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

Answer:

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

Step-by-step 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:

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

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:

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 $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 12, \sigma = 3.2, n = 36, s = \frac{3.2}{\sqrt{36}} = 0.533$

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

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

By the Central Limit Theorem

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

$Z = \frac{10-12}{0.533}$

$Z = -3.75$

$Z = -3.75$ 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.