Question

Almost all medical schools require applicants to take the Medical College Admission Test (MCAT). To estimate the mean score of those who took the MCAT at WSSU, you will obtain the scores of an SRS of students. The scores follow a Normal distribution, and from published information you know that the standard deviation is 6.5. Suppose that (unknown to you) the mean score of those taking the MCAT at WSU is 25.0. You sampled 25 students. What is the probability that the mean score of your sample is between 22 and 28

Answer 1

Answer:

97.92% probability that the mean score of your sample is between 22 and 28

Step-by-step 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 can be 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 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 $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 25, \sigma = 6.5, n = 25, s = \frac{6.5}{\sqrt{25}} = 1.3$

What is the probability that the mean score of your sample is between 22 and 28

This is the pvalue of Z when X = 28 subtracted by the pvalue of Z when X = 22. So

X = 28

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

By the Central Limit theorem

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

$Z = \frac{28 - 25}{1.3}$

$Z = 2.31$

$Z = 2.31$ has a pvalue of 0.9896

X = 22

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

$Z = \frac{22 - 25}{1.3}$

$Z = -2.31$

$Z = -2.31$ has a pvalue of 0.0104

0.9896 - 0.0104 = 0.9792

97.92% probability that the mean score of your sample is between 22 and 28