Question 49·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

The median is the middle value after the integers are put in order. With $19$ integers, nine are on each side of the middle, so the median is the $10$th value.

The first bar’s $4$ values fill positions $1$ through $4$. The second bar’s $5$ values fill positions $5$ through $9$. So the 10th value is in the third bar. Call it $m$; because the values are integers, $30\le m\le39$.

For any fixed $m$, put the other values at the largest integers their bars allow: four $19$s, five $29$s, three $39$s, and six $49$s. The third bar holds four values, but one is $m$, so only its other three can be set to $39$. The right endpoints $20,30,40,50$ are excluded.

If $m$ rises by $1$, the total rises by only $1$, so the mean rises by $\frac{1}{19}$. But the median rises by $1$. That makes mean minus median smaller, so choose the lowest allowed median: $m=30$. Push the other values up, but keep the middle value down.

Make a Desmos table with $x_1$ for each chosen value and $y_1$ for its frequency, or number of times it occurs. Desmos displays the five pairs $(19,4),(29,5),(30,1),(39,3),(49,6)$. They give 19 values, and the $10$th is $30$, so this arrangement fits the histogram.

Each value contributes its value times its frequency. Type $\operatorname{total}(x_1y_1)$; Desmos adds those products and shows $662$. That is the total for the arrangement you built.

The mean is the total divided by $19$. Type $\frac{\operatorname{total}(x_1y_1)}{19}-30$; Desmos shows about $4.842105$, and its fraction button gives $\frac{92}{19}$. Since the table reaches the limits you found, the greatest possible mean minus median is $\frac{92}{19}$. Choice C.