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:

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \sqrt{\frac{p(1-p)}{n}}$.

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:

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

The distribution has mean and standard deviation:

$\mu = p = 0.1667$

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

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

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.1 - 0.1667}{0.0471}$

$Z = -1.39$

$Z = -1.39$ has a pvalue of 0.0823

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

Answer 2

Answer: 0.082

Step-by-step explanation: