The composite scores of students on the ACT college entrance examination in a recent year had a Normal distribution with mean µ = 20.4 and standard deviation = 5.8.

1) What is the probability that a randomly chosen student scored 24 or higher on the ACT?

2) What are the mean and standard deviation of the average ACT score for an SRS of 30 students?

3) What is the probability that the average ACT score of an SRS of 30 students is 24 or higher?

4) Would your answers to 1, 2, or 3 be affected if the distribution of ACT scores in the population were distinctly non-Normal? Explain.

1) To find the probability that a randomly chosen student scored 24 or higher on the ACT, we need to calculate the area under the Normal distribution curve to the right of the score 24.

To do this, we can use a standard Normal table or a statistical calculator. However, in this case, we need to adjust our calculation to fit the specific mean and standard deviation provided.

First, we calculate the z-score for the score 24:
z = (x - µ) / σ = (24 - 20.4) / 5.8 = 0.62

Next, we find the corresponding area under the standard Normal curve for this z-score. Using a standard Normal table or calculator, we find that the area to the right of z = 0.62 is approximately 0.267.

Therefore, the probability that a randomly chosen student scored 24 or higher on the ACT is 0.267, or 26.7%.

2) The mean of the average ACT score for an SRS (simple random sample) of 30 students is the same as the mean of the population, which is µ = 20.4.

The standard deviation of the average ACT score for an SRS of 30 students, also known as the standard error, can be calculated using the formula:
σx = σ / √n
where σ is the population standard deviation and n is the sample size.

Plugging in the values, we have:
σx = 5.8 / √30 ≈ 1.060

Therefore, the mean of the average ACT score for an SRS of 30 students is 20.4 and the standard deviation (standard error) is approximately 1.060.

3) To find the probability that the average ACT score of an SRS of 30 students is 24 or higher, we can use the same approach as in question 1, but with the adjusted mean and standard deviation.

First, we calculate the z-score for the score 24, using the mean and standard deviation of the average ACT score for an SRS of 30 students:
z = (x - µ) / σx = (24 - 20.4) / 1.060 ≈ 3.396

Next, we find the corresponding area under the standard Normal curve for this z-score. Using a standard Normal table or calculator, we find that the area to the right of z = 3.396 is nearly 0.0003.

Therefore, the probability that the average ACT score of an SRS of 30 students is 24 or higher is approximately 0.0003, or 0.03%.

4) Yes, the answers to questions 1, 2, and 3 would be affected if the distribution of ACT scores in the population were distinctly non-Normal.

In question 1, we assumed that the scores follow a Normal distribution. If the distribution were different, such as skewed or bimodal, we would need to use different techniques to calculate probabilities.

In question 2, we used the formula for the standard error assuming a Normal distribution. If the distribution were non-Normal, we would need to use different methods, such as bootstrapping or resampling, to estimate the standard deviation of the sample mean.

In question 3, we used the z-score and standard Normal distribution assumption to calculate the probability. If the distribution were non-Normal, the z-score approach and standard Normal table would not be applicable. We would need to use different techniques, such as the t-distribution, to calculate probabilities for non-Normal distributions.