The number of customers in a line at a supermarket express checkout counter is a random variable with the following probability distribution.

$\begin{array}{l}
\begin{array} { c | c | c | c | c | c | c }
\boldsymbol { x } & 0 & 1 & 2 & 3 & 4 & 5 \
\hline \boldsymbol { P } ( \boldsymbol { x } ) & 0.11 & 0.22 & 0.31 & 0.16 & 0.09 & 0.11
\end{array}\\
\text { Compute the standard deviation } \sigma _ { x } \text {. }
\end{array}$

A) 2.12
B) 2.23
C) 4.48
D) 1.46

Question

The number of customers in a line at a supermarket express checkout counter is a random variable with the following probability distribution.

$\begin{array}{l}
\begin{array} { c | c | c | c | c | c | c } 
\boldsymbol { x } & 0 & 1 & 2 & 3 & 4 & 5 \
\hline \boldsymbol { P } ( \boldsymbol { x } ) & 0.11 & 0.22 & 0.31 & 0.16 & 0.09 & 0.11
\end{array}\\
\text { Compute the standard deviation } \sigma _ { x } \text {. }
\end{array}$

A) 2.12
B) 2.23
C) 4.48
D) 1.46

Quizplus · Accepted Answer

To compute the standard deviation, we first need to find the mean ($\mu$) of the distribution. The mean is calculated as $\mu = \sum xP(x)$. Then, we calculate the variance ($\sigma^2$) using the formula $\sigma^2 = \sum (x - \mu)^2P(x)$. Finally, the standard deviation ($\sigma$) is the square root of the variance.Mean ($\mu$) calculation:$\mu = (0)(0.11) + (1)(0.22) + (2)(0.31) + (3)(0.16) + (4)(0.09) + (5)(0.11) = 0 + 0.22 + 0.62 + 0.48 + 0.36 + 0.55 = 2.23$Variance ($\sigma^2$) calculation:$\sigma^2 = (0-2.23)^2(0.11) + (1-2.23)^2(0.22) + (2-2.23)^2(0.31) + (3-2.23)^2(0.16) + (4-2.23)^2(0.09) + (5-2.23)^2(0.11)$$\sigma^2 = 4.9729(0.11) + 1.5129(0.22) + 0.0529(0.31) + 0.5929(0.16) + 3.1329(0.09) + 7.6729(0.11)$$\sigma^2 = 0.547 + 0.3328 + 0.0164 + 0.0949 + 0.2819 + 0.844 = 2.116$Standard deviation ($\sigma$) calculation:$\sigma = \sqrt{\sigma^2} = \sqrt{2.116} \approx 1.46$Therefore, the standard deviation of the number of customers in line at the supermarket express checkout counter is approximately 1.46.

The Number of Customers in a Line at a Supermarket