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

Explanation:

Step1: Sort first period data

Sorted data: $0, 0, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 9$

Step2: Identify outliers in data

Calculate quartiles: $Q_1=1$, $Q_3=3$, $IQR=3-1=2$. Outlier bounds: $1-1.5\times2=-2$, $3+1.5\times2=6$. Value 9 > 6, so 9 is an outlier.

Step3: Choose robust center measure

Outliers skew the mean, so median is better.

Answer:

Median, because there is 1 outlier that affects the center