Question

PLEASE HELP ASAP!!!! 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 second period, and why? Mean, because there are no outliers that affect the center Median, because there is one outlier that affects the center Interquartile range, because there is one outlier that affects the center Standard deviation, because there are no outliers that affect the center

Answer 1

Answer: D : Median, because there is 1 outlier that affects the center

Explanation:

Mean and median terms are use to measure the central tendency in a data set.

Mean is the best measure of center if there is no outlier present in the data otherwise median is the best measure of center if there is outlier present because outliers effects the means value of a data but not the median value.

In the third period, there is an outlier present as 8 that affects the center, then the best measure of center for third period must be Median.