Question 603·Hard·Math

Let $n$ be the original number of students. A mean is total hours divided by student count, so total hours equals mean times count. The original students' hours of sleep per night add to $6.4n$.

The four added students contribute $4(8.5)$ hours per night in all. The updated study has $n+4$ students because the count includes the four newcomers. Match the total built from both groups to the updated mean times the updated count:

Type $6.4n+4(8.5)\sim6.8(n+4)$ in Desmos. The $\sim$ tells Desmos to use a regression to find $n$; under PARAMETERS, it reports $n=17$. So there were 17 students in the original group. Choice C.

The original group's average is $6.8-6.4=0.4$ hours below the new mean. Each added student slept $8.5-6.8=1.7$ hours above it. These are per-student gaps from the same mean.

For the combined mean to be $6.8$, hours below that mean must balance hours above it. If $n$ counts the original students, their total shortfall is $0.4n$ hours, and the four newcomers' excess is $4(1.7)$ hours. So the two amounts match:

Divide by $0.4$ to isolate $n$:

Type $\frac{4(1.7)}{0.4}$ in Desmos; it prints $17$. So the original group had $17$ students. Choice C.