Question

Needing help with the rest of the questions in the photo. Please help ! Thank you

Answer 1

The mean of a discrete random variale is given by the formula:

$\mu=E(X)=\sum xP(x)$

Thus, we have:

$\begin{gathered} \mu=(1\times0.07)+(2\times0.07)+(9\times0.11)\times(12\times0.52)\times(14\times0.12)+(19\times0.11) \\ \end{gathered}$$\begin{gathered} \mu=0.07+0.14+0.99+6.24+1.68+2.09 \\ \mu=11.21 \end{gathered}$

The variance of a discrete random variable is given by the formula:

$\begin{gathered} \sigma^2=\sum (x-\mu)^2P(x) \\ \end{gathered}$

Thus, we have:

$\sigma^2=\lbrack(1-11.21)^2\times0.07\rbrack+\lbrack(2-11.21)^2\times0.07\rbrack+\lbrack(9-11.21)^2\times0.11\rbrack+\lbrack(12-11.21)^2\times0.52\rbrack+\lbrack(14-11.21)^2\times0.12\rbrack+\lbrack(19-11.21)^2\times0.11\rbrack$$\begin{gathered} \sigma^2=7.297+5.938+0.5373+0.3245+0.9341+6.6753 \\ \sigma^2=21.706 \end{gathered}$

The standard deviation is given as:

$\begin{gathered} \sigma=\sqrt[]{\sigma^2} \\ \sigma=\sqrt[]{21.706} \\ \sigma=4.659 \end{gathered}$

The Expected value of X is the mean, = 11.21