Find the mean and variance of number of message sent per hour.

We must have

$\begin{align} \sum{pi} = 1\\ 0.08 + 3k + 6k+ 4k + 4k + 0.07 = 1 \\ 17k + 0.15 = 1 \\ 17k = 1-0.15 \\ k= 0.005 \\ \end{align}$

put k = 0.005 in table, we get

Now,

$ \begin{align} mean = E(z) \\ mean= \sum{p_i x_i} \\ mean = (0.08 \times 10) + (0.15 \times 11) + (0.3 \times 12) + (0.2 \times 13) + (0.2 \times 14) + (0.07 \times 15) \\ = 0.8 + 1.65 + 3.6 + 2.6+ 2.8 + 1.05 \\ E(x) = mean = 12.5 \\ \\ Variance = \sigma ^2 = E(x^2) - [E(x)]^2 - - - - 1 \\ Now E(x^2) = \sum P_iX_i^2 \\ = (0.08 \times 10^2) + (0.15 \times 11^2) + (0.3 \times 12^2) + (0.2 \times 13^2) + (0.2 \times 14^2) + (0.07 \times 15^2) \\ = 8+18.15+43.2+33.8+39.2+15.75 \\ E(x^2) = 158.1 \\ \\ Variance = \sigma ^2 = E(x^2) - [E(x)]^2 \\ = 158.1 - (12.5)^2 \\ Variance = 1.85 \end{align}$