How to find a data value using only standard deviation?

Question

I have a homework problem with the following data values: x, 17, 24, 13, 21. I need to find x, but I do not have the mean. I only have the standard deviation, which is square root of 14, or 3.745. How do I find x, or at least just the mean, using only the standard deviation?

Answer 1

Hint (not intended to be a complete solution): usually when people refer to the "standard deviation," they mean $$S = \sqrt{\dfrac{1}{n-1}\sum\limits_{i=1}^{n}\left(x_i - \bar{x}\right)^2}\text{.}$$ Notice here that $$\sum\limits_{i=1}^{n}\left(x_i - \bar{x}\right)^2 = \sum\limits_{i=1}^{n}x_i^2-\sum\limits_{i=1}^{n}2x_i\bar{x}+\sum\limits_{i=1}^{n}\bar{x}^2=\sum\limits_{i=1}^{n}x_i^2 - 2n\bar{x}^2+n\bar{x}^2=\sum\limits_{i=1}^{n}x_i^2-n\bar{x}^2\text{.}$$ So you have $$\sqrt{\dfrac{1}{n-1}\sum\limits_{i=1}^{n}\left(x_i - \bar{x}\right)^2} = \sqrt{14}$$ so that $$\dfrac{1}{n-1}\sum\limits_{i=1}^{n}\left(x_i - \bar{x}\right)^2 = 14$$ and as shown above, $$\sum\limits_{i=1}^{n}\left(x_i - \bar{x}\right)^2 = \sum\limits_{i=1}^{n}x_i^2-n\bar{x}^2$$ which is $$x^2 + 17^2 + 24^2 + 13^2 + 21^2 - n\left[\dfrac{x + 17 + 24 + 13 + 21 }{n}\right]^2$$ so $$\dfrac{1}{n-1}\left\{x^2 + 17^2 + 24^2 + 13^2 + 21^2 - n\left[\dfrac{x + 17 + 24 + 13 + 21}{n}\right]^2\right\} = 14\text{.}$$ You know the value of $n$; now solve for $x$.