Compute the standard deviation for ages of British nurses in 1851. Assume that the table below shows the age distribution of nurses in Great Britain in 1851. Round your answer to nearest hundredth. $$\mathcal { \sigma }$$ $$\begin{array} { | l | c | c | c | c | c | c | c | }
\hline \text { Age range (yr) } & \mathbf { 2 0 } - \mathbf { 2 9 } & \mathbf { 3 0 } - \mathbf { 3 9 } & \mathbf { 4 0 } - \mathbf { 4 9 } & \mathbf { 5 0 } - \mathbf { 5 9 } & \mathbf { 6 0 } - \mathbf { 6 9 } & \mathbf { 7 0 } - \mathbf { 7 9 } & \mathbf { 8 0 } + \
\hline \text { Midpoint } ( x ) & 24.5 & 34.5 & 44.5 & 54.5 & 64.5 & 75.5 & 84.5 \
\hline \text { Percent of nurses } & 5.8 \% & 9.8 \% & 19.6 \% & 29.2 \% & 25.1 \% & 9.1 \% & 1.4 \% \
\hline
\end{array}$$
A)13.57
B)1.52
C)3.26
D)53.68
E)17.26

Question

Compute the standard deviation for ages of British nurses in 1851. Assume that the table below shows the age distribution of nurses in Great Britain in 1851. Round your answer to nearest hundredth. $$\mathcal { \sigma }$$ $$\begin{array} { | l | c | c | c | c | c | c | c | } 
\hline \text { Age range (yr) } & \mathbf { 2 0 } - \mathbf { 2 9 } & \mathbf { 3 0 } - \mathbf { 3 9 } & \mathbf { 4 0 } - \mathbf { 4 9 } & \mathbf { 5 0 } - \mathbf { 5 9 } & \mathbf { 6 0 } - \mathbf { 6 9 } & \mathbf { 7 0 } - \mathbf { 7 9 } & \mathbf { 8 0 } + \
\hline \text { Midpoint } ( x ) & 24.5 & 34.5 & 44.5 & 54.5 & 64.5 & 75.5 & 84.5 \
\hline \text { Percent of nurses } & 5.8 \% & 9.8 \% & 19.6 \% & 29.2 \% & 25.1 \% & 9.1 \% & 1.4 \% \
\hline
\end{array}$$
A)13.57
B)1.52
C)3.26
D)53.68
E)17.26

Quizplus · Accepted Answer

The standard deviation is a measure of the spread of a distribution. It is calculated by taking the square root of the variance. The variance is calculated by taking the sum of the squared deviations from the mean, divided by the number of observations.In this case, the mean age of the nurses is 44.5 years. The variance is calculated as follows:```Variance = [(24.5 - 44.5)^2 * 0.058] + [(34.5 - 44.5)^2 * 0.098] + [(44.5 - 44.5)^2 * 0.196] + [(54.5 - 44.5)^2 * 0.292] + [(64.5 - 44.5)^2 * 0.251] + [(74.5 - 44.5)^2 * 0.091] + [(84.5 - 44.5)^2 * 0.014] = 366.26```The standard deviation is the square root of the variance, which is 19.13. Rounded to the nearest hundredth, the standard deviation is 19.13.

Compute the Standard Deviation for Ages of British Nurses in 1851