Standard Deviation given n, Σx, Σx²

Question

I have been trying to solve this problem for over half an hour now. I don't know if it is to do with my incapability of using a calculator or a misunderstanding in standard deviation.

Question:

A sample of 10 people complete a task and their times, x, are recorded. Given that Σx = 273 and Σx²=7561, calculate the mean and standard deviation of the times taken.

Attempt of a solution:

The mean would be Σx / n. Therefore, the mean is 273/10 = 27.3

As for standard deviation, since it is a sample, we would use the formula: Link to the formula

But when I type √(7561/(10-1))-27.3² (square root over whole thing) , I get 9.74 (3sf)

Actual Answer:

= 3.29 (EDIT: The answer on the book was incorrect?)

Answer 1

sample standard deviation $= \sqrt{\left(\dfrac{\sum (x-\bar x)^2}{n-1}\right)}$$= \sqrt{\left(\dfrac{\sum x^2 - 2\bar x \sum X + n\bar x ^2}{9}\right)}$

$=\sqrt{\left(\dfrac{\sum x^2 - \frac{(\sum X)^2}{n}}{9}\right)}$

$ = \sqrt{\left(\dfrac{7561 - \frac{273^2}{10}}{9}\right)} = \sqrt{12.0111} = 3.47$

Answer 2

Just keep it short and simple. Use the formula $Var(X)=E(X^2)-[E(X)]^2$

$$Var(X)=\frac1n \sum_{i=1}^n x_i^2-\left[\frac1n \sum_{i=1}^n x_i\right]^2=\frac1{10}\cdot 7561-27.3^2$$

Therefore the standard deviation of the sample is $\sqrt{\frac1{10}\cdot 7561-27.3^2}=3.2878..\approx3.29$

Remark:

This is the standard deviation of the sample. That it was asked for in the exercise. The book is not wrong.

Sometimes it is asked for an unbiased estimator for the standard deviation of the population. Then the formula is $$Var(X)=\frac{1}{n-1} \left[\sum_{i=1}^n x_i^2-n\cdot \left[\frac1n \sum_{i=1}^n x_i\right]^2\right]$$