NAEP Grade 12 Data Analysis, Statistics, and Probability. Practice it free.

$P = \dfrac{\text{favorable}}{\text{total}} = \dfrac{5}{8} = \dfrac{5}{8}$.

Add bin frequencies: $4+7+5=16$.

$C(5,5) = \dfrac{5!}{5! \cdot 0!} = \dfrac{120}{120 \cdot 1} = 1\textsf{.}$

$P(6,1) = \dfrac{6!}{5!} = \dfrac{720}{120} = 6\textsf{.}$

$9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880\textsf{.}$

Set C has mean $9$ with consecutive values from $7$ to $11$, each only one or two units from the center, giving the smallest standard deviation, about $1.4$. Sets A, B, and D spread farther from their means.

The original mean is $\dfrac{10 + 16 + 14 + 8 + 13 + 15 + 8}{7} = 12$. The set $6\textsf{,}$ $8\textsf{,}$ $10\textsf{,}$ $12\textsf{,}$ $14\textsf{,}$ $16\textsf{,}$ $18$ also has mean $12$, so the means match.

In order, the original values have middle value $82$, so the median is $82$. The set $63\textsf{,}$ $65\textsf{,}$ $72\textsf{,}$ $82\textsf{,}$ $88\textsf{,}$ $88\textsf{,}$ $99$ also has median $82$, so the medians match.