Expected value homework question 
Im not sure why the expected value that I got is incorrect. The method that I did is: (0)(0.568)+(7)(0.101)+(11)(0.218)+(9)(0.113)
 A: I think there's a typo, either when entering the answer or making the answer key.
$$\begin{align}
E(a) &= 0 \times 0.568 
     + 7 \times 0.101 
     + 11 \times 0.218 
     + 9 \times 0.113 \\
     &= 0 + 0.707 + 2.398 + 1.017 \\
&= 4.122
\end{align}$$
But the screenshot says 4.112.
A: Using R as a calculator, I got the following:
(0)*(0.568)+(7)*(0.101)+(11)*(0.218)+(9)*(0.113)
[1] 4.122

I'm guessing there's a typo in your answer or in your answer key, depending on what "4.112" is. (It certainly
wouldn't be the first time an answer key was wrong.)
With the following vectors in R, I get $\mu = \sum_{i=1}^4 p_ik_i = 4.122.$
k = c(0, 7, 11, 9)
p = c(0.568, 0.101, 0.218, 0.113)
mu = sum(p*k);  mu
[1] 4.122

Then the population variance is $\sigma^2 = \sum_{i=1}^4 p_i(k_i-\mu)^2 = 23.48912$ and the population standard deviation is
$\sigma = \sqrt{\sigma^2} = 4.846557.$ [You may have been
instructed to round to four decimal places.]
vr = sum(p*(k-mu)^2);  vr
[1] 23.48912
sqrt(vr)
[1] 4.846557

Note: In R, I can take a random sample of ten million observations from
this distribution. Then the mean and standard deviation
of this huge sample should agree closely with the the population mean and standard deviations (to two or three significant digits).
set.seed(220)
x = sample(k, 10^7, rep=T, p)
table(x)/10^7
x
        0         7         9        11 
0.5681086 0.1009472 0.1131462 0.2177980 
                # aprx 0.568, 0.101, 0.113, 0218
                # note different order of listing
mean(x);  sd(x) 
[1] 4.120724    # aprx 4.122
[1] 4.846075    # aprx 4.847

