Download Unit 2, Quiz 1 - Union Academy Charter School

Name: ________________________ Class: ___________________ Date: __________
ID: A
Unit 2, Quiz 1 (Practice Quiz)
____
1. Researchers randomly choose two groups from 15 volunteers. Over a period of 9 weeks, one group
watches television before going to sleep, and the other does not. Volunteers wear monitoring devices while
sleeping, and researchers record dream activity. Which type of study method is described in each situation?
a. observational study
c. controlled experiment
b. survey
Identify the type of sampling method used.
2. A trucking company places a card with their office phone number on the door step of every home within 5
miles of their office.
3. A candidate for the Senate creates an automated message that calls every third listed phone number and
reminds them to vote for him in the upcoming election.
Find the mean, median, and mode of the data set. Round to the nearest tenth.
____
4. 2, 10, 6, 9, 1, 15, 11, 10, 15, 13, 15
a. mean = 8.9,
b. mean = 9.7,
median = 10,
median = 8,
mode = 15
mode =15
c.
mean = 9.7,
median = 10,
mode = 15
d.
mean = 8.9,
median = 10,
mode = 8
5. 3.4, 4.8, 3.1, 0.2, 6.9, 5.5, 6.6, 5.1
a. 3.1
b. 5.1
c.
0.2
d.
5.1
6. 17, 13, 16, 18, 38, 14, 21, 24, 30, 8, 61, 2
a. no outliers
b. 2
c.
61
d.
38 and 61
Find the outlier(s) in the set of data.
____
____
____
7. Over the first five years of owning her car, Gina drove about 12,700 miles the first year, 15,478 miles the
second year, 12,675 the third year, 11,850 the fourth year, and 13,075 the fifth year.
a. Find the mean, median, and mode of this data.
b. Explain which measure of central tendency will best predict how many miles Gina will drive in the sixth
year.
a. mean = 13,156; median = 12,700; no mode; the median is the best choice because it is
not skewed by the high outlier.
b. mean = 13,156; median = 12,700; mode = 3,628; the median is the best choice because it
is not skewed by the high outlier.
c. mean = 12,700; median = 13,156; no mode; the mean is the best choice because it is
representative of the entire data set.
d. mean = 13,156; median = 12,700; no mode; the mean is the best choice because it is
representative of the entire data set.
1
Name: ________________________
ID: A
Make a box-and-whisker plot of the data.
____
8. 24, 18, 29, 21, 16, 23, 13, 11
a.
b.
c.
d.
9. Average daily temperatures in Tucson, Arizona, in December:
67, 57, 52, 51, 64, 58, 67, 58, 55, 59, 66, 50, 57, 62, 58, 50, 58, 50, 60, 63
What are the mean, variance, and standard deviation of these values? Round to the nearest tenth.
____ 10. 1, 9, 4, 12, 13, 13
a. mean = 10.5
variance = 2.2;
standard deviation = 4.6
b. mean = 8.7
variance = 2.2;
standard deviation = 5.1
c.
d.
11. 92, 97, 53, 90, 95, 98
2
mean = 10.5
variance = 21.6;
standard deviation = 5.1
mean = 8.7
variance = 21.6;
standard deviation = 4.6
Name: ________________________
ID: A
____ 12.
a.
b.
x
x−x
53
51
48
49
37
5.4
3.4
0.4
1.4
–10.6
(x − x) 2
29.2
11.6
0.2
2
112.4
mean = 208.4
variance = 31
standard deviation = 5.6
mean = 47.6
variance = 65.3
standard deviation = 5.6
c.
d.
mean = 47.6
variance = 31
standard deviation = 5.6
mean = 47.6
variance = 31
standard deviation = 963.5
Use a calculator to find the mean and standard deviation of the data. Round to the nearest tenth.
____ 13. 946, 726, 956, 519, 104, 415, 428, 457, 614, 201, 772, 801
a. mean = 566.5;
c. mean = 566.5;
standard deviation = 68572.4
standard deviation = 261.9
b. mean = 578.3;
d. mean = 578.3;
standard deviation = 68572.4
standard deviation = 261.9
14. The height (in feet) of a sample of trees in the school playground:
12.5, 9.8, 13.5, 11.2, 12.3, 14.2, 11.7, 9.8, 12.6, 10.4
____ 15. A stem and leaf display is shown below. The stem is the tens place and the leaf is
the units place.
a.
b.
