Given data:
First we will find :
![x\mathrm{}p(x)andx^2p(x)](https://img.qammunity.org/2023/formulas/mathematics/college/jqsluswhq8ict6h1w1ajs6zy4wz07ettfo.png)
X P(x) x.p(x) x^2.p(x)
4 0.13 0.52 2.08
8 0.08 0.64 5.12
11 0.13 1.43 15.73
12 0.12 1.44 17.28
20 0.54 10.8 216
total 14.83 256.21
For x.p(x), we will multiply x and p(x).
Similarly, x^2.p(x) we will multiply x twice and p(x).
Now,
(a) mean is given as:
![\operatorname{mean}(\mu)=\Sigma x.p(x)]()
Hence, from the calculation above, we have:
![\text{Mean}=14.83](https://img.qammunity.org/2023/formulas/mathematics/college/ajg5g18e8jrscg7zwfrqkam3w332k92i6s.png)
(b) Variance is given by:
![\begin{gathered} \text{Variance(}\sigma^2)=\Sigma.x^2p(x)^{}-\mu^2 \\ =256.21-219.9289 \\ =36.2811 \\ \approx36.281 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/1ecvoyckymmf4e2zh99moj8b4lb1i7b2c7.png)
(c) Standard deviation is given by:
![\begin{gathered} \text{Standard deviation(}\sigma)=\sqrt[]{36.2811} \\ =6.02337 \\ \approx6.023 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/xkxs6hn0doirx3ort7sm0d6pv6bwf9gnci.png)
(d) Expectation value of X is given by:
![\begin{gathered} E(X)=\mu \\ =14.83 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kugegote3zh6dhlegmymn4fnpq9zxikmgb.png)