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

Answer:

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.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \frac{\sigma}{\sqrt{n}}$.

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:

$\mu = 502, \sigma = 100, n = 90, s = \frac{100}{\sqrt{90}} = 10.54$

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

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{512 - 502}{10.54}$

$Z = 0.95$

$Z = 0.95$ has a pvalue of 0.8289

X = 492

$Z = \frac{X - \mu}{s}$

$Z = \frac{492 - 502}{10.54}$

$Z = -0.95$

$Z = -0.95$ 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.