Question

The composite scores of individual students on the ACT college entrance examination in 2009 followed a normal distribution with mean 21.1 and standard deviation 5.1. What is the probability that a single student randomly chosen from all those taking the test scores 23 or higher? P(X > 0.3548) = P(z > 0.3725) = 0.3725 What is the probability that a simple random sample of 50 students chosen from all those taking the test has an average score of 23 or higher? p(X >) = p(z >) = Why is it more likely that a single student would score this high instead of the sample of students?

Answer 1

35.57% probability that a single student randomly chosen from all those taking the test scores 23 or higher.

0.41% probability that a simple random sample of 50 students chosen from all those taking the test has an average score of 23 or higher.

The lower the standard deviation, the higher the z-score, which means that the higher the pvalue of X = 23, which means there is a lower probability of scoring above 23. By the Central Limit Theorem, as the sample size increases, the standard deviation decreases, which means that Z increases.

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 = 21.1, \sigma = 5.1`

What is the probability that a single student randomly chosen from all those taking the test scores 23 or higher?

This is the pvalue of Z when X = 23.

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

`Z = (23 - 21.1)/(5.1)`

`Z = 0.37`

`Z = 0.37` has a pvalue of 0.6443

1 - 0.6443 = 0.3557

35.57% probability that a single student randomly chosen from all those taking the test scores 23 or higher.

What is the probability that a simple random sample of 50 students chosen from all those taking the test has an average score of 23 or higher?

Now we use the central limit theorem, so
`n = 50, s = (5.1)/(√(50)) = 0.72`

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

`Z = (23 - 21.1)/(0.72)`

`Z = 2.64`

`Z = 2.64` has a pvalue of 0.9959

1 - 0.9959 = 0.0041

0.41% probability that a simple random sample of 50 students chosen from all those taking the test has an average score of 23 or higher.

Why is it more likely that a single student would score this high instead of the sample of students?

The lower the standard deviation, the higher the z-score, which means that the higher the pvalue of X = 23, which means there is a lower probability of scoring above 23. By the Central Limit Theorem, as the sample size increases, the standard deviation decreases, which means that Z increases.