Question

The scores of individual students on the American College Testing (ACT) composite college entrance examination have a Normal Distribution with mean 18.6 and standard deviation 5.9.
a. What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?
b. Now take an SRS of 50 students who took the test. What are the mean and standard deviation of the average (sample mean) score for the 50 students? Do your results depend on the fact that individual scores have a Normal Distribution?
c. What is the probability that the mean score x of these students is 21 or higher?

Answer 1

Using the normal distribution and the central limit theorem, it is found that:

a) There is a 0.3409 = 34.09% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The mean is the population mean of $\mu = 5.9$, while considering the sample of n = 50, the standard deviation is of $s = \frac{5.9}{\sqrt{50}} = 0.8344$. Since the sample size is greater than 30, these results do not depend on the fact that individual scores have a Normal Distribution.

c) There is a 0.002 = 0.20% probability that the mean score x of these students is 21 or higher.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

• By the Central Limit Theorem, the sampling distribution of sample means of size n is approximately normal and has standard deviation $s = \frac{\sigma}{\sqrt{n}}$, as long as the underlying distribution is normal or the sample size is greater than 30.

In this problem:

• The mean is of $\mu = 18.6$.

• The standard deviation is of $\sigma = 5.9$.

Item a:

The probability is 1 subtracted by the p-value of Z when X = 21, hence:

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

$Z = \frac{21 - 18.6}{5.9}$

$Z = 0.41$

$Z = 0.41$ has a p-value of 0.6591.

1 - 0.6591 = 0.3409.

There is a 0.3409 = 34.09% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

Item b:

The mean is the population mean of $\mu = 5.9$, while considering the sample of n = 50, the standard deviation is of $s = \frac{5.9}{\sqrt{50}} = 0.8344$. Since the sample size is greater than 30, these results do not depend on the fact that individual scores have a Normal Distribution.

Item c:

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

By the Central Limit Theorem:

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

$Z = \frac{21 - 18.6}{0.8344}$

$Z = 2.88$

$Z = 2.88$ has a p-value of 0.998.

1 - 0.998 = 0.002

There is a 0.002 = 0.20% probability that the mean score x of these students is 21 or higher.

More can be learned about the normal distribution and the central limit theorem at https://brainly.com/question/24663213