Maine MTA High School (grade 10) Statistics and Probability. Practice it free.

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

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

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.

$P(A\mid B) = \dfrac{P(A \text{ and } B)}{P(B)} = \dfrac{5}{12} = \dfrac{5}{12}$.

Weights are $1$, $1$, and $3$ (total $5$), so probabilities are $\dfrac{1}{5}$, $\dfrac{1}{5}$, and $\dfrac{3}{5}$. Then $E = \sum p_i v_i$, which means the sum of each probability times its value: $E = \dfrac{1}{5}(20) + \dfrac{1}{5}(10) + \dfrac{3}{5}(0) = \text{\char36}6$.

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