Question

amber and chloe recorded the number of minutes they spent studying for ten days. their data are listed below.
amber: 25, 35, 30, 28, 27, 22, 65, 20, 33, 22
chloe: 55, 60, 45, 53, 44, 57, 64, 51, 45, 58
which statements are correct? check all that apply.
amber’s data set contains an outlier.
chloe’s data set contains an outlier.
the median describes chloe’s data more accurately than the mean.
the median describes amber’s data more accurately than the mean.
the difference between the mean and median of amber’s data set is greater than the difference between the mean and median of chloe’s data set.

Explanation:

Step1: Sort Amber's data

$20, 22, 22, 25, 27, 28, 30, 33, 35, 65$

Step2: Find Amber's median

Since $n = 10$ (even), median is $\frac{27 + 28}{2}=27.5$

Step3: Calculate Amber's mean

$\frac{20 + 22+22 + 25+27+28+30+33+35+65}{10}=\frac{317}{10} = 31.7$

Step4: Check for Amber's outlier

$Q_1=\frac{22+22}{2}=22$, $Q_3=\frac{33 + 35}{2}=34$
$IQR = 34 - 22=12$
Lower - bound $=22-1.5\times12=22 - 18 = 4$
Upper - bound $=34 + 1.5\times12=34+18 = 52$
$65$ is an outlier for Amber.

Step5: Sort Chloe's data

$44, 45, 45, 51, 53, 55, 57, 58, 60, 64$

Step6: Find Chloe's median

Since $n = 10$ (even), median is $\frac{53+55}{2}=54$

Step7: Calculate Chloe's mean

$\frac{44 + 45+45+51+53+55+57+58+60+64}{10}=\frac{532}{10}=53.2$

Step8: Check for Chloe's outlier

$Q_1=\frac{45+45}{2}=45$, $Q_3=\frac{58+60}{2}=59$
$IQR=59 - 45 = 14$
Lower - bound $=45-1.5\times14=45 - 21 = 24$
Upper - bound $=59+1.5\times14=59 + 21 = 80$
No outlier for Chloe.

Step9: Analyze median and mean

For Amber, due to outlier, median ($27.5$) describes data better than mean ($31.7$)
For Chloe, mean ($53.2$) and median ($54$) are close, so mean also describes data well.
Difference for Amber: $|31.7 - 27.5|=4.2$
Difference for Chloe: $|53.2 - 54| = 0.8$

Answer:

A. Amber’s data set contains an outlier.
D. The median describes Amber’s data more accurately than the mean.
E. The difference between the mean and median of Amber’s data set is greater than the difference between the mean and median of Chloe’s data set.