We have se following data set expressed in the for (x,P(X=x)):
(7,0.1), (8,0.1), (9,0.3), (10,0.1), (11,0.4)
This is equivalent to a data set (7,8,9,9,9,10,11,11,11,11)
The general formula for the standard deviation of a variable x and a set of n values is given by
![Sd=\sqrt[]{\frac{\Sigma(x_i-\bar{x})^2}{n}}](https://img.qammunity.org/2023/formulas/mathematics/college/6p1f2h0ivhlffgzx6iwj1gd233o7zagqcs.png)
For the given data set, the mean is 9.6 and n = 10
Therefore, we have:
![\begin{gathered} Sd=\sqrt[]{((7-9.6)^2+(8-9.6)^2+3(9-9.6)^2+(10-9.6)^2+4(11-9.6)^2)/(10)} \\ Sd=\sqrt[]{((-2.6)^2+(-1.6)^2+3(-0.6)^2+0.4^2+4\cdot1.4^2)/(10)} \\ Sd=\sqrt[]{\frac{6.76+2.56+3\cdot0.36+0.16+4\cdot1.96_{}}{10}}=\sqrt[]{(18.4)/(10)}=1.4\text{ (rounded)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qtyynyb3sevq5jkf9sfq8ui1jbo0jkg1md.png)