# Calculate standard deviation given tabled number of occurrences

I am going back through my old homework assignments to study for an upcoming Statistics test and one of the questions gives a table with number of children and number of women who have that number of children:

Children        |   0    |    1    |    2    |    3    |    4    |    5
----------------+--------+---------+---------+---------+---------+--------
Number of Women |  27    |   22    |   30    |   12    |    7    |    2


The question is:

Find the sample standard deviation of the number of children.

As I understand it, the sample standard deviation is

$s=\sqrt{\frac{1}{n-1}(\sum\limits_{i=1}^n{X^2_i-n\bar{X}^2})}$

And in order to do this, I would need to sum up each of the individual values given within the set of $n$ values. However, since there are 100 values (collapsed as the table shows), How might I do this by hand, that is, without the aid of a program on a test?

I was able to (I think) calculate $\bar{X}$ by summing the product of the number of children by the number of women and dividing by $n$:

$\bar{X}=\frac{1}{n}(\sum\limits_{i=1}^n{X_i})$

$\bar{X}=\frac{1}{100}\big[(0*27)+(1*22)+(2*30)+(3*12)+(4*7)+(5*2)\big] = 1.56$

But I'm unsure how to do the same with the sample standard deviation calculation.

I tried searching and looking online for the answer to this, but I'm obviously a novice at this, so I might just be searching for the wrong things.

Your formula for the mean is correct.

Presumably you can work out the mean from the variance.

Have you seen the formula for the population variance $\text{Var}(X)=E(X^2)-E(X)^2$?

The usual sample variance is the same, with Bessel's correction -
$\text{Var}(x)=\frac{n}{n-1} [\overline{x^2}-\overline{x}^2]$.

(Where $\overline{x^2}$ means 'take the mean of the squared observations')

From that you can use exactly the same method you did for the mean to work out the mean of the $x^2$ values and hence obtain the variance and then the standard deviation.

The correct formula for the unbiased standard deviation is

$$\sigma = \sqrt\frac{\sum_{i=1}^n(X_i - \overline{X})^2}{n-1}$$

Edit

To compute the standard deviation of the children the right formula to apply is:

$$\sigma = \sqrt{E[(X-\mu)^2]} = \sqrt{E[X^2] - (E[X])^2}$$

So to compute the mean, the procedure you did could be rewritten as:

$$\overline{X} = E[X] = \sum{P(X)X} = (0.27\times0) + (0.22\times1) + (0.30\times2) + (0.12\times3) +(0.07\times4) + (0.02\times5) = 1.56$$

I let you complete the exercise.

• Can you be more specific? How might I actually calculate the number, given that I have a large $n$ (100)? Nov 2, 2014 at 0:08
• Excuse me, I misunderstood the question. I've updated the answer. Nov 2, 2014 at 10:52

What I ended up doing (which kinda goes along the lines of giuseppe's answer) is to look at the calculation for $s$ as:

$s=\sqrt{\frac{1}{n-1}\big((\sum\limits_{i=1}^n{X^2_i})-n\bar{X}^2\big)}$

Note the parentheses surrounding the summation. This can be done because $n\bar{X}^2$ can be seen as a constant term which can be removed from the summation.

This then lends to the calculation easily as before:

$s^s=\frac{1}{99}\Big(\big[(0*27)^2+(1*22)^2+(2*30)^2+(3*12)^2+(4*7)^2+(5*2)^2\big] - 1.56\Big)$

$s^2 = 1.7034$

$s = \sqrt{s^2} = 1.30515$