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. $\begin{array}{c|ccccc}
x & 2 & 3 & 5 & 8 & 10 \
\hline p(x) & 0.10 & 0.20 & 0.30 & 0.30 & 0.10
\end{array}$

A) $1.845$
B) $5.7$
C) $6.41$
D) $2.532$

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. $\begin{array}{c|ccccc}
x & 2 & 3 & 5 & 8 & 10 \
\hline p(x) & 0.10 & 0.20 & 0.30 & 0.30 & 0.10
\end{array}$

A) $1.845$
B) $5.7$
C) $6.41$
D) $2.532$

Quizplus · Accepted Answer

The standard deviation of a discrete random variable is calculated using the formula $\sqrt{\sum (x_i - \mu)^2 \cdot p(x_i)}$, where $\mu$ is the mean of the distribution. First, calculate the mean: $\mu = (2)(0.10) + (3)(0.20) + (5)(0.30) + (8)(0.30) + (10)(0.10) = 5.7$. Then, calculate the variance: $\sigma^2 = (2-5.7)^2(0.10) + (3-5.7)^2(0.20) + (5-5.7)^2(0.30) + (8-5.7)^2(0.30) + (10-5.7)^2(0.10) = 6.41$. Finally, the standard deviation is the square root of the variance: $\sigma = \sqrt{6.41} = 2.532$.

A Discrete Random Variable X Can Assume Five Possible Values x235810p(x)0.100.200.300.300.10\begin{array}{c|ccccc}x & 2 & 3 & 5 & 8 & 10 \\\hline p(x) & 0.10 & 0.20 & 0.30 & 0.30 & 0.10\end{array}xp(x)​20.10​30.20​50.30​80.30​100.10​​

A Discrete Random Variable X Can Assume Five Possible Values $\begin{array}{c|ccccc}x & 2 & 3 & 5 & 8 & 10 \\\hline p(x) & 0.10 & 0.20 & 0.30 & 0.30 & 0.10\end{array}$