Label the seats 1 to $n$ going clockwise around the table. Let $X_i$ be the Bernoulli random variable with value 1 if the person in seat $i$ is shorter than his or her neighbors. Then $X=\sum_{i=1}^n X_i$ represents the total number of people who are shorter than both of their neighbors, and
$$
E(X)=E\left(\sum_{i=1}^n X_i\right)=\sum_{i=1}^n E\left(X_i\right)
$$
by linearity of expected value. Recall that this property of expected values holds even when the $X_i$ are dependent, as is the case here!

Among 3 random people the probability that the middle one is the shortest is $1 / 3$. Therefore $X_i \sim \operatorname{Bernoulli}(1 / 3)$, which implies $E\left(X_i\right)=1 / 3$. Therefore the expected number of people shorter than both their neighbors is
$$
E(X)=\sum_{i=1}^n E\left(X_i\right)=\frac{n}{3}
$$

We have $P(A)=P(B)=P(C)=1 / 2$. Writing the outcome of die 1 first, we can easily list all outcomes in the following intersections.
$$
\begin{aligned}
& A \cap B={(1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)} \
& A \cap C={(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)} \
& B \cap C={(2,1),(4,1),(6,1),(2,3),(4,3),(6,3),(2,5),(4,5),(6,5)}
\end{aligned}
$$
By counting we see
$$
P(A \cap B)=\frac{1}{4}=P(A) P(B)
$$
Likewise,
$$
P(A \cap C)=\frac{1}{4}=P(A) P(C) \quad \text { and } \quad P(B \cap C)=\frac{1}{4}=P(B) P(C) .
$$
So, we see that $A, B$, and $C$ are pairwise independent.
However, $A \cap B \cap C=\emptyset$, since if we roll an odd on die 1 and an odd on die 2 , then the sum of the two will be even. So, in this case,
$$
P(A \cap B \cap C)=0 \neq P(A) P(B) P(C),
$$
and we conclude that $A, B$ and $C$ are not mutually independent.

(a) We know that $E(X)=a E(Z)+b=b$ and
$$
\operatorname{Var}(X)=\operatorname{Var}(a Z+b)=a^2 \operatorname{Var}(Z)=a^2 .
$$
(b) Let $x$ be any real number. We will first compute $F_X(x)=P(X \leq x)$. Since $X=a Z+b$, we see that
$$
F_X(x)=P(X \leq x)=P(a Z+b \leq x)=P\left(Z \leq \frac{x-b}{a}\right)=\Phi\left(\frac{x-b}{a}\right) .
$$
So $F_X(x)=\Phi\left(\frac{x-b}{a}\right)$. Differentiating this with respect to $x$, we find
$$
f_X(x)=\frac{1}{a} \Phi^{\prime}\left(\frac{x-b}{a}\right)=\frac{1}{a} \phi\left(\frac{x-b}{a}\right)=\frac{1}{\sqrt{2 \pi} a} \mathrm{e}^{-\frac{(x-b)^2}{2 a^2}}
$$
(c) From (b), we see that $f_X(x)$ is the pdf of $N\left(b, a^2\right)$ distribution
(d) From (b) and (c), we see that if $Z$ is standard normal, then $\sigma Z+\mu$ follows a $N\left(\mu, \sigma^2\right)$ distribution. From (a), we know that $E(\sigma Z+\mu)=\mu$ and $\operatorname{Var}(\sigma Z+\mu)=\sigma^2$.