Question

analyzing two data sets
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.
□ ambers data set contains an outlier.
□ chloes data set contains an outlier.
□ the median describes chloes data more accurately than the mean.
□ the median describes ambers data more accurately than the mean.
□ the difference between the mean and median of ambers data set is greater than the difference between the mean and median of chloes data set.

Explanation:

Step1: Sort Amber's data

Sorted data: $20, 22, 22, 25, 27, 28, 30, 33, 35, 65$

Step2: Check Amber's outliers

Calculate quartiles: $Q_1=22$, $Q_3=33$, $IQR=33-22=11$.
Lower bound: $22 - 1.5\times11=5.5$, Upper bound: $33 + 1.5\times11=49.5$.
$65>49.5$, so it is an outlier.

Step3: Sort Chloe's data

Sorted data: $44, 45, 45, 51, 53, 55, 57, 58, 60, 64$

Step4: Check Chloe's outliers

Calculate quartiles: $Q_1=45$, $Q_3=58$, $IQR=58-45=13$.
Lower bound: $45 - 1.5\times13=25.5$, Upper bound: $58 + 1.5\times13=77.5$.
All values are within bounds, no outliers.

Step5: Calculate Amber's mean/median

Mean: $\frac{25+35+30+28+27+22+65+20+33+22}{10}=\frac{307}{10}=30.7$
Median: $\frac{27+28}{2}=27.5$
Difference: $|30.7-27.5|=3.2$
Outliers skew mean, so median is more accurate.

Step6: Calculate Chloe's mean/median

Mean: $\frac{55+60+45+53+44+57+64+51+45+58}{10}=\frac{532}{10}=53.2$
Median: $\frac{53+55}{2}=54$
Difference: $|53.2-54|=0.8$
No outliers, mean/median are close; mean is accurate.

Step7: Compare mean-median differences

$3.2>0.8$, so Amber's difference is larger.

Answer:

• Amber's data set contains an outlier.
• 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.