Find the variance for the given data. Round your answer to one more decimal place than the original data.
-The manager of an electrical supply store measured the diameters of the rolls of wire in the inventory. The diameters of the rolls (in m)are listed below. Compute the standard deviation s. $$\begin{array} { l l l l l l l }
0.189 & 0.518 & 0.665 & 0.568 & 0.149 & 0.573 & 0.202
\end{array}$$
A)1.1718
B)0.568
C)0.2193
D)1.4602

Question

Find the variance for the given data. Round your answer to one more decimal place than the original data.
-The manager of an electrical supply store measured the diameters of the rolls of wire in the inventory. The diameters of the rolls (in m)are listed below. Compute the standard deviation s. $$\begin{array} { l l l l l l l } 
0.189 & 0.518 & 0.665 & 0.568 & 0.149 & 0.573 & 0.202
\end{array}$$
A)1.1718
B)0.568
C)0.2193
D)1.4602

Quizplus · Accepted Answer

The variance is calculated by finding the mean of the data, subtracting the mean from each data point, squaring the result, finding the mean of those squared differences, and rounding to one more decimal place than the original data. The calculation steps are as follows: 1. Calculate the mean (average) of the data: $\frac{0.189 + 0.518 + 0.665 + 0.568 + 0.149 + 0.573 + 0.202}{7} = 0.409$2. Subtract the mean from each data point and square the result:    - $(0.189 - 0.409)^2 = 0.0484$   - $(0.518 - 0.409)^2 = 0.0119$   - $(0.665 - 0.409)^2 = 0.0656$   - $(0.568 - 0.409)^2 = 0.0253$   - $(0.149 - 0.409)^2 = 0.0676$   - $(0.573 - 0.409)^2 = 0.0268$   - $(0.202 - 0.409)^2 = 0.0428$3. Find the mean of those squared differences (variance):    $\frac{0.0484 + 0.0119 + 0.0656 + 0.0253 + 0.0676 + 0.0268 + 0.0428}{7} = 0.0412$4. Round to one more decimal place than the original data: $0.0412$ rounded to four decimal places is $0.0412$, which is not an option provided. However, the calculation seems to have been intended to find the variance, not the standard deviation. The correct calculation for variance seems to have been misunderstood in the options provided. Given the options and the process, it appears there might have been a misunderstanding in the calculation or the interpretation of the question. The correct variance calculation based on the provided steps should yield a result that matches the options, but none of the options accurately reflect the variance calculated from the given steps. The correct variance calculation involves squaring the differences from the mean, summing those squared differences, and dividing by the number of data points. The mistake might be in the interpretation of the final step or a miscalculation in the variance. Without recalculating with correct values or having the correct option listed, it's challenging to match the provided options.

Find the Variance for the Given Data 0.1890.5180.6650.5680.1490.5730.202\begin{array} { l l l l l l l } 0.189 & 0.518 & 0.665 & 0.568 & 0.149 & 0.573 & 0.202\end{array}0.189​0.518​0.665​0.568​0.149​0.573​0.202​

Find the Variance for the Given Data $\begin{array} { l l l l l l l } 0.189 & 0.518 & 0.665 & 0.568 & 0.149 & 0.573 & 0.202\end{array}$