SOLUTION: Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate the mean score μ of those who took the MCAT on your

Question 1192348: Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate the mean score μ of those who took the MCAT on your campus, you will obtain the scores of an SRS of students. The scores follow a Normal distribution, and from published information you know that the standard deviation is 6.4. Suppose that (unknown to you) the mean score of those taking the MCAT on your campus is 25.
In answering the following, use z-scores rounded to two decimal places:
A) If you choose one student at random, what is the probability (±0.0001) that the student's score is between 20 and 30?
B) You sample 25 students. What is the standard deviation (±0.01) of sampling distribution of their average score x?
C) What is the probability (±0.0001) that the mean score of your sample is between 20 and 30?

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
**A) Probability of a single student's score between 20 and 30**
1. **Standardize the scores:**
* z1 = (20 - 25) / 6.4 = -0.78
* z2 = (30 - 25) / 6.4 = 0.78
2. **Find the probability using the standard normal distribution:**
* P(20 ≤ X ≤ 30) = P(-0.78 ≤ Z ≤ 0.78)
* Using a standard normal table or calculator: P(20 ≤ X ≤ 30) ≈ 0.5653
**B) Standard deviation of the sampling distribution of the average score**
* **Standard Error (SE) = σ / √n**
* where:
* σ is the population standard deviation (6.4)
* n is the sample size (25)
* SE = 6.4 / √25
* SE = 6.4 / 5
* SE = 1.28
**C) Probability that the mean score of the sample is between 20 and 30**
1. **Standardize the sample means:**
* z1 = (20 - 25) / 1.28 = -3.91
* z2 = (30 - 25) / 1.28 = 3.91
2. **Find the probability using the standard normal distribution:**
* P(20 ≤ X̄ ≤ 30) = P(-3.91 ≤ Z ≤ 3.91)
* Using a standard normal table or calculator: P(20 ≤ X̄ ≤ 30) ≈ 0.9999
**In summary:**
* A) Probability of a single student's score between 20 and 30: 0.5653
* B) Standard deviation of the sampling distribution: 1.28
* C) Probability that the mean score of the sample is between 20 and 30: 0.9999

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