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.

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The subject is Statistics

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. Since the scores are normally distributed with a mean of µ = 20.4 and a standard deviation of σ = 5.8, we can use a standard normal distribution table or a statistical calculator to find this probability.

Using a standard normal distribution table, we can convert the ACT score of 24 to a z-score by using the formula:

z = (x - µ) / σ

where x is the ACT score, µ is the mean, and σ is the standard deviation. Plugging in the values, we have:

z = (24 - 20.4) / 5.8 ≈ 0.62

Now, we can look up the area under the standard normal distribution curve to the right of 0.62 in the z-table. The z-table provides the probability corresponding to the z-score. In this case, the probability is approximately 0.2676.

Therefore, the probability that a randomly chosen student scored 24 or higher on the ACT is approximately 0.2676 or 26.76%.

2) The mean of the average ACT scores for a Simple Random Sample (SRS) of 30 students can be found using the Central Limit Theorem. According to the Central Limit Theorem, the mean of the sampling distribution of the sample mean is equal to the population mean.

In this case, the mean of the ACT scores for the population is given as µ = 20.4. Since the sampling distribution follows a Normal distribution, the mean of the sampling distribution of the sample mean is also µ = 20.4.

Now, to find the standard deviation (σ) of the average ACT score for an SRS of 30 students, we divide the population standard deviation (σ = 5.8) by the square root of the sample size (n = 30). This gives us:

σ_average = σ / √n = 5.8 / √30 ≈ 1.059

Therefore, the mean of the average ACT scores for an SRS of 30 students is 20.4 and the standard deviation is approximately 1.059.

3) Similar to question 1, to find the probability that the average ACT score of an SRS of 30 students is 24 or higher, we need to calculate the area under the normal distribution curve to the right of the score 24.

Using the same methodology as in question 1, we convert the ACT score of 24 to a z-score:

z = (24 - 20.4) / 1.059 ≈ 3.39

We look up the area under the standard normal distribution curve to the right of 3.39 in the z-table. The z-table provides the probability corresponding to the z-score. The probability in this case is extremely small and approaches 0.

Therefore, the probability that the average ACT score of an SRS of 30 students is 24 or higher is practically 0.

4) If the distribution of ACT scores in the population were distinctly non-Normal, it would affect the answers to questions 1, 2, and 3. The probabilities calculated in questions 1 and 3 rely on the assumption that the scores follow a Normal distribution. If the distribution is non-Normal, the probabilities would be different.

Similarly, the mean and standard deviation calculated in question 2 assume that the scores follow a Normal distribution. If the distribution is non-Normal, these values may not be accurate representations of the population.

In such cases, alternative methods or assumptions specific to the non-Normal distribution would need to be used to calculate the probabilities or summarize the data accurately.