A manager at a local company asked his employees how many times they had given blood in the last year. Theresults of the survey are given below. The random variable x represents the number of times a person gaveblood and P(x) represents the probability of selecting an employee who had given blood that percent of thetime. What is the standard deviation for the number of times a person gave blood based on this survey? $$\begin{array} { | c | c | c | c | c | c | c | c | }
\hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \
\hline P ( x ) & 0.30 & 0.25 & 0.20 & 0.12 & 0.07 & 0.04 & 0.02 \
\hline
\end{array}$$
A) 1.54
B) 1.82
C) 1.16
D) 2.23

Question

A manager at a local company asked his employees how many times they had given blood in the last year. Theresults of the survey are given below. The random variable x represents the number of times a person gaveblood and P(x) represents the probability of selecting an employee who had given blood that percent of thetime. What is the standard deviation for the number of times a person gave blood based on this survey? $$\begin{array} { | c | c | c | c | c | c | c | c | } 
\hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \
\hline P ( x ) & 0.30 & 0.25 & 0.20 & 0.12 & 0.07 & 0.04 & 0.02 \
\hline
\end{array}$$
A) 1.54
B) 1.82
C) 1.16
D) 2.23

Quizplus · Accepted Answer

The standard deviation is the square root of the variance. The variance is calculated as the sum of the squared deviations from the mean, weighted by their probabilities. The mean is calculated as the sum of the products of the values and their probabilities.In this case, the mean is:E(X) = 0 * 0.30 + 1 * 0.25 + 2 * 0.20 + 3 * 0.12 + 4 * 0.07 + 5 * 0.04 + 6 * 0.02 = 1.5The variance is:Var(X) = (0 - 1.5)^2 * 0.30 + (1 - 1.5)^2 * 0.25 + (2 - 1.5)^2 * 0.20 + (3 - 1.5)^2 * 0.12 + (4 - 1.5)^2 * 0.07 + (5 - 1.5)^2 * 0.04 + (6 - 1.5)^2 * 0.02 = 2.37The standard deviation is:SD(X) = sqrt(Var(X)) = sqrt(2.37) = 1.54Therefore, the standard deviation is 1.54.

A Manager at a Local Company Asked His Employees How