As given by the question
There are given that the probability distribution and the value of x
Now,
First find the value of mean:
So,
The value of mean will be:
[tex]\operatorname{mean}=0\times0.193+1\times0.354+2\times0.304+3\times0.112+4\times0.034+5\times0.003[/tex]
Then,
Solve the above expression:
[tex]\begin{gathered} \operatorname{mean}=0\times0.193+1\times0.354+2\times0.304+3\times0.112+4\times0.034+5\times0.003 \\ \operatorname{mean}=0+0.354+0.608+0.336+0.136+0.015 \\ \operatorname{mean}=1.449 \end{gathered}[/tex]
Hence, the value of the mean is 1.449.
Now,
Find the standard deviation:
So, from the formula of standard deviation
[tex]\text{Standard deviation=}\sqrt[]{\sum^{\infty}_{n\mathop=0}(x-mean})^2\times P(x)[/tex]
Then,
[tex]\begin{gathered} \text{Standard deviation=}\sqrt[]{\sum^{\infty}_{n\mathop{=}0}(x-mean})^2\times P(x) \\ =\sqrt[]{(0-1.449)^2\times0.193+(1-1.449)^2\times0.354+(2-1.449)^2\times0.304+(3-1.449)^2\times0.112+(4-1.449)^2\times0.034+(5-1.449)^2\times0.003} \end{gathered}[/tex]
Then,
[tex]\begin{gathered} =\sqrt[]{(0-1.449)^2\times0.193+(1-1.449)^2\times0.354+(2-1.449)^2\times0.304+(3-1.449)^2\times0.112+(4-1.449)^2\times0.034+(5-1.449)^2\times0.003} \\ =0.405+2.123+0.092+0.27+0.221+0.0378 \\ =3.1488 \end{gathered}[/tex]
Hence, the value of the standard deviation is 3.1488.
