Question

a dance team sold candy bars for 8 weeks as a fund - raiser. the number of boxes of candy bars sold during the 8 weeks was 8, 8, 9, 10, 10, 12, 15, 17. the following year, the team sold 2 more boxes every week than they sold for that same week during the previous year. how do the mean and median number of boxes sold in the second year compare to those same statistics from the previous year? both the mean and the median increased. the mean increased, but the median stayed the same. the mean stayed the same, but the median increased. both the mean and the median stayed the same.

Explanation:

Step1: Calculate the mean of the first - year data

The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. Here, $n = 8$, and $x=\{8,8,9,10,10,12,15,17\}$. So, $\sum_{i=1}^{8}x_{i}=8 + 8+9+10+10+12+15+17=89$, and the mean $\bar{x}_1=\frac{89}{8}=11.125$.

Step2: Calculate the median of the first - year data

Since $n = 8$ (an even number), the median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th ordered data values. The ordered data is $\{8,8,9,10,10,12,15,17\}$. The $\frac{n}{2}=4$th value is $10$ and the $(\frac{n}{2}+1) = 5$th value is $10$. So the median $M_1=\frac{10 + 10}{2}=10$.

Step3: Analyze the effect on the mean and median when each data value is increased by 2

If we add 2 to each data value, the new sum of data values $\sum_{i = 1}^{n}(x_{i}+2)=\sum_{i = 1}^{n}x_{i}+2n$. For $n = 8$ and $\sum_{i=1}^{n}x_{i}=89$, the new sum is $89+2\times8=89 + 16=105$. The new mean $\bar{x}_2=\frac{105}{8}=13.125$. The new ordered data set is $\{10,10,11,12,12,14,17,19\}$. Since $n = 8$ (even), the new median $M_2=\frac{12 + 12}{2}=12$.

Answer:

A. Both the mean and the median increased.