Download Alg 1 Unit 3 Formative Assessments Items

Algebra I Items to Support Formative Assessment
Unit 4: Descriptive Statistics
S.ID.A.2 Use statistics appropriate to the shape of the data distribution to compare center
(median, mean) and spread (interquartile range, standard deviation) of two or more
different sets
S.ID.A.2 Task
1. Complete the table for each company.
Tiny Tech Corporate Structure
Job Title
Number of
Employees
Salary (In Thousands) Total (In Thousands)
CEO
1
75
CFO
2
60
Office Manager
2
60
Tech Support Tech
4
45
Office Staff
8
30
Production Staff
12
25
Cleaners
3
15
Total
Adventure Web Corporate Structure
Job Title
Number of
Employees
Salary (In Thousands) Total (In Thousands)
CEO
1
85
CFO
1
70
Office Manager
2
65
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Tech Support Tech
2
50
Office Staff
4
40
Production Staff
8
27
Cleaners
2
17
Total
Tiny Tech Corporate Structure
Job Title
Number of
Employees
Salary (In Thousands) Total (In Thousands)
CEO
1
75
75
CFO
2
60
120
Office Manager
2
60
120
Tech Support Tech
4
45
180
Office Staff
8
30
240
Production Staff
12
25
300
Cleaners
3
15
45
Total
32
1,080
Adventure Web
Job Title
Number of
Employees
Salary
Total (In Thousands)
CEO
1
85
85
CFO
1
70
70
Office Manager
2
65
130
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Tech Support Tech
2
50
100
Office Staff
4
40
160
Production Staff
8
27
216
Cleaners
2
17.5
35
Total
20
796
2. Calculate the mode, median, and mean for the Tiny Tech Company’s salary (in thousands).
Mode - 25
Median - 30
Mean - 31.875
3. Calculate the mode, median, and mean for the Adventure Web Company’s salary (in
thousands).
Mode - 27
Median - 33.5
Mean - 39.8
4. For Tiny Tech, which measure of central tendency best represents the salary structure of the
company? Use words and/or numbers to justify your answer.
Answers may vary. The Mean represents the salary structure the best. If you remove the
extremes the mean changes very little.
5. For Adventure Web, which measure of central tendency best represents the salary structure of
the company? Use words and/or numbers to justify your answer.
Answers may vary. The Mean represents the salary structure the best. If you remove the
extremes the mean changes very little.
S.ID.A.2 Item
Ms. Wilson’s Period 4 Algebra Class 25 point Unit 3 Exam scores.
Name
Score
Name
Score
Name
Score
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Adams, B.
12
Evans, A.
25
Jacobs, T.
18
Baker, K.
20
Frazier, L.
23
Terry, S.
22
Brown, L.
22
Grey, C.
24
Weng, Z.
24
Daniels, M.
18
Hamil, F.
19
Zoug, J.
25
Drake, L.
18
Iras, G.
18
Zough, Y.
22
Ms. Wilson’s Period 4 Algebra Class 25 point Unit 4 Exam scores.
Name
Score
Name
Score
Name
Score
Adams, B.
17
Evans, A.
25
Jacobs, T.
22
Baker, K.
23
Frazier, L.
25
Terry, S.
25
Brown, L.
25
Grey, C.
25
Weng, Z.
25
Daniels, M.
22
Hamil, F.
23
Zoug, J.
25
Drake, L.
22
Iras, G.
22
Zough, Y.
25
To analyze the data, Ms. Wilson developed a box and whisker plot for her unit 4 exam score.
Ms. Wilson’s Period 4 Algebra Class Unit 4 Exam scores.
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under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.
a. Use the plot and identify the minimum, maximum, 1st quartile, median, 3rd quartile, range,
and interquartile range.
Minimum
Maximum
1st quartile
Median
3rd quartile
Range
Interquartile range
-
17
25
22
25
25
8
3
b. Use the data for the Unit 3 Exam and sketch a box and whisker plot to display the data.
Graph
c. Use the plot and identify the minimum, maximum, 1st quartile, median, 3rd quartile, range,
and interquartile range.
Minimum
Maximum
1st quartile
Median
3rd quartile
Range
Interquartile range
-
12
25
18
22
24
13
6
d. Using the data in the Box & Whisker plots write complete sentences to compare and contrast
the results from the two exams.
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The students performed better on the Unit 4 exam. 50 % of the students scored 25 out of 25 on
the Unit 4 exam. 75% of the students scored 22 out of 25 on the Unit 4 exam.
S.ID.A.2 Item
The Hair Shop surveyed 25 female and 25 male college students to learn the cost (in dollars) of
his or her most recent haircut. The survey data is summarized in the following table.
Female
Male
Minimum
$5
$ 10
Maximum
$ 225
$ 70
Quartile 1
$ 40
$ 18
Median
$ 60
$ 35
Quartile 3
$ 85
$ 60
Mean
$ 65
$ 40
a. Create two box plots to display the costs of haircuts by gender.
Male Students
Female Students
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b. Compare the two sets of data using critical points, measures of central tendency, and variance
to justify conclusions.
Male Students
Range
Interquartile Range
Median
Mean
-
60
42
35
40
The data does not have a large range. The mean and median cost are very close showing that the
data is grouped close to the middle.
