Expectation and Variance of a Binomial sum of Poisson random variables 
Before a bank opens, customers begin to arrive from 8 a.m. onwards. They arrive independently such that the number that arrive in any 1 minute interval is a Poisson random variable with mean $3$. A bank employee randomly checks the line-up $M$ minutes after 8 a.m, where $M$ is a binomial r.v. with $p =0.8$ and $n = 20$.   a) Find the expected number of customers in the queue when it is checked.
  b) What is the standard deviation of the number of customers in the queue when it is checked?

My thoughts: 
Since the $E[ X]$ of a Poission random variable is just $\lambda$, I feel like the answer for a) is just $3$. However, this seems way to easy.
Also, for b), the SD for a Poission is $\sqrt{\lambda}$, so is it just $\sqrt{3}$ ?
 A: The number of customers $X$ can be expressed as a sum of the customers that arrive before the bank employee checks the line after $M$ minutes. In other words, we can write $X$ as
\begin{align*}
X = \sum_{i=1}^M Y_i
\end{align*}
where $Y_i$ is distributed Pois$(\lambda = 3)$. We wish to compute the expected value of $X$. Using the tower property of conditional expectations, we find
\begin{align*}
E(X) &= E(E(X \mid M)) \\
&= E(ME(Y_i)) \\
&= E(M \cdot 3) \\
&= 3 E(M) \\
&= 3 np = 3 \cdot 20 \cdot 0.8 = 48
\end{align*}
So the expected number of customers in line by the time the banker checks is 48.
The standard deviation $\sigma$ is the square root of the variance of $X$. Recall the law of total variance,
\begin{align*}
Var(X) &= E(Var(X \mid M)) + Var(E(X \mid M)) \\
&= E\Big(\sum_{i=1}^M Var(Y_i)\Big) + Var(3M) \\
&= E(M Var(Y_1)) + 9 Var(M) \\
&= E(M \cdot 3) + 9 n p (1-p) \\
&= 3np + 9 np(1-p) \\
&= 3 \cdot 20 \cdot 0.8 + 9 \cdot 20 \cdot 0.8 \cdot 0.2 \\
&= 48 + 28.8 = 76.8
\end{align*}
Taking the square root, we recover the standard deviation of $X$ to be $\sigma = \sqrt{76.8} \approx 8.76$
A: "Random sum of Random Variables":  $X = \sum_{i=1}^M Y_i,$ where 
$Y_i \sim Pois(\lambda = 3),\,$ $M \sim Binom(n=20,p=.8),$ with all
random variables independent.
Let's simulate a million realizations of $X$ using R statistical software.
Results agree to three significant digits with the Answer of @DanielXiang (+1).
Notice in particular that the there are two components to $Var(X),$ so it
is larger than just $np\lambda.$
 B = 10^6;  lam = 3;  n = 20;  p = .8;  x = numeric(B)
 for (i in 1:B) {
   M = rbinom(1, n, p);  y = rpois(M, lam)  # if M=0, then y=0
   x[i] = sum(y)  }
 mean(x);  sd(x)
 ## 48.01645  # aprx E(X) = 48.0
 ## 8.771066  # aprx SD(X) = 8.77

Below is a histogram of the simulated distribution, compared with the PDF
of $Norm(48, 8.77).$ The distribution of $X$ is slightly right-skewed, but nearly normal.

