Question 48·Hard·One-Variable Data Distributions; Measures of Center and Spread

The mean is the total divided by the count. Since all $19$ days have a mean of $42$, their total is $s=19(42)$. Type $s=19\cdot42$ in Desmos; it shows $s=798$ visitors.

Excluding the greatest day leaves $18$ days with a mean of $38$, so those days total $18(38)$. Subtract their total from $s$ to recover the one removed value. Add $b=s-18\cdot38$ in Desmos; it shows $b=114$, the maximum, or greatest count.

Excluding the least day leaves $18$ days with a mean of $44$. Subtract their total from $s$: add $a=s-18\cdot44$ in Desmos. It shows $a=6$, the minimum, or least count. Don't pair $44$ with the greatest day: removing a low value raises the mean.

The range is the greatest value minus the least value. Type $b-a$ in Desmos; it shows $108$. So the range of the original daily visitor counts is $108$. Choice C.

You can find the range without calculating the original total. Let $T$ be that total, $H$ the greatest day's count, and $L$ the least. Each shortened group has $18$ days, so removing the greatest gives $T-H=18(38)$, while removing the least gives $T-L=18(44)$.

Subtract the greatest-day equation from the least-day equation. The original total cancels, leaving the difference between the endpoints:

Factor out the common $18$:

$H-L$ is the range. Type $18(44-38)$ in Desmos; it shows $108$. So the original visitor counts have a range of $108$. Choice C.