If $x_1\le \dots \le x_{15}$, which value equals the median? Set that value equal to 7.

To make the mean of the 5 largest as big as possible, make the other 10 numbers as small as allowed (but still consistent with the median and positivity). Аniк‌o.⁠ai -‌ ‌S‌АТ ‌Prеp

In Desmos, enter 15*8 to compute the total sum of all 15 integers. Anіkо ⁠- Freе ‌SAТ Prер

Enter 7*1 + 3*7 to represent the minimum possible sum of $x_1$ through $x_{10}$ given $x_8=7$ and positive integers.

Enter (15*8 - (7*1 + 3*7))/5. The value shown is the maximum possible mean; match it to the answer choices.

Since the mean of 15 integers is 8, the total sum is $15\cdot 8=120$. Ѕ⁠our‍ce: аniк‌o.⁠аi

Order the data as $x_1\le x_2\le \dots \le x_{15}$. With 15 values, the median is $x_8$, so $x_8=7$.

To maximize the sum of the 5 largest values ($x_{11}$ through $x_{15}$), minimize $x_1$ through $x_{10}$.

• Because the integers are positive, the smallest possible values for $x_1$ through $x_7$ are all 1.
• Since $x_8=7$ and the list is nondecreasing, $x_9$ and $x_{10}$ must be at least 7.

So the minimum possible sum of the first 10 values is

That leaves a maximum possible sum for the 5 largest integers of $120-28=92$.

So the maximum possible mean of data set B is

Therefore, the correct choice is $\frac{92}{5}$.