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:
![\operatorname{mean}=0*0.193+1*0.354+2*0.304+3*0.112+4*0.034+5*0.003]()
Then,
Solve the above expression:
![\begin{gathered} \operatorname{mean}=0*0.193+1*0.354+2*0.304+3*0.112+4*0.034+5*0.003 \\ \operatorname{mean}=0+0.354+0.608+0.336+0.136+0.015 \\ \operatorname{mean}=1.449 \end{gathered}]()
Hence, the value of the mean is 1.449.
Now,
Find the standard deviation:
So, from the formula of standard deviation
![\text{Standard deviation=}\sqrt[]{\sum^(\infty)_(n\mathop=0)(x-mean})^2* P(x)](https://img.qammunity.org/2023/formulas/mathematics/college/x2w06liu8fpupxuwyn1tif5tl6biuga4hq.png)
Then,
![\begin{gathered} \text{Standard deviation=}\sqrt[]{\sum^(\infty)_{n\mathop{=}0}(x-mean})^2* P(x) \\ =\sqrt[]{(0-1.449)^2*0.193+(1-1.449)^2*0.354+(2-1.449)^2*0.304+(3-1.449)^2*0.112+(4-1.449)^2*0.034+(5-1.449)^2*0.003} \end{gathered}]()
Then,
![\begin{gathered} =\sqrt[]{(0-1.449)^2*0.193+(1-1.449)^2*0.354+(2-1.449)^2*0.304+(3-1.449)^2*0.112+(4-1.449)^2*0.034+(5-1.449)^2*0.003} \\ =0.405+2.123+0.092+0.27+0.221+0.0378 \\ =3.1488 \end{gathered}]()
Hence, the value of the standard deviation is 3.1488.