Question

In a game of luck, a turn consists of a player rolling 121212 fair 666-sided dice. Let X=X=X, equals the number of dice that land showing "111" in a turn. Find the mean and standard deviation of XXX. You may round your answers to the nearest tenth.

Answer 1

Answer:

The mean of X is 2 and the standard deviation is 1.3.

Step-by-step explanation:

For each dice rolled, there are only two possible outcomes. Either it lands on one, or it does not. The probability of a dice rolled landing on one is independent of any other dice. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Probability of rolling a 1.

Six sides, so:

$p = \frac{1}{6} = 0.1667$

12 dices are rolled:

This means that $n = 12$

Find the mean and standard deviation of X

$E(X) = np = 12*0.1667 = 2$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{12*0.1667*0.8333} = 1.3$

The mean of X is 2 and the standard deviation is 1.3.