57
59
c.
d.
3
62
78
ID: A
Unit 2, Quiz 1 (Practice Quiz)
Answer Section
1. ANS: C
PTS: 1
DIF: L2
NAT: CC S.IC.1| CC S.IC.3| CC S.IC.4| CC S.IC.6| D.3.a| D.3.b| D.3.c| D.3.d| D.5.a
TOP: 1-3 Problem 2 Analyzing Study Methods
KEY: observational study | controlled experiment | survey
2. ANS:
convenience sample
PTS: 1
DIF: L3
NAT: CC S.IC.1| CC S.IC.3| CC S.IC.4| CC S.IC.6| D.3.a| D.3.b| D.3.c| D.3.d| D.5.a
TOP: 1-3 Problem 1 Analyzing Sampling Methods
KEY: sample | bias | self-selected sample
3. ANS:
systematic sample
4.
5.
6.
7.
8.
9.
PTS:
NAT:
TOP:
ANS:
NAT:
TOP:
KEY:
ANS:
NAT:
TOP:
ANS:
NAT:
TOP:
ANS:
NAT:
KEY:
ANS:
NAT:
TOP:
KEY:
ANS:
PTS:
TOP:
KEY:
10. ANS:
TOP:
1
DIF: L3
CC S.IC.1| CC S.IC.3| CC S.IC.4| CC S.IC.6| D.3.a| D.3.b| D.3.c| D.3.d| D.5.a
1-3 Problem 1 Analyzing Sampling Methods
KEY: sample | bias | systematic sample
C
PTS: 1
DIF: L2
CC S.IC.6| D.1.a| D.1.b| D.2.c| D.1.e| D.2.a
1-1 Problem 1 Finding Measures of Central Tendency
measure of central tendency | mean | median | mode
C
PTS: 1
DIF: L2
CC S.IC.6| D.1.a| D.1.b| D.2.c| D.1.e| D.2.a
1-1 Problem 2 Identifying an Outlier
KEY: outlier
C
PTS: 1
DIF: L3
CC S.IC.6| D.1.a| D.1.b| D.2.c| D.1.e| D.2.a
1-1 Problem 2 Identifying an Outlier
KEY: outlier
A
PTS: 1
DIF: L3
CC S.IC.6| D.1.a| D.1.b| D.2.c| D.1.e| D.2.a
TOP: 1-1 Problem 3 Comparing Data Sets
measure of central tendency | mean | median | mode | range of a set of data | interquartile range
A
PTS: 1
DIF: L2
CC S.IC.6| D.1.a| D.1.b| D.2.c| D.1.e| D.2.a
1-1 Problem 4 Using a Box-and-Whisker Plot
median | quartile | box-and-whisker plot
1
DIF: L3
NAT: CC S.IC.6| D.1.a| D.1.b| D.2.c| D.1.e| D.2.a
1-1 Problem 4 Using a Box-and-Whisker Plot
median | quartile | box-and-whisker plot
D
PTS: 1
DIF: L3
NAT: CC S.ID.4| CC S.IC.6| D.1.c
1-2 Problem 1 Finding Variance and Standard Deviation KEY: standard deviation | variance
1
ID: A
11. ANS:
mean = 87.5
variance = 245.6;
standard deviation = 15.7
PTS: 1
DIF: L3
NAT: CC S.ID.4| CC S.IC.6| D.1.c
TOP: 1-2 Problem 1 Finding Variance and Standard Deviation KEY: standard deviation | variance
12. ANS: C
PTS: 1
DIF: L2
NAT: CC S.ID.4| CC S.IC.6| D.1.c
TOP: 1-2 Problem 1 Finding Variance and Standard Deviation KEY: standard deviation | variance
13. ANS: D
PTS: 1
DIF: L3
NAT: CC S.ID.4| CC S.IC.6| D.1.c
TOP: 1-2 Problem 2 Using a Calculator to Find Standard Deviation
KEY: mean | standard deviation
14. ANS:
mean = 11.8 ft;
standard deviation = 1.43 ft
PTS:
TOP:
KEY:
15. ANS:
1
DIF: L4
NAT: CC S.ID.4| CC S.IC.6| D.1.c
1-2 Problem 2 Using a Calculator to Find Standard Deviation
mean | standard deviation
B
PTS: 1
2