chucky grabbed 12 items in the grocery store that each had a different price and had a mean cost of about…

Brief Explanations:

• Mean:
  • The mean is the sum of all values divided by the number of values. The original mean is (\bar{x}=\frac{\sum_{i = 1}^{12}x_{i}}{12}=7.41), so (\sum_{i = 1}^{12}x_{i}=12\times7.41 = 88.92). After removing the cheese ((x = 39.99)), the new sum is (\sum_{i = 1}^{11}x_{i}=88.92 - 39.99=48.93), and the new mean is (\bar{y}=\frac{48.93}{11}\approx4.45). Since the removed value ((39.99)) is much larger than the original mean ((7.41)), the mean will decrease.
• Median:
  • The median of (n = 12) (even number of data points) is the average of the (6^{th}) and (7^{th}) ordered - values. When we remove the largest value (the cheese), for (n = 11) (odd number of data points), the median is the (6^{th}) ordered - value. If the original data is ordered (x_1\leq x_2\leq\cdots\leq x_{12}), the original median (M_1=\frac{x_6 + x_7}{2}), and the new median (M_2=x_6). Since (x_6\leq\frac{x_6 + x_7}{2}) (assuming (x_6\leq x_7)), the median will decrease.
  • The mean is affected by the magnitude of the out - lier (the expensive cheese). The median is only affected by the position of the data points. The mean is more sensitive to extreme values. The change in the mean: (\Delta\bar{x}=7.41-4.45 = 2.96). The change in the median: if we assume (x_7\geq x_6), (\Delta M=\frac{x_6 + x_7}{2}-x_6=\frac{x_7 - x_6}{2}). Since (x_{12}=39.99) is an extreme value, the mean decreases more than the median.

Answer:

B. Both the mean and median will decrease, but the mean will decrease by more than the median.