A recent analysis of the number of rainy days per month found the following outcomes and probabilities. $$\begin{array} { | c | c | }
\hline \text { Namber of Raining Days } ( x ) & \mathrm { P } ( x ) \
\hline \hline 3 & .40 \
\hline 4 & .20 \
\hline 5 & .40 \
\hline
\end{array}$$ The standard deviation of this distribution is _____________.
A).800
B).894
C).400
D)4.00
E).457

Question

A recent analysis of the number of rainy days per month found the following outcomes and probabilities. $$\begin{array} { | c | c | } 
\hline \text { Namber of Raining Days } ( x ) & \mathrm { P } ( x ) \
\hline \hline 3 & .40 \
\hline 4 & .20 \
\hline 5 & .40 \
\hline
\end{array}$$ The standard deviation of this distribution is _____________.
A).800
B).894
C).400
D)4.00
E).457

Quizplus · Accepted Answer

To find the standard deviation of this distribution, first, calculate the mean ($\mu$):$\mu = (3 \times .40) + (4 \times .20) + (5 \times .40) = 1.2 + 0.8 + 2 = 4$Then, calculate the variance ($\sigma^2$):$\sigma^2 = [(3-4)^2 \times .40] + [(4-4)^2 \times .20] + [(5-4)^2 \times .40] = [1 \times .40] + [0 \times .20] + [1 \times .40] = .40 + .40 = .80$Finally, find the standard deviation ($\sigma$) by taking the square root of the variance:$\sigma = \sqrt{.80} \approx .894$Therefore, the standard deviation of this distribution is approximately .894.

A Recent Analysis of the Number of Rainy Days Per