# If we were to toss a single die (singular for dice) 1000 times, how many 6s could we expect over the long run? What is the standard deviation?

## Suppose you did this and got 200 6s. What is the z score of 200?

Jul 24, 2017

$\mu = \frac{500}{3}$, $\sigma = \frac{25}{3} \sqrt{2}$, $z = 2 \sqrt{2}$.

#### Explanation:

Let $X$ denote the number of 6's thrown in 1000 trials. Then $X$ has a binomial distribution as it satisfies the 4 conditions for a random variable to have a binomial distribution.

• There is a fixed number of trials. (1000)
• The probability of the event occurring is constant. (1/6)
• Each throw of the die is independent.
• There are only two outcomes. (a 6 is thrown or it is not)

In general if we had a random variable $Z$ distributed binomially we would write Z~B(n,p) where $n$ is the number of trials and $p$ is the probability of the event occurring.

So we have X~B(1000,1/6). We are asked to find the expectation of $X$ and the standard deviation.

It can be shown that the expectation of the general binomial random variable $Z$ is $n p$ and its variance is $n p \left(1 - p\right)$. See a proof here.

So,
$\text{E} \left(X\right) = 1000 \cdot \frac{1}{6}$,
$\text{E} \left(X\right) = \frac{500}{3}$,
$\text{E} \left(X\right) \approx 166.67$ to 2 decimal places.

and
$\text{Var} \left(X\right) = 1000 \cdot \frac{1}{6} \cdot \frac{5}{6}$,
$\text{Var} \left(X\right) = \frac{1250}{9}$.

As $\sigma = \sqrt{\text{Var} \left(X\right)}$,

$\sigma = \frac{25}{3} \sqrt{2}$,
$\sigma \approx 11.79$ to 2 decimal places.

The $z$ score is the number of standard deviations away from the mean. A result of 200 gives a z score of

$z = \frac{200 - \frac{500}{3}}{\frac{25}{3} \sqrt{2}}$,
$z = \left(\frac{\frac{100}{3}}{\frac{25}{3} \sqrt{2}}\right)$,
$z = \frac{4}{\sqrt{2}}$,
$z = 2 \sqrt{2}$,
$z = 2.83$ to 2 decimal places.