The Medical College Admission Test. 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 25.0.
(a) If you choose one student at random, what is the probability that the student’s score is between 20 and 30?
(b) You sample 25 students. What is the sampling distribution of their average score ?
(c) What is the probability that the mean score of your sample is between 20 and 30?
The correct answer and explanation is :
To address the given problem, we’ll analyze each part step-by-step.
(a) Probability that a randomly selected student’s score is between 20 and 30:
Given:
- Mean (μ) = 25.0
- Standard deviation (σ) = 6.4
The MCAT scores are normally distributed. To find the probability that a student’s score falls between 20 and 30, we calculate the Z-scores for these values:
[ Z = \frac{X – \mu}{\sigma} ]
For X = 20:
[ Z_{20} = \frac{20 – 25}{6.4} = -0.78125 ]
For X = 30:
[ Z_{30} = \frac{30 – 25}{6.4} = 0.78125 ]
Using standard normal distribution tables or a calculator, the cumulative probabilities corresponding to these Z-scores are approximately:
- P(Z ≤ -0.78125) ≈ 0.2177
- P(Z ≤ 0.78125) ≈ 0.7823
Therefore, the probability that a student’s score is between 20 and 30 is:
[ P(20 < X < 30) = P(Z ≤ 0.78125) – P(Z ≤ -0.78125) ]
[ P(20 < X < 30) = 0.7823 – 0.2177 = 0.5646 ]
So, there’s approximately a 56.46% chance that a randomly selected student’s score is between 20 and 30.
(b) Sampling distribution of the average score for a sample of 25 students:
When drawing a simple random sample (SRS) of size n from a normally distributed population with mean μ and standard deviation σ, the sampling distribution of the sample mean (x̄) is also normally distributed with:
- Mean (μₓ̄) = μ
- Standard deviation (σₓ̄) = σ / √n
Given:
- μ = 25.0
- σ = 6.4
- n = 25
The standard deviation of the sampling distribution (standard error) is:
[ \sigma_{x̄} = \frac{6.4}{\sqrt{25}} = \frac{6.4}{5} = 1.28 ]
Thus, the sampling distribution of the average score for a sample of 25 students is N(25.0, 1.28).
(c) Probability that the mean score of the sample is between 20 and 30:
Using the sampling distribution N(25.0, 1.28), we calculate the Z-scores for the sample mean range of 20 to 30.
For x̄ = 20:
[ Z_{20} = \frac{20 – 25}{1.28} = -3.90625 ]
For x̄ = 30:
[ Z_{30} = \frac{30 – 25}{1.28} = 3.90625 ]
The cumulative probabilities corresponding to these Z-scores are:
- P(Z ≤ -3.90625) ≈ 0.00005
- P(Z ≤ 3.90625) ≈ 0.99995
Therefore, the probability that the sample mean is between 20 and 30 is:
[ P(20 < \bar{X} < 30) = P(Z ≤ 3.90625) – P(Z ≤ -3.90625) ]
[ P(20 < \bar{X} < 30) = 0.99995 – 0.00005 = 0.9999 ]
So, there’s approximately a 99.99% chance that the mean score of a sample of 25 students falls between 20 and 30.
Explanation:
The Medical College Admission Test (MCAT) scores on your campus are normally distributed with a mean (μ) of 25.0 and a standard deviation (σ) of 6.4. When selecting one student at random, the probability that their score lies between 20 and 30 is found by converting these scores to Z-scores, which standardize the values relative to the mean and standard deviation. The resulting probability of approximately 56.46% indicates that over half of the students score within this range.
When considering the average score from a sample of 25 students, the Central Limit Theorem tells us that the distribution of the sample mean will be approximately normal, regardless of the population distribution, provided the sample size is sufficiently large. In this case, the sample mean’s distribution has the same mean as the population (25.0) but a reduced standard deviation (standard error) of 1.28, reflecting the decreased variability in sample means compared to individual scores.
Calculating the probability that this sample mean falls between 20 and 30 involves determining the Z-scores for these values within the sampling distribution. The resulting probability of approximately 99.99% indicates that it’s highly likely for the average score of a sample of 25 students to fall within this range, due to the reduced variability (smaller standard error) in the sampling distribution.
Understanding these probabilities is crucial for making inferences about the population based on sample data, which is a fundamental aspect of statistical analysis in educational testing and many other fields.