Question

A population has mean 187 and standard deviation 32. If a random sample of 64 observations is selected at random from this population, what is the probability that the sample average will be less than 182

Answer 1

11.51% probability that the sample average will be less than 182

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 = 187, \sigma = 32, n = 64, s = (32)/(√(64)) = 4`

What is the probability that the sample average will be less than 182

This is the pvalue of Z when X = 182. So

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

By the Central Limit Theorem

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

`Z = (182 - 187)/(4)`

`Z = -1.2`

`Z = -1.2` has a pvalue of 0.1151

11.51% probability that the sample average will be less than 182