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.