Calculating the Probability of a Class Average Score Using the Central Limit Theorem

Given that the marks on a midterm test are normally distributed with a mean of 69 and a standard deviation of 10, we are tasked with finding the probability that a class of 27 students has an average score less than 67. This problem can be solved using the Central Limit Theorem (CLT).

Introduction to the Central Limit Theorem

The Central Limit Theorem (CLT) states that, given a sufficiently large sample size, the distribution of the sample mean will be approximately normally distributed, regardless of the underlying population distribution. This theorem is instrumental in solving problems involving sample means.

Steps to Calculate the Probability

Identify Parameters

Mean (μ): 69 Standard Deviation (σ): 10 Sample Size (n): 27

Calculate the Standard Error (SE)

The standard error of the mean is calculated using the formula:

SE frac{sigma}{sqrt{n}} frac{10}{sqrt{27}} approx 1.9245

Calculate the Z-Score

The Z-score for the sample mean can be calculated with the formula:

Z frac{bar{x} - mu}{SE}

where bar{x} is the sample mean, which is 67 in this case:

Z frac{67 - 69}{1.9245} approx frac{-2}{1.9245} approx -1.039

Find the Probability

We can use the Z-score to find the probability from the standard normal distribution table or a calculator. We want to find P(Z leq -1.039):

From the standard normal distribution table, we find:

P(Z leq -1.039) approx 0.1492

Round to Three Decimal Places

Therefore, the probability that a class of 27 students has an average score less than 67 is approximately:

boxed{0.149}

This result indicates that there is approximately a 14.9% chance that the average score of the class will be less than 67.

Alternative Approach Using a Z-Test

If you initially misread the question and treated 10 as the sample standard deviation instead, the Z-test approach can be used to find the same probability. Let's go through this step by step.

Calculate the Z-Score

The Z-score formula for the sample mean is:

Z frac{bar{x} - mu}{frac{sigma}{sqrt{n}}}

Plugging in the values, we get:

Z frac{67 - 69}{frac{10}{sqrt{27}}} frac{-2}{1.9245} approx -1.04

Find the Probability Using a Z-Table

From the standard normal distribution table, we find:

P(Z leq -1.04) approx 0.1492

This confirms the earlier result, showing that the probability is approximately 0.1492 or 14.9%.

Additional Context

This problem can also be approached by considering the sum of 27 independent and identically distributed (i.i.d.) normal random variables. The variance of the sum is:

variance 27 times 100 2700

The standard deviation is:

sqrt{2700} approx 52

The probability that the mean is at least 54 below the expected value:

frac{54}{52} approx 1.038

This calculation also confirms a probability of approximately 0.149.