Question

A normal population has a mean of 58 and a standard deviation of 13. You select a random sample of 25. Compute the probability the sample mean is: (Round your z values to 2 decimal places and final answers to 4 decimal places): Greater than 60.

Answer 1

0.2206 = 22.06% probability that the sample mean is greater than 60.

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 = 58, \sigma = 13, n = 25, s = (13)/(√(25)) = 2.6`

Compute the probability the sample mean is greater than 60:

This probability is 1 subtracted by the pvalue of Z when X = 60. So

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

By the Central Limit Theorem

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

`Z = (60 - 58)/(2.6)`

`Z = 0.77`

`Z = 0.77` has a pvalue of 0.7794

1 - 0.7794 = 0.2206

0.2206 = 22.06% probability that the sample mean is greater than 60.