Question

In a game of luck, a turn consists of a player rolling [12] fair [6]-sided dice. Let [x=] the number of dice that land showing "[1]" in a turn. Find the mean and standard deviation of [x]. You may round your answers to the nearest tenth.

Answer 1

In a game where 12 dice are rolled, the mean of the number of dice showing "1" is 2, and the standard deviation is approximately 3.2.

Step-by-step explanation:

In the context of a game of chance, where a player rolls 12 fair six-sided dice, the variable x denotes the number of dice that land showing "1" in a single turn. To find the mean (expected value) of x, consider that each die is independent, and the probability of landing a "1" on any die is 1/6. Since there are 12 dice, the mean is calculated as:

Mean (expected value) = Number of dice × Probability of rolling a "1" = 12 × 1/6 = 2

To calculate the standard deviation, we use the formula for the standard deviation of a binomial distribution, which is:

Standard deviation = \(\sqrt{n × p × (1 - p)}\), where n is the number of trials (dice), and p is the probability of success (rolling a "1"). Plugging in the numbers, we get:

Standard deviation = \(\sqrt{12 × 1/6 × (1 - 1/6)}\) = \(\sqrt{12 × 1/6 × 5/6}\) = \(\sqrt{10}\) which is approximately 3.2.

Therefore, the mean of x is 2, and the standard deviation is approximately 3.2 when rounded to the nearest tenth.