 11
REF: Lesson 5.3
OBJ: 1.2 Calculate, using technology, the population standard deviation of a data set. | 1.3 Explain, using
examples, the properties of a normal curve, including the mean, median, mode, standard deviation, symmetry
and area under the curve.
TOP: Standard deviation
KEY: standard deviation
9. ANS: C
PTS: 1
DIF: Grade 11
REF: Lesson 5.3
OBJ: 1.2 Calculate, using technology, the population standard deviation of a data set. | 1.3 Explain, using
examples, the properties of a normal curve, including the mean, median, mode, standard deviation, symmetry
and area under the curve.
TOP: Standard deviation
KEY: standard deviation
10. ANS: B
PTS: 1
DIF: Grade 11
REF: Lesson 5.4
OBJ: 1.3 Explain, using examples, the properties of a normal curve, including the mean, median, mode,
standard deviation, symmetry and area under the curve.
TOP: The normal distribution
KEY: normal distribution | mean | standard deviation
11. ANS: D
PTS: 1
DIF: Grade 11
REF: Lesson 5.4
OBJ: 1.3 Explain, using examples, the properties of a normal curve, including the mean, median, mode,
standard deviation, symmetry and area under the curve.
TOP: The normal distribution
KEY: normal curve
12. ANS: D
PTS: 1
DIF: Grade 11
REF: Lesson 5.4
OBJ: 1.7 Solve a contextual problem that involves the interpretation of standard deviation. | 1.9 Solve a
contextual problem that involves normal distribution.
TOP: The normal distribution
KEY: normal distribution | mean | standard deviation
13. ANS: D
PTS: 1
DIF: Grade 11
REF: Lesson 5.4
OBJ: 1.4 Determine if a data set approximates a normal distribution, and explain the reasoning.
TOP: The normal distribution
KEY: normal distribution
14. ANS: B
PTS: 1
DIF: Grade 11
REF: Lesson 5.5
15.
16.
17.
18.
19.
20.
OBJ: 1.8 Determine, with or without technology, and explain the z-score for a given value in a normally
distributed data set. TOP: Applying the normal distribution: z-scores
KEY: z-score | standard normal distribution
ANS: B
PTS: 1
DIF: Grade 11
REF: Lesson 5.5
OBJ: 1.8 Determine, with or without technology, and explain the z-score for a given value in a normally
distributed data set. TOP: Applying the normal distribution: z-scores
KEY: z-score | standard normal distribution
ANS: D
PTS: 1
DIF: Grade 11
REF: Lesson 5.6
OBJ: 2.1 Explain, using examples, how confidence levels, margin of error and confidence intervals may vary
depending on the size of the random sample. | 2.2 Explain, using examples, the significance of a confidence
interval, margin of error or confidence level.
TOP:
Confidence intervals
KEY: margin of error | confidence interval | confidence level
ANS: C
PTS: 1
DIF: Grade 11
REF: Lesson 5.6
OBJ: 2.1 Explain, using examples, how confidence levels, margin of error and confidence intervals may vary
depending on the size of the random sample. | 2.2 Explain, using examples, the significance of a confidence
interval, margin of error or confidence level.
TOP:
Confidence intervals
KEY: margin of error | confidence interval | confidence level
ANS: C
PTS: 1
DIF: Grade 11
REF: Lesson 5.6
OBJ: 2.1 Explain, using examples, how confidence levels, margin of error and confidence intervals may vary
depending on the size of the random sample. | 2.2 Explain, using examples, the significance of a confidence
interval, margin of error or confidence level.
TOP:
Confidence intervals
KEY: margin of error | confidence interval | confidence level
ANS: C
PTS: 1
DIF: Grade 11
REF: Lesson 5.6
OBJ: 2.1 Explain, using examples, how confidence levels, margin of error and confidence intervals may vary
depending on the size of the random sample. | 2.2 Explain, using examples, the significance of a confidence
interval, margin of error or confidence level.
TOP:
Confidence intervals
KEY: margin of error | confidence interval | confidence level
ANS: D
PTS: 1
DIF: Grade 11
REF: Lesson 5.4
OBJ: 1.7 Solve a contextual problem that involves the interpretation of standard deviation. | 1.9 Solve a
contextual problem that involves normal distribution.
