amber and chloe recorded the number of minutes they spent studying for ten days. their data are listed…

amber and chloe recorded the number of minutes they spent studying for ten days. their data are listed below.\namber: 25, 35, 30, 28, 27, 22, 65, 20, 33, 22\nchloe: 55, 60, 45, 53, 44, 57, 64, 51, 45, 58\nwhich statements are correct? check all that apply.\n□ ambers data set contains an outlier.\n□ chloes data set contains an outlier.\n□ the median describes chloes data more accurately than the mean.\n□ the median describes ambers data more accurately than the mean.\n□ 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.

amber and chloe recorded the number of minutes they spent studying for ten days. their data are listed below.\namber: 25, 35, 30, 28, 27, 22, 65, 20, 33, 22\nchloe: 55, 60, 45, 53, 44, 57, 64, 51, 45, 58\nwhich statements are correct? check all that apply.\n□ ambers data set contains an outlier.\n□ chloes data set contains an outlier.\n□ the median describes chloes data more accurately than the mean.\n□ the median describes ambers data more accurately than the mean.\n□ 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.

Answer

Explanation:

Step1: Arrange Amber's data in ascending order

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

Step2: Calculate Amber's median

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

Step3: Calculate Amber's mean

$\frac{20 + 22+22 + 25+27+28+30+33+35+65}{10}=\frac{317}{10} = 31.7$ The inter - quartile range (IQR) for Amber's data: First quartile $Q_1$ is the median of the lower half. Lower half: $20, 22, 22, 25, 27$, $Q_1 = 22$ Third quartile $Q_3$ is the median of the upper half. Upper half: $28, 30, 33, 35, 65$, $Q_3=33$ $IQR = Q_3 - Q_1=33 - 22 = 11$ Lower fence $=Q_1-1.5\times IQR=22-1.5\times11=22 - 16.5 = 5.5$ Upper fence $=Q_3 + 1.5\times IQR=33+1.5\times11=33 + 16.5 = 49.5$ $65$ is an outlier in Amber's data set.

Step4: Arrange Chloe's data in ascending order

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

Step5: Calculate Chloe's median

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

Step6: Calculate Chloe's mean

$\frac{44 + 45+45+51+53+55+57+58+60+64}{10}=\frac{532}{10}=53.2$ The inter - quartile range (IQR) for Chloe's data: First quartile $Q_1$ is the median of the lower half. Lower half: $44, 45, 45, 51, 53$, $Q_1 = 45$ Third quartile $Q_3$ is the median of the upper half. Upper half: $55, 57, 58, 60, 64$, $Q_3 = 58$ $IQR=Q_3 - Q_1=58 - 45 = 13$ Lower fence $=Q_1-1.5\times IQR=45-1.5\times13=45 - 19.5 = 25.5$ Upper fence $=Q_3+1.5\times IQR=58+1.5\times13=58 + 19.5 = 77.5$ Chloe's data set has no outlier.

Step7: Analyze the median - mean relationship

For Amber, difference between mean and median is $|31.7 - 27.5|=4.2$ For Chloe, difference between mean and median is $|53.2 - 54| = 0.8$

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.