The following table presents the probability distribution of the number of vacations taken last year for a randomly chosen family. Compute the standard deviation $\sigma$.
$\begin{array}{c|ccccc}
x & 0 & 1 & 2 & 3 & 4 \
\hline P(x) & 0.06 & 0.67 & 0.18 & 0.07 & 0.02
\end{array}$

A) 1.38
B) 1.32
C) 0.77
D) 0.60

Question

The following table presents the probability distribution of the number of vacations taken last year for a randomly chosen family. Compute the standard deviation $\sigma$.
$\begin{array}{c|ccccc}
x & 0 & 1 & 2 & 3 & 4 \
\hline P(x) & 0.06 & 0.67 & 0.18 & 0.07 & 0.02
\end{array}$

A) 1.38
B) 1.32
C) 0.77
D) 0.60

Quizplus · Accepted Answer

To find the standard deviation, first calculate the mean $\mu = \sum xP(x) = 0(0.06) + 1(0.67) + 2(0.18) + 3(0.07) + 4(0.02) = 1.32$. Then, calculate the variance $\sigma^2 = \sum (x - \mu)^2 P(x) = (0-1.32)^2(0.06) + (1-1.32)^2(0.67) + (2-1.32)^2(0.18) + (3-1.32)^2(0.07) + (4-1.32)^2(0.02) = 0.5929$. Finally, the standard deviation $\sigma = \sqrt{0.5929} \approx 0.77$.

The Following Table Presents the Probability Distribution of the Number σ\sigmaσ

The Following Table Presents the Probability Distribution of the Number $\sigma$