Question 567·Hard·Math

A frequency is the number of times a value appears: M has four $5$s, so that row adds $5(4)$ to its total. Enter the values as $x_1$, M’s frequencies as $y_1$, and N’s as $y_2$ in a Desmos table. Type $\frac{\operatorname{total}(x_1y_1)}{\operatorname{total}(y_1)}$. Desmos shows about $9.78$ (or $\frac{88}{9}$ with the fraction button). This is M’s mean: the sum of every value-frequency product divided by all $36$ observations.

Use N’s frequency column in the same weighted-mean calculation. Type $\frac{\operatorname{total}(x_1y_2)}{\operatorname{total}(y_2)}$. Desmos shows about $11.94$ (or $\frac{215}{18}$), so M’s mean is lower. That settles the means, but not the medians.

For 36 ordered values, the median lies halfway between positions $18$ and $19$. Type $4+6+8$ for M’s frequencies at $5$, $7$, and $9$; Desmos shows $18$. So position $18$ holds $9$, and the next position holds $11$. The middle rows alone can’t tell you this; their frequencies determine the positions.

Because positions $18$ and $19$ hold different values, average both: type $\frac{9+11}{2}$. Desmos shows $10$, M’s median. Taking $9$ alone would leave out position $19$.

Type $1+3+5+7$ for N’s frequencies through $11$. Desmos shows $16$, so N has $16$ values before its nine $13$s. Both middle positions, $18$ and $19$, fall among those $13$s. N’s median is $13$.

M’s mean, $\frac{88}{9}$, is below N’s, $\frac{215}{18}$; M’s median, $10$, is below N’s, $13$. Both centers of data set M are less than the corresponding centers of data set N. Choice C.