Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The textbook gives an example of testing a null hypothesis that rolling a die 100 times will give you a value of $6$, $\frac{1}{6}$ times. In the experiment, a die was rolled 100 times and 30 of them were $6$'s.

The book obtains a $z$ score for this with the formula $$\frac{\bar{x}- \mu}{ \sqrt{ \frac{p(1-p)}{100}}} = \frac{.30- .167}{ \sqrt{ \frac{.167(1-.167)}{100}}} $$.

I understand that $\sqrt{ \frac{0.167(1-0.167)}{100}}$ must be the standard deviation of the sample mean (tell me if I'm wrong), but how did they get $0.167(1-0.167)$ as the variance?

Where did that formula come from?

share|cite|improve this question
in your case $p=0.167$ is just the probability that a particular number on the dice shows up. For Bernoulli trials the variance is simply $p(1-p)$. Hope that helps. – jay-sun Nov 26 '12 at 23:32
+1 for the gravatar – alancalvitti Nov 27 '12 at 3:34
up vote 1 down vote accepted

Toss a fair die once, and let $X=1$ if the result is a $6$, and let $X=0$ otherwise. Then $E(X)=\frac{1}{6}$, and $E(X^2)=\frac{1}{6}$, and therefore $$\text{Var}(X)=E(X^2)-(E(X))^2=\frac{1}{6}-\frac{1}{6^2}=\frac{1}{6}\left(1-\frac{1}{6}\right).$$

In general, let us repeat an experiment independently $n$ times, and suppose the probability of success on any trial is $p$. Let $Y$ be the number of successes. Then $Y$ has binomial distribution. You probably have seen the fact that $Y$ has variance $npq=np(1-p)$. This can be proved in various ways. For example, let $X_i=1$ if we get a success on the $i$-th trial, and $0$ otherwise. Note that $Y=X_1+X_2+\cdots+X_n$. By an argument virtually identical to the argument of the first paragraph, each $X_i$ has variance $p(1-p)$.

But the variance of a sum of independent random variables is the sum of the variances, so $Y$ has variance $np(1-p)$.

Let $\overline{Y}$ be the sample mean. Then $\overline{Y}=\dfrac{Y}{n}$. It follows that $$\text{Var}(\overline{Y})=\frac{1}{n^2}\text{Var}(Y)=\frac{np(1-p)}{n^2}=\frac{p(1-p)}{n}.$$

share|cite|improve this answer
Ah it's a Bernoulli equation - I didn't think of it like that. But why is $E[X^2] = \frac{1}{6}$? Shouldn't it be $\frac{1}{216}$? (Since $X^2 = \frac{1}{36}$, divided by $6, = \frac{1}{216}$) – CodyBugstein Nov 26 '12 at 23:35
Note that $X^2=X$, since $X$ only takes on values $1$ and $0$. So if $W=X^2$, then $W=1$ with probability $\frac{1}{6}$, $0$ with probability $\frac{5}{6}$. – André Nicolas Nov 26 '12 at 23:43
Thanks for the awesome explanation! – CodyBugstein Nov 26 '12 at 23:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.