IM Commentary
In high school, students build on their experience with the mean absolute deviation (MAD) in middle school. The standard deviation is introduced as a measure of variability, and students calculate and interpret the standard deviation in context. The purpose of this task is to develop students’ understanding of standard deviation (SID.2).
In the discussion of question 1, make sure that students can explain the differences in the way the standard deviation and the MAD are calculated (squaring instead of absolute value and division by n1 instead of n).
Other discussion points include the following.

Students are familiar with MAD (mean absolute difference) and should be able to discuss how the standard deviation is related to the MAD.

Students should recognize which of the data points contribute the most to the size of the standard deviation.

Students should recognize the purpose of squaring each deviation. Note that the sum of the deviations will always sum to zero.

Note that the score of 84 on Jim’s first exam contributes the most to the size of the standard deviation and 92 contributes the least. These correspond to the scores that are furthest away from and closest to the mean respectively.
Solution
1. The standard deviation is a measure of spread about the mean and is defined as
Standard Deviation = \(\sqrt{{1 \over{n1}} \sum {{(x  \bar{x})}^2}}\)
MAD = \({1 \over{n}} \sum {{ x  \bar{x}}}\)
While both measures rely on the deviations from the mean (\(x  \bar{x}\)), the MAD uses the absolute values of the deviations and the standard deviation uses the squares of the deviations. Both methods result in nonnegative differences. The MAD is simply the mean of these nonnegative (absolute) deviations. The standard deviation is the square root of the sum of the squares of the deviations, divided by (n1). This measure also results in a value that in some sense represents the “typical” difference between each data point and the mean.
2. To calculate the standard deviation, we must first find the mean of the test scores. \(\bar{x} = {{92+84+96+100} \over {4}} = {{372} \over {4}} = 93\). Then, we find the deviations from the mean, find the sum of the squares of the deviations, divide by one less than the number of scores and take the square root of the result.
Exam

Score

\(x \bar{x}\)

\((x  \bar{x})^2\)

1

92

92  93 = 1

1

2

84

84  93 = 9

81

3

96

96  93 = 3

9

4

100

100  93 = 7

49

\(\sqrt{{1 \over{n1}} {\sum{(x  \bar{x})}^2}} = \sqrt{{1 \over {3}}(1+81+9+49)}= \sqrt{{140} \over{3}}=\sqrt{46.67} = 6.83\).
This value is larger than the mean absolute difference between the mean and each test score, which is MAD = \({1 \over{n}} \sum{x \bar{x}} = {1 \over{4}}(1+9+3+7)=5.\). Because the standard deviation is based on the squared deviations from the mean, it will be large when the values in a data set are spread out around the mean and small when the values are tightly clustered.
3. Sally’s scores and Jim’s scored both have a mean score of 93. Since Sally’s scores are clustered more closely around 93 than Jim’s scores, Sally’s scores are less variable and the standard deviation of her scores will be smaller than the standard deviation of Jim’s scores.
4. In order for the standard deviation to be zero, all scores must be the same. So Tom must have scored 91 on all four exams.