Question

The mean number of words per minute (WPM) read by sixth graders is 89 with a standard deviation of 16 WPM. If 66 sixth graders are randomly selected, what is the probability that the sample mean would differ from the population mean by more than 5.09 WPM?

Answer 1

0.98% probability that the sample mean would differ from the population mean by more than 5.09 WPM.

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

In this problem, we have that:

`\mu = 89, sigma = 16, n = 66, s = (16)/(√(66)) = 1.97`

What is the probability that the sample mean would differ from the population mean by more than 5.09 WPM?

Either it differs less than 5.09 or equal, or it differs by more than 5.09. The sum of these probabilities is 100%.

So we first find the probability that the sample mean would differ from the population mean by less than 5.09 WPM, and then we subtract 100 by this probability.

Differ less than 5.09

pvalue of Z when X = 89 + 5.09 = 94.09 subtracted by the pvalue of Z when X = 89 - 5.09 = 83.91

X = 94.09

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

By the Central Limit Theorem

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

`Z = (94.09 - 89)/(1.97)`

`Z = 2.58`

`Z = 2.58` has a pvalue of 0.9951.

X = 83.91

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

`Z = (83.91 - 89)/(1.97)`

`Z = -2.58`

`Z = -2.58` has a pvalue of 0.0049.

0.9951 - 0.0049 = 0.9902 = 99.02% probability that it differs less than 5.09

Differ more than 5.09

100 - 99.02 = 0.98% probability that the sample mean would differ from the population mean by more than 5.09 WPM.