Find the standard deviation for the given sample data. Round your answer to one more decimal place than is present in
the original data.
-Listed below are the amounts of time (in months) that the employees of a restaurant have been working at the restaurant. $$\begin{array} { l l l l l l l l l l }
2 & 3 & 6 & 17 & 22 & 40 & 54 & 73 & 101 & 122
\end{array}$$
A) 41.5 months
B) 42.7 months
C) 43.9 months
D) 40.5 months

Question

Find the standard deviation for the given sample data. Round your answer to one more decimal place than is present in
the original data.
-Listed below are the amounts of time (in months) that the employees of a restaurant have been working at the restaurant. $$\begin{array} { l l l l l l l l l l } 
2 & 3 & 6 & 17 & 22 & 40 & 54 & 73 & 101 & 122
\end{array}$$
A) 41.5 months
B) 42.7 months
C) 43.9 months
D) 40.5 months

Quizplus · Accepted Answer

1. Find the mean: 
Mean = (2 + 3 + 6 + 17 + 22 + 40 + 54 +73 + 101 + 122) / 10 = 42.8 months

2. Find the deviations from the mean:
2- 42.8 = -40.8 
3 - 42.8 = -39.8 
6 - 42.8 = -36.8 
17 - 42.8 = -25.8 
22 - 42.8 = -20.8 
40 - 42.8 = -2.8 
54 - 42.8 = 11.2 
73 - 42.8 = 30.2 
101 - 42.8 = 58.2 
122 - 42.8 = 79.2

3. Find the sum of squared deviations from the mean: 
(-40.8)^2 + (-39.8)^2 + (-36.8)^2 + (-25.8)^2 + (-20.8)^2 + (-2.8)^2 + (11.2)^2 + (30.2)^2 + (58.2)^2 + (79.2)^2 = 41,004.72

4. Divide the sum by the sample size minus 1, then take the square root: 
s = sqrt(41,004.72 / 9) = 42.65

Rounded to one more decimal place than present in the data, the standard deviation is 42.7 months.

Find the Standard Deviation for the Given Sample Data 2361722405473101122\begin{array} { l l l l l l l l l l } 2 & 3 & 6 & 17 & 22 & 40 & 54 & 73 & 101 & 122\end{array}2​3​6​17​22​40​54​73​101​122​

Find the Standard Deviation for the Given Sample Data $\begin{array} { l l l l l l l l l l } 2 & 3 & 6 & 17 & 22 & 40 & 54 & 73 & 101 & 122\end{array}$