Question 651·Hard·Math

Enter the weights in Desmos's $x_1$ table column and their frequencies in $y_1$. Type $\frac{\operatorname{total}(x_1y_1)}{\operatorname{total}(y_1)}$; Desmos shows about $16.549$ pounds. Each weight is multiplied by how often it occurs, so the calculation counts all 71 tortoises rather than the eight table rows.

The added weight, $39$ pounds, is above the old mean of about $16.549$ pounds. An added value pulls the mean toward itself, so the new mean is greater. Don't decide from the increased total alone: the number of tortoises increases too.

The median is the middle weight in order. With $71$ weights, it's at position $36$. Adding frequencies from the lightest weight, the running counts through $13$, $14$, $15$, $16$, and $17$ pounds are $12$, $20$, $25$, $32$, and $41$. So positions $33$ through $41$ all hold $17$ pounds, including position $36$.

Since $39$ exceeds every weight in the table, it goes at the end; the earlier positions don't shift. With $72$ weights, the median averages positions $36$ and $37$. Both still hold $17$ pounds, so the medians are equal. An added value pulls the mean toward itself, but the median must be checked at the new middle positions. The new mean is greater, and the medians are equal. Choice B.