TOP: The normal distribution
KEY: normal distribution | mean | standard deviation
SHORT ANSWER
21. ANS:
83.0
PTS: 1
DIF: Grade 11
REF: Lesson 5.4
OBJ: 1.7 Solve a contextual problem that involves the interpretation of standard deviation. | 1.9 Solve a
contextual problem that involves normal distribution.
TOP: The normal distribution
KEY: normal distribution | mean | standard deviation
22. ANS:
±7.2%
PTS: 1
DIF: Grade 11
REF: Lesson 5.6
OBJ: 2.1 Explain, using examples, how confidence levels, margin of error and confidence intervals may vary
depending on the size of the random sample. | 2.2 Explain, using examples, the significance of a confidence
interval, margin of error or confidence level.
TOP:
Confidence intervals
KEY: margin of error | confidence level | confidence interval
23. ANS:
Khamid has the lower standard deviation.
PTS: 1
DIF: Grade 11
REF: Lesson 5.3
OBJ: 1.2 Calculate, using technology, the population standard deviation of a data set. | 1.3 Explain, using
examples, the properties of a normal curve, including the mean, median, mode, standard deviation, symmetry
and area under the curve.
TOP: Standard deviation
KEY: standard deviation
PROBLEM
24. ANS:
a)
Interval
100–119
120–139
140–159
160–179
180–199
Frequency
4
6
15
13
2
b)
PTS: 1
DIF: Grade 11
REF: Lesson 5.2
TOP: Frequency tables, histograms, and frequency polygons
25. ANS:
Determine the two z-scores:
KEY: frequency distribution | histogram
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.
PTS: 1
DIF: Grade 11
REF: Lesson 5.5
OBJ: 1.7 Solve a contextual problem that involves the interpretation of standard deviation. | 1.8 Determine,
with or without technology, and explain the z-score for a given value in a normally distributed data set. | 1.9
Solve a contextual problem that involves normal distribution.
TOP: Applying the normal distribution: z-scores
KEY: normal distribution | mean | standard deviation | z-score
26. ANS:
a) The confidence level is 19 times out of 20 or 95%.
16.5% – 5.2% = 11.3%
16.5% + 5.2% = 21.7%
The confidence interval is 11.3% to 21.7%.
b) 0.113(425 000) = 48 025
0.217(425 000) = 92 225
The number of people who are undecided should be in the range of 48 025 to 92 225.
PTS: 1
DIF: Grade 11
REF: Lesson 5.6
OBJ: 1.6 Explain, using examples that represent multiple perspectives, the application of standard deviation
for making decisions in situations such as warranties, insurance or opinion polls. | 1.9 Solve a contextual
problem that involves normal distribution. | 2.3 Make inferences about a population from sample data, using
given confidence intervals, and explain the reasoning.
TOP: Confidence intervals
KEY: mean | range | standard deviation | margin of error | confidence level | confidence interval
27. 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.
PTS: 1
DIF: Grade 11
REF: Lesson 5.3
OBJ: 1.2 Calculate, using technology, the population standard deviation of a data set. | 1.3 Explain, using
examples, the properties of a normal curve, including the mean, median, mode, standard deviation, symmetry
and area under the curve. | 1.5 Compare the properties of two or more normally distributed data sets.
TOP: Standard deviation
KEY: mean | median | range | standard deviation
28. 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.
PTS: 1
DIF: Grade 11
REF: Lesson 5.4
OBJ: 1.3 Explain, using examples, the properties of a normal curve, including the mean, median, mode,
standard deviation, symmetry and area under the curve. | 1.7 Solve a contextual problem that involves the
interpretation of standard deviation. | 1.9 Solve a contextual problem that involves normal distribution.
TOP: The normal distribution
KEY: normal distribution | mean | standard deviation
29. 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.
PTS: 1
DIF: Grade 11
REF: Lesson 5.5
OBJ: 1.7 Solve a contextual problem that involves the interpretation of standard deviation. | 1.8 Determine,
with or without technology, and explain the z-score for a given value in a normally distributed data set. | 1.9
Solve a contextual problem that involves normal distribution.
TOP: Applying the normal distribution: z-scores
KEY: normal distribution | mean | standard deviation | z-score