# In a series of 2n observations, half of them equal a and remaining half -a.If the standard deviation of the observations is 2,then |a| equals ?

Jan 12, 2018

$\left\mid a \right\mid = 2$

#### Explanation:

Recall the definition of standard deviation:

$\sigma = \sqrt{\frac{\sum {\left({x}_{i} - \mu\right)}^{2}}{N}}$

Where $\mu$ is the mean, ${x}_{i}$ are the observations, $\sum$ is the sum, and $N$ is the number of observations.

Let's first find the mean. By definition:

$\mu = \frac{\sum {x}_{i}}{N}$

Recall that $\sum {x}_{i}$ represents the sum of all the observations. Since we have $n$ observations that are $a$ and $n$ observations that are $- a$, we write:

$\mu = \frac{{\overbrace{a + a + \ldots + a}}^{n \text{ times")+overbrace(-a -a-...-a)^(n " times}}}{2 n}$

Which is equivalent to:

$\mu = \frac{n a + n \left(- a\right)}{2 n}$

And, simplifying:

$\mu = 0$

This makes our calculation for the standard deviation much simpler as well:

$\sigma = \sqrt{\frac{\sum {\left({x}_{i} - 0\right)}^{2}}{N}} = \sqrt{\frac{\sum {x}_{i}^{2}}{2 n}}$

Recall that $\sum {x}_{i}^{2}$ means to take the sum of the square of every observation we have. This translates into:

$\sigma = \sqrt{\frac{{\overbrace{{a}^{2} + {a}^{2} + \ldots + {a}^{2}}}^{n \text{ times")+overbrace((-a)^2+(-a)^2+...+(-a)^2)^(n" times}}}{2 n}}$

Which we can rewrite with more mathematical precision as:

$\sigma = \sqrt{\frac{n \left({a}^{2}\right) + n \left({a}^{2}\right)}{2 n}}$

Then, we simplify:

$\sigma = \sqrt{\frac{2 n {a}^{2}}{2 n}} = \sqrt{{a}^{2}} = \left\mid a \right\mid$

We are told that $\sigma = 2$, so

$\left\mid a \right\mid = 2$