Suppose that a population is known to be normally distributed with £ = 2400 and € = 210. Of a random sample of size n = 8 is selected, calculate the probability that the sample mean will exceed 2,500.

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 normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

In this question:

[tex]\mu = 2400, \sigma = 210, n = 8, s = \frac{210}{\sqrt{8}} = 74.25[/tex]

Calculate the probability that the sample mean will exceed 2,500.

This is 1 subtracted by the pvalue of Z when X = 2500. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{2500 - 2400}{74.25}[/tex]

[tex]Z = 1.35[/tex]

[tex]Z = 1.35[/tex] has a pvalue of 0.9115

1 - 0.9115 = 0.0885

0.0885 = 8.85% probability that the sample mean will exceed 2,500.