A discrete random variable x can assume five possible values: 2, 3, 5, 8, 10. Its probability distribution is shown below. Find the standard deviation of the distribution. $$\begin{array} { c | c c c c c }
\mathrm { x } & 2 & 3 & 5 & 8 & 10 \
\hline \mathrm { p } ( \mathrm { x } ) & 0.10 & 0.20 & 0.30 & 0.30 & 0.10
\end{array}$$
A) 2.532
B) 5.7
C) 6.41
D) 1.845

Question

A discrete random variable x can assume five possible values: 2, 3, 5, 8, 10. Its probability distribution is shown below. Find the standard deviation of the distribution. $$\begin{array} { c | c c c c c } 
\mathrm { x } & 2 & 3 & 5 & 8 & 10 \
\hline \mathrm { p } ( \mathrm { x } ) & 0.10 & 0.20 & 0.30 & 0.30 & 0.10
\end{array}$$
A) 2.532
B) 5.7
C) 6.41
D) 1.845

Quizplus · Accepted Answer

The standard deviation is the square root of the variance. The variance is the expected value of the squared deviation from the mean. The mean is the expected value of the random variable. The expected value of the random variable is:$$E(X) = 2(0.10) + 3(0.20) + 5(0.30) + 8(0.30) + 10(0.10) = 5.4$$The variance is:$$Var(X) = E(X^2) - [E(X)]^2$$$$E(X^2) = 2^2(0.10) + 3^2(0.20) + 5^2(0.30) + 8^2(0.30) + 10^2(0.10) = 36.6$$$$Var(X) = 36.6 - 5.4^2 = 16.24$$The standard deviation is:$$\sigma = \sqrt{Var(X)} = \sqrt{16.24} = 4.03$$

A Discrete Random Variable X Can Assume Five Possible Values