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 !
 A: 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 $\frac{181}{5} - 25 = \frac{56}{5}$.
A: You have the right formula for $(a)$ but you evaluated it wrongly
$$1 \cdot (\frac{1}{5} ) + 3 \cdot (\frac{2}{5} )+ 9 \cdot (\frac{2}{5} ) = 5$$
For part $(b)$.
$$\frac{181}5-\color{red}{5^2}$$
The second term can't be $625$ as it is bigger than the first term and we would get a negative variance otherwise.
A: 
So for the E(x) of (a) I got $1 \cdot \tfrac 15 + 3 \cdot \tfrac25+ 9 \cdot\tfrac25 = 9.8$
Why is the answer showing E(x)= 5?

Because $\tfrac {1+ 6+18}5 =\tfrac {25}5$  not $\tfrac {49}5$ (...how did you even?)

Also To compute the Variance, according to the formula, E($X^2$) - E$(X)^2$ shouldn't the answer be $\frac{181}{5} $ - $\frac{625}{5}$ ?

$$\dfrac{1+2\cdot 3^2+2\cdot 9^2}{5}-\left(5\right)^2=\dfrac{181}5-\dfrac {125}5=\dfrac{56}5$$
