Find the variance for the given data. Round your answer to one more decimal place than the original data.
-Christine is currently taking college astronomy. The instructor often gives quizzes. On the past seven quizzes, Christine got the following scores: $$\begin{array} { l l l l l l l }
50 & 12 & 38 & 25 & 16 & 48 & 58
\end{array}$$ Compute the standard deviation s.
A)10,637
B)8715.6
C)38
D)17.9

Question

Find the variance for the given data. Round your answer to one more decimal place than the original data.
-Christine is currently taking college astronomy. The instructor often gives quizzes. On the past seven quizzes, Christine got the following scores: $$\begin{array} { l l l l l l l } 
50 & 12 & 38 & 25 & 16 & 48 & 58
\end{array}$$ Compute the standard deviation s.
A)10,637
B)8715.6
C)38
D)17.9

Quizplus · Accepted Answer

To find the standard deviation, we first calculate the mean of the scores, then use this mean to find the variance, and finally take the square root of the variance to get the standard deviation. Mean = (50 + 12 + 38 + 25 + 16 + 48 + 58) / 7 = 247 / 7 = 35.3Variance = [(50-35.3)^2 + (12-35.3)^2 + (38-35.3)^2 + (25-35.3)^2 + (16-35.3)^2 + (48-35.3)^2 + (58-35.3)^2] / 7= [217.16 + 540.89 + 7.29 + 106.09 + 372.49 + 161.29 + 513.69] / 7= 1918.9 / 7= 274.1Standard deviation = sqrt(variance) = sqrt(274.1) ≈ 16.6However, the correct answer is not listed directly in the options, which suggests a possible mistake in the calculation or interpretation of the options. Given the options, the closest relevant calculation to the process described would be the standard deviation, but the final calculation provided does not match any of the options exactly. The correct process involves calculating the mean, variance, and then the standard deviation, but a mistake in the final step or a misunderstanding of the options may have occurred. The correct standard deviation calculation should involve finding the square root of the variance, which was calculated as 274.1, and the square root of 274.1 is approximately 16.6, not 17.9 as listed in any of the options. Therefore, it seems there was a mistake in the final interpretation of the answer.

Find the Variance for the Given Data 50123825164858\begin{array} { l l l l l l l } 50 & 12 & 38 & 25 & 16 & 48 & 58\end{array}50​12​38​25​16​48​58​

Find the Variance for the Given Data $\begin{array} { l l l l l l l } 50 & 12 & 38 & 25 & 16 & 48 & 58\end{array}$