Understanding the Problem
In this scenario, we’re dealing with a class where the final exam scores follow a normal distribution. The mean score for the exam is 73, with a standard deviation of 78. We want to find out the probability that the average score of a randomly selected group of 24 students is less than 70.
Using the Central Limit Theorem
According to the Central Limit Theorem, when we take a sample from a population, the distribution of the sample mean will be normally distributed if the sample size is large enough. In this case, we have a sample size (n) of 24 students, which is generally considered adequate for the Central Limit Theorem to apply, provided that the original distribution is not highly skewed.
Calculating the Mean and Standard Error
The mean of the sample means will be the same as the population mean:
- Population Mean (μ): 73
Next, we find the standard error (SE) of the mean, which can be calculated using the formula:
SE = σ / √n
Here, σ (the population standard deviation) is 78, and n (the sample size) is 24:
SE = 78 / √24 ≈ 15.96
Finding the Z-score
To find the probability of the sample mean being less than 70, we calculate the Z-score using the formula:
Z = (X̄ - μ) / SE
Where:
- X̄ is the sample mean (70 in this case)
- μ is the population mean (73)
- SE is the standard error (approximately 15.96)
Substituting the known values:
Z = (70 - 73) / 15.96 ≈ -0.19
Finding the Probability
Now that we have the Z-score, we can use the standard normal distribution table or a calculator to find the corresponding probability. A Z-score of -0.19 indicates a percentile in the distribution. Checking a Z-table or using statistical software, we find:
P(Z < -0.19) ≈ 0.4247
This means there is approximately a 42.47% chance that the mean test score of a random sample of 24 students is less than 70.
Conclusion
In summary, the probability that the mean score of 24 randomly selected students from the Math 160 final exam is less than 70 is approximately 42.47%.