Question

Suppose that the distribution of IQ's of North Catalina State University's students can be approximated by a normal model with mean 130 and standard deviation 8 points. Also suppose that the distribution of IQ's of Chapel Mountain University's students can be approximated by a normal model with mean 120 and standard deviation 10 points.

Required:
What is the probability that the mean IQ of the 3 North Catalina students is at least 5 points higher than the mean IQ of the 3 Chapel Mountain students?

Answer 1

0.7517 = 75.17% probability that the mean IQ of the 3 North Catalina students is at least 5 points higher than the mean IQ of the 3 Chapel Mountain students

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Subtraction of normal variables:

When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

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.

North Catalina:

Population has mean 130 and standard deviation 8 points. Sample of 3. This means that:

`\mu_(N) = 130, s_(N) = (8)/(√(3)) = 4.6188`

Chapel Mountain:

Population has mean 120 and standard deviation 10 points. Sample of 3. This means that:

`\mu_(C) = 120, s_(C) = (10)/(√(3)) = 5.7735`

What is the probability that the mean IQ of the 3 North Catalina students is at least 5 points higher than the mean IQ of the 3 Chapel Mountain students?

We want that:
`N - C > 5`

Distribution N - C:

The mean is:

`\mu = \mu_N - \mu_C = 130 - 120 = 10`

The standard deviation is:

`s = √(s_N^2+s_C^2) = √(4.6188^2+5.7735^2) = 7.3937`

This probability is 1 subtracted by the pvalue of Z when X = 5. So

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

By the Central Limit Theorem

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

`Z = (5 - 10)/(7.3937)`

`Z = -0.68`

`Z = -0.68` has a pvalue of 0.2483

1 - 0.2483 = 0.7517

0.7517 = 75.17% probability that the mean IQ of the 3 North Catalina students is at least 5 points higher than the mean IQ of the 3 Chapel Mountain students