Female Students
Range
Interquartile Range
Median
Mean
-
220
45
60
80
The data has a large range. The mean and median cost are varied due to the range of data. The
data is not grouped close to the median and is spread throughout the 1st and 3rd quartiles.
S.ID.A.2 Item
Donna and Gerry’s classes collected signatures for a new petition. By the end, 12 students had
collected the following amount of signatures:
90, 125, 130, 92, 96, 145, 80, 120, 110, 66, 97, and 125
a.
Use the data to sketch a box and whisker plot for the data. (remove numbers)
Graph
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under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.
b.
Identify the minimum, maximum, quartiles, median, mean, range and interquartile range
of the data.
Minimum
Maximum
1st quartile
Median
3rd quartile
Range
Interquartile range
Mean
-
66
145
91
103.5
125
79
34
106.3
S.ID.A.2 & S.ID.A.3 Task
Using the following link, find the data for the Total Gross Money for months January 2012 and
January 2013. Compare the center (mean and median) and spread (IQR and standard deviation)
of the total grosses of the top 10 movies released in those months.
Write a brief report summarizing what you found and what might account for the differences you
observe.
http://boxofficemojo.com/yearly/chart/past365.htm - Past 365 days, Top 100 movies, date
released and total gross
Solution:
When comparing the mean values of January 2012 and January 2013, we can see that the mean
value of January 2012 is significantly higher than January 2013. The exact difference is
$10,126,856.9. The median gross income of January 2012 is $48,746,722, whereas the median
gross income of January 2013 is $27,021,554.5. This would again be used to show that January
2012 grossed a lot more money than January 2013. The IQR of 2012 is 26,877,417 and the IQR
of 2013 is 28,384,262. The standard deviation of 2012 is 16,674,096.17 and the standard
deviation of 2013 is 19,683,249.95. These values show that the values in January 2013 are more
spread out than the values in January 2012. There could possibly have been a movie in January
2013 that did not gross a lot of money, and than also a movie that did gross a lot of money. This
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may account for the larger spread of the data. Overall, these values show that January 2012 had
higher grossing movies than January 2013. Other reasons for differences may have been the
release of a long awaited movie and/or sequel being released.
S.ID.A.3 Interpret differences in shape, center, and spread in the context of the data sets,
accounting for possible effects of extreme data points (outliers).
S.ID.A.3 Item
The stem and leaf plots show the ages of 20 randomly surveyed people on two different days at
an amusement park. Compare and contrast the typical age of a person at the amusement park on
the two different days.
Possible Solution:
On day one, the average age is 25 years and the median age is 23 years. On day two, the
average age is 30.2 years and the median age is 29 years. We can conclude that attendees
are older on day 2 by analyzing the mean or the median values. Using standard deviation,
we can conclude that a typical attendee on Day one is between 12.6 and 37.4 years old, and
on Day two, a typical attendee is between 15.2 and 45.2 years old.
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under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.
S.ID.A.3 Item
A survey to determine the typical age of a sky-diver was taken at two different sky-diving
companies. The data is shown in the lists below. Create a data representation to show how
the two companies compare. Compare and contrast the typical age of a sky-diver at each of
the companies using words, numbers, and/or symbols.
Free Falling
Ace in the Sky
21
21
25
41
27
28
33
34
18
18
22
34
24
50
22
49
36
19
27
29
18
19
21
23
36
23
22
21
19
18
19
32
33
40
38
35
36
32
22
21
Possible Solution:
The youngest skydiver is the same age at both companies. The average age at Ace in the
Sky is lower than at Free Falling (29.4 years at Free Falling and 26.4 years at Ace in the
Sky). Also, the median age is lower at Ace in the Sky than at Free Falling.
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S.ID.A.3 Item
The graphs below show the typical incomes of 30 randomly selected people from Howard
County, Maryland and 30 randomly selected people from Montgomery County, Maryland. The
median income in Howard County is $82,500. The median income in Montgomery County is
$84,000. Compare and contrast the incomes of the two counties discussing measures of central
tendencies in your response.
Howard County
Montgomery County
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Possible Solution:
The two counties have similar median incomes, as noted on the graphs. The
Montgomery County incomes are skewed right, implying a high extreme value (outlier).
This extreme value would raise the mean value of Montgomery County to be higher than
that of Howard County.
S.ID.A.3 Item
The average monthly rainfall (measured in inches) for two cities was recorded in the table below.
Month
Baltimore
San Diego
January
3.47
2.28
February
3.02
2.04
March
3.93
2.26
April
3
0.75
May
3.89
.20
June
3.43
.09
July
3.85
.03
August
3.74
.09
September
3.98
.21
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October
3.16
.44
November
3.12
1.07
December
3.35
1.31
http://average-rainfall-cities.findthedata.org/
www.NOAA.org
Create and analyze a graphical display that compares the average monthly rainfall of the two
cities.
Sample Response: San Diego has significantly less precipitation than Baltimore. The average
monthly precipitation in Baltimore is 3.495 inches, where as San Diego is .8975 inches. The
values for San Diego also have a larger range than the Baltimore values. The median values for
the cities are also very different. Baltimore’s median is 3.14 inches and San Diego’s median
is .595 inches. Baltimore has much more precipitation than San Diego.
S.ID.A.3 TASK
Create a context that would have a median value of 23.5 and be skewed right when represented
graphically.
Answers will vary.
Possible context: Ages of people at a restaurant for young adults (where the median age is
23.5) and one senior citizen is present.
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