<s>[INST] How to get the expected value of these numbers? 
So for the E(x) of (a)  I got 1 $\cdot$ ($\frac{1}{5} $) + 3 $\cdot$ ($\frac{2}{5} $)+ 9 $\cdot$ ($\frac{2}{5} $) = 9.8
Why is the answer showing E(x)= 5? 
Also To compute the Variance, according to the formula, E($X^2$) - E$(X)^2$ shouldn't the answer be $\frac{181}{5} $ - $\frac{625}{5}$ ?
Many thanks in advance !
 [/INST] For the expectation, it looks like you made a calculation error:
$$1 \cdot \frac{1}{5} + 3 \cdot \frac{2}{5} + 9 \cdot \frac{2}{5} = \frac{1}{5} + \frac{6}{5} + \frac{18}{5} = \frac{25}{5} = 5.$$
It's worth noting that the expectation is interpreted as the average, so saying you get an expected value of $9.8$ when none of the items are larger than $9$ should ring alarm bells in your head.
For the variance, $E(X^2) = \frac{181}{5}$ and $E(X) = 5$ so $E(X)^2 = 25$, hence the variance is $\frac{181}{5} - 25 = \frac{56}{5}$.
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