Name:_________________________________________ Date:______________________ Class:_____
Summative Assessment:
Two students, Marcos and Shantae, had the test scores shown below for the first grading period
in their Algebra 1 class.
Use their data for Problems 1 - 5.
Marcos:
Shantae:
94, 80, 60, 78, 90, 94, 98, 100,
84, 88, 80, 86, 90, 82, 86, 84,
1. Compare the mean score for both students. Justify your answer.
Marcos: (
94+80+60+78+90+94+98+100
694
8
8
Shantae: (
)=
84+88+80+86+90+82+86+84
680
8
8
)=
= 86.75
= 85
Marcos has a mean score that is 1.75 points higher than Shantae.
2. Compare the median score for both students. Show all work.
90+94
184
Marcos: 60, 78, 80, 90, 94, 94, 98, 100 = 2 = 2 = 92 because the set of data is
even you take the mean of the middle two numbers.
Shantae: 80, 82, 84, 84, 86, 86, 99, 90 =
84+86
2
=
170
2
= 85
Marcos’ median score is higher than Shantae’s.
3. Create a box plot to display each student’s test scores.
Name:_________________________________________ Date:______________________ Class:_____
4
Find the interquartile range for both students. Show calculations and answer.
a. Marcos: 60, 78, 80, 90, 92 94, 94, 98, 100
Q1 =
78+80
2
= 79
Q3 =
94+98
2
= 96
IQR = 96 -79 = 17
b. Shantae: 80, 82, 83 84, 84, 85 86, 86, 87 88, 90
Q1 =
82+84
2
= 83
Q3 =
86+88
2
= 87
IQR = 87 – 83 = 4
5. Calculate the standard deviation for both students’ data. Show all work.
a. Marcos: 86.75 -60 =26.75; 86.75 – 78 = 8.75; 86.75 - 80 = 6.75;
86.75 – 100 = -13.25;
Name:_________________________________________ Date:______________________ Class:_____
86.75- 90= -3.25; 86.75 – 94 = -7.25; 86.75 – 94 = -7.25; 86.75
– 98 = -11.25
Variance = [(26.75)2 + (8.75)2 + (6.75)2 + (-13.25)2 + (-3.25)2 + (-7.25)2 + (-11.25)2 +
(-7.25)2] ÷ 8 = 157.01
Standard Deviation = 157.01 = 12.53 (rounded to nearest hundredth)
Shantae: 85 -80 =5; 85 – 82 = 3; 85 - 84 = 1; 85 – 84 = 1;
85- 86= -1; 85 – 86 = -1; 85 – 88 = -3; 85 – 90 = -5
Variance = [52 + 32 + 12 + 12 + (-1)2 + (-1)2 + (-3)2 + (-5)2] ÷ 8 = 8.875
Standard Deviation = 8.875 = 2.98 (rounded to nearest hundredth)
6. The following data represent a summary of the grades on a previous Algebra 1 test for a
different class consisting of 19 students: minimum = 65, Q1 = 70, median =80, Q3 = 82,
maximum = 95.
Is there an outlier for this test? Justify your answer
Outlier must be 1.5IQR above Q3 or below Q1.
IQR = 82 – 70 =12
1.5(12) = 18
Q1 – 18 = 70 -18 = 52
Q3 + 18 = 82 + 18 = 100
The Minimum is 65 and the Maximum is 95 so there are no outliers.
Name:_________________________________________ Date:______________________ Class:_____
7. The results of the class scores are normally distributed. The mean test score is 70.
Sixteen percent of the scores are under 60. What would be the value of the standard
deviation? Justify your answer.
In a normal distribution of data the mean is always in the center so 50 percent of data is above the
mean and 50 percent below. Sixty eight percent of the data lies within one Standard Deviation of the
mean, 34 percent above and 34 percent below. This implies that beyond one Standard Deviation there is
16 percent of data above and below one Standard Deviation. Given that the mean score is 70 and 16
percent of the scores fall below 60 one can conclude that a score of 60 is one Standard Deviation below
the mean. The Standard Deviation is therefore equal to 70 minus 60 which is 10.
8. Graph A has less of a spread indicating that the Spurs scores were more clustered around the
median; giving the team a smaller variance as well as a smaller Standard Deviation of scores.
Graph B has a wider distribution of scores because the graph has a wider spread so there are
more variances and as a result the Heat has a Standard Deviation that is larger than the Spurs.