Question

The College Board reported the following mean scores for the three parts of the SAT: Critical reading 502 Mathematics 515 Writing 494 Assume that the population standard deviation on each part of the test is 100. What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test?

Answer 1

65.78% probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test.

Explanation:

To solve this question, we need 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 normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

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

In this problem, we have that:

`\mu = 502, \sigma = 100, n = 90, s = (100)/(√(90)) = 10.54`

What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test?

This is the pvalue of Z when X = 502+10 = 512 subtracted by the pvalue of Z when X = 502-10 = 492.

X = 512

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

By the Central Limit Theorem

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

`Z = (512 - 502)/(10.54)`

`Z = 0.95`

`Z = 0.95` has a pvalue of 0.8289

X = 492

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

`Z = (492 - 502)/(10.54)`

`Z = -0.95`

`Z = -0.95` has a pvalue of 0.1711

0.8289 - 0.1711 = 0.6578

65.78% probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test.