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Answers Measures Handout Your teacher has a problem and needs your input. She has to give one math award this year to a deserving student, but she can’t make a decision. Here are the test grades for her two best students: Bryce: 90, 90, 80, 100, 99, 81, 98, 82 Brianna: 90, 90, 91, 89, 91, 89, 90, 90 1) Find the five-­β€number summary and the IQR for each student. Bryce: min = 80, Q1 = 81.5, median = 90, Q3 = 98.5, max = 100, IQR = 17 Briana: min = 89, Q1 = 89.5, median = 90, Q3 = 90.5, max = 91, IQR = 1 2) Based on your display, write down which of the two students should get the math award and discuss why they should be the one to receive it. Your students will probably first calculate the average of each student. They will soon discover that both students have an average of 90. They will then either say that both deserve the award or go on to say that Bryce should get it because he had higher A’s or Brianna should get it because she was more consistent. This should open up a discussion that it is very important to use a measure of spread (or variability) to describe a distribution. Many times we only look at a measure of center to describe a distribution. 3) Calculate the mean (π‘₯) of Bryce’s grade distribution. 90 Calculate the mean absolute deviation, variance, and standard deviation of Bryce’s distribution. In Coordinate Algebra, students calculated the mean absolute deviation. This is probably the first time for them to calculate the variance and the standard deviation. Point out that the sum of the column X i βˆ’ X will always be zero and that is why they have to take the absolute value or square those values before they average them to get the mean deviation or the variance. They will probably discover that there is a huge discrepancy between the mean deviation and the variance. That should lead into the discussion of why they take the square root of the variance to get the standard deviation. The formulas for mean absolute deviation, variance, and standard deviation are below. !
!
mean absolute deviation: 𝑀𝐴𝐷 =
π‘₯! βˆ’ π‘₯ variance: 𝜎 ! =
(π‘₯! βˆ’ π‘₯)! !
standard deviation: 𝝈 = !
!
(π‘₯! βˆ’ π‘₯)! which is the square root of the variance 4) Fill out the table to help you calculate them by hand. Scores for Bryce Mean Deviation Mean Absolute (π’™π’Š ) π’™π’Š βˆ’ 𝒙 Deviation π’™π’Š βˆ’ 𝒙 90 0 0 90 0 0 80 -­β€10 10 100 10 10 99 9 9 81 -­β€9 9 98 8 8 82 -­β€8 8 Total 0 54 MAD for Bryce: !"
!
Variance for Bryce: = 6.75 !"#
!
!
= 61.25 Variance (π’™π’Š βˆ’ 𝒙)𝟐 0 0 100 100 81 81 64 64 490 Standard deviation for Bryce: 61.25 = 7.826 5) What do these measures of spread tell you about Bryce’s grades? All of these values tell you how your data deviates from the mean. The variance is much larger than the mean deviation or the standard deviation because the deviations from the mean were squared. That’s why you take the square root of the variance to get the standard deviation. The standard deviation and mean deviation are pretty close in value. 6) Calculate the mean of Brianna ’s distribution. 90 7) Calculate the mean deviation, variance, and standard deviation of Brianna’s distribution. Scores for Brianna Mean Deviation Mean Absolute Variance (π’™π’Š ) π’™π’Š βˆ’ 𝒙 Deviation π’™π’Š βˆ’ 𝒙 (π’™π’Š βˆ’ 𝒙)𝟐 90 0 0 0 90 0 0 0 91 1 1 1 89 -­β€1 1 1 91 1 1 1 89 -­β€1 1 1 90 0 0 0 90 0 0 0 Total 0 4 4 !
MAD for Brianna: = 0.5 !
!
Variance for Brianna: = 0.5 !
Standard deviation for Brianna: 0.5 = 0.707 8) What do these measures of spread tell you about Brianna’s grades? All of these values tell you how your data deviates from the mean. In this case, the mean deviation and the variance are the same. Unlike Bryce, and probably surprising to students, the standard deviation is larger than the variance. 9) Based on this information, write down which of the two students should get the math award and discuss why they should be the one to receive it. Students will have different opinions of who should win. Students must be able to justify their answer using an analysis of the statistics.