Given that the total of 16 values is 528 and the sum of the squares of deviation from 33 is 9158. The variance is
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Always double check if the numerical value given for deviations matches the actual mean of the dataset. Since $\frac{528}{16} = 33$, you are dealing with standard deviations about the mean, allowing you to bypass any complex correction formulas and solve the problem using a single division step: $\frac{9158}{16} = 572.375$.
Step 1: List what is known. There are $n = 16$ values, their total is $\sum x_i = 528$, and the sum of squared deviations from 33 is $\sum (x_i - 33)^2 = 9158$. Step 2: Find the mean. $\bar{x} = \dfrac{\sum x_i}{n} = \dfrac{528}{16} = 33$. Step 3: Match the reference point to the mean. The deviations were taken about 33, which is exactly the mean. So $\sum (x_i - 33)^2$ is literally $\sum (x_i - \bar{x})^2$, the quantity variance needs. Step 4: Write the variance formula. $\sigma^2 = \dfrac{1}{n}\sum (x_i - \bar{x})^2$. Step 5: Substitute. $\sigma^2 = \dfrac{9158}{16}$. Step 6: Divide. $\dfrac{9158}{16} = 572.375$. \[ \boxed{\text{Variance} = 572.375} \]