Question

A fair six-sided number cube is rolled 60 times. What is the probability that fewer than 10% of the rolls are a five?

Find the z-table here.

Answer 1

Answer: 0.082

Explanation:

Answer 2

0.0823 = 8.23% probability that fewer than 10% of the rolls are a five

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`.

A fair six-sided number cube is rolled 60 times.

This means that
`n = 60`

Rolls that are a five:

For each roll, there are 6 possible outcomes: 1, 2, 3, 4, 5 or 6. So the probability of rolling a five is:

`p = (1)/(6) = 0.1667`

The distribution has mean and standard deviation:

`\mu = p = 0.1667`

`s = \sqrt{(p(1-p))/(n)} = \sqrt{(0.1667*(1-0.1667))/(60)} = 0.0481`

What is the probability that fewer than 10% of the rolls are a five?

This is the pvalue of Z when X = 0.1. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (0.1 - 0.1667)/(0.0471)`

`Z = -1.39`

`Z = -1.39` has a pvalue of 0.0823

0.0823 = 8.23% probability that fewer than 10% of the rolls are a five