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Please Help! Multiple Choice Question!
A survey was taken of students in math classes to find out how many hours per day students spend on social media. The survey results for the first-, second-, and third-period classes are as follows:

First period: 2, 4, 3, 1, 0, 2, 1, 3, 1, 4, 9, 2, 4, 3, 0
Second period: 3, 2, 3, 1, 3, 4, 2, 4, 3, 1, 0, 2, 3, 1, 2
Third period: 4, 5, 3, 4, 2, 3, 4, 1, 8, 2, 3, 1, 0, 2, 1, 3

Which is the best measure of center for first period and why?

Mean, because there are no outliers that affect the center
Median, because there is 1 outlier that affects the center
Interquartile range, because there is 1 outlier that affects the center
Standard deviation, because there are no outliers that affect the center

Respuesta :

GoddessofDork

So immediately, we can eliminate the third and fourth options since they are not measures of center, but rather they are measures of spread.

Now, it appears that 9 could be a possible outlier. However, we cannot judge on appearances alone. The equation to find the outliers is [tex] Q_1-1.5* IQR \\ Q_3+1.5*IQR [/tex] (IQR = Interquartile range, which is the difference between the upper and lower quartile.) This sets a range for the set of data, and anything outside of this range is considered an outlier.

Now I've gone ahead and found that the IQR is 3, the Upper quartile is 4, and the lower Quartile is 1. Now using the prior equation above, we can solve to see if there's an outlier:

[tex] 1-1.5*3 \\ 4+1.5*3\\ \\ -3.5\\ 8.5 [/tex]

Now interpreting this, any number that is less than -3.5 and greater than 8.5 is considered an outlier. Looking back on our data, we see that there is one outlier in this set, which is 9. Because there is an outlier, the best measure of center is the median, or the second option.

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