Question

Find the variance and standard deviation of the probability distribution of the random variable X, which can take only the values 1,2, and 3, given that P(1)= 10/33 P(2)=1/3 and P(3) = 22/33

Answer 1

The probability of each of these events and the corresponding value of X, can be summarized below.

$$\begin{matrix} x & 1 & 2 & 3\\ P(x) & \frac{10}{33} & \frac{1}{3} & \frac{12}{33}. \end{matrix}$$

The mean (also known as the expected value) of a discrete random variable X is the number given by

$$\mu =E(X)=\sum x P(x)=1\left(\frac{10}{33}\right)+2\left(\frac{1}{3}\right)+3\left(\frac{12}{33}\right),\\ =\frac{10}{33}+\frac{2}{3}+\frac{36}{33}=\frac{68}{33}\approx2.060606$$

Therefore, mean of the random variable x is $\dfrac{68}{33}.$

$$E(X^2)=\sum x P(x)=1^2\left(\frac{10}{33}\right)+2^2\left(\frac{1}{3}\right)+3^2\left(\frac{12}{33}\right),\\$$

$$=\frac{10}{33}+\frac{4}{3}+\frac{108}{33}=\frac{162}{33}.$$

$$Var(X)=\sigma^2=E(X^2)-(E(X))^2$$

$$=\frac{162}{33}-(\frac{68}{33})^2=\frac{722}{1089}$$

$$\sigma=\sqrt{\frac{722}{1089}}=\dfrac{19\sqrt{2}}{33}\approx0.814244$$