Question

The average number of moves a person makes in his or her lifetime is 12. If the standard deviation is 3.2, find the probability that the mean of a sample of 36 people is greater than 10.

Answer 1

99.99% probability that the mean of a sample of 36 people is greater than 10.

Explanation:

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

Normal probability distribution:

Problems of normally distributed samples are solved using 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:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, a large sample size can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 12, \sigma = 3.2, n = 36, s = (3.2)/(√(36)) = 0.533`

Find the probability that the mean of a sample of 36 people is greater than 10.

This probability is 1 subtracted by the pvalue of Z when X = 10. So

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

By the Central Limit Theorem

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

`Z = (10-12)/(0.533)`

`Z = -3.75`

`Z = -3.75` has a pvalue of 0.0001

1 - 0.0001 = 0.9999

99.99% probability that the mean of a sample of 36 people is greater than 10.