Answer to Question #295621 in Statistics and Probability for Alyssa

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 $\dfrac{68}{33}.$

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

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

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

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

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