14 donna and andrew compared their math final exam scores from grade 8 through grade 12. their scores are…

Explanation:

Step1: Calculate Donna's mean

Sum of Donna's scores: $90 + 92+87 + 94+95=458$. Mean of Donna's scores: $\frac{458}{5}=91.6$.

Step2: Calculate Andrew's mean

Sum of Andrew's scores: $78 + 96+87 + 94+93=448$. Mean of Andrew's scores: $\frac{448}{5}=89.6$. So Andrew does not have a higher mean than Donna.

Step3: Find Donna's median

Arrange Donna's scores in ascending - order: $87,90,92,94,95$. The median is the middle value, which is $92$.

Step4: Find Andrew's median

Arrange Andrew's scores in ascending - order: $78,87,93,94,96$. The median is the middle value, which is $93$. So they do not have the same median.

Step5: Find Donna's quartiles

Donna's scores in ascending - order: $87,90,92,94,95$. The first quartile $Q_1$ is the median of the lower half. The lower half is $87,90$, so $Q_1 = 88.5$. The third quartile $Q_3$ is the median of the upper half. The upper half is $94,95$, so $Q_3=94.5$. The inter - quartile range $IQR_D=Q_3 - Q_1=94.5 - 88.5 = 6$.

Step6: Find Andrew's quartiles

Andrew's scores in ascending - order: $78,87,93,94,96$. The first quartile $Q_1$ is the median of the lower half. The lower half is $78,87$, so $Q_1 = 82.5$. The third quartile $Q_3$ is the median of the upper half. The upper half is $94,96$, so $Q_3 = 95$. The inter - quartile range $IQR_A=Q_3 - Q_1=95 - 82.5 = 12.5$. So Andrew has a larger inter - quartile range than Donna.

Step7: Compare third quartiles

Donna's $Q_3 = 94.5$ and Andrew's $Q_3 = 95$. So the third quartile for Donna is not greater than the third quartile for Andrew.

Answer:

(3) Andrew has a larger interquartile range than Donna.