Answer :
The probability that one student's score is between 26 and 30 is approximately 0.2198.The probability that the mean score of a sample of 25 students is between 26 and 30 is approximately 0.2165.
(a) An MCAT score is a random variable because it can take on different values based on the performance of individual students. It is a quantitative random variable because it represents a numerical measurement – in this case, the numerical value of a student's performance on the MCAT exam.
(b) To find the probability that the mean score of a sample of 25 students is between 26 and 30, we need to use the properties of the sampling distribution of the sample mean. The distribution of sample means tends to follow a normal distribution, even if the underlying population distribution is not normal, when the sample size is sufficiently large (thanks to the Central Limit Theorem).
Given that the mean (μ) of the MCAT scores is 25 and the standard deviation (σ) is 6.4, the standard error of the mean (SE) can be calculated as:
SE = σ / √n,
where n is the sample size (25 in this case).
SE = 6.4 / √25 = 6.4 / 5 = 1.28.
Now, we need to standardize the values 26 and 30 using the calculated standard error:
Z(26) = (26 - μ) / SE = (26 - 25) / 1.28 ≈ 0.78,
Z(30) = (30 - μ) / SE = (30 - 25) / 1.28 ≈ 3.91.
Using a standard normal distribution table or calculator, we can find the corresponding probabilities:
P(0.78 ≤ Z ≤ 3.91) ≈ P(Z ≤ 3.91) - P(Z ≤ 0.78) ≈ 0.9999 - 0.7834 ≈ 0.2165.
So, the probability that the mean score of a sample of 25 students is between 26 and 30 is approximately 0.2165.
(c) To find the probability that one student's score is between 26 and 30, we can directly use the properties of the normal distribution for a single observation. We'll use the same standardization process as above:
Z(26) = (26 - μ) / σ = (26 - 25) / 6.4 ≈ 0.16,
Z(30) = (30 - μ) / σ = (30 - 25) / 6.4 ≈ 0.78.
Again, using a standard normal distribution table or calculator, we find:
P(0.16 ≤ Z ≤ 0.78) ≈ P(Z ≤ 0.78) - P(Z ≤ 0.16) ≈ 0.7834 - 0.5636 ≈ 0.2198.
So, the probability that one student's score is between 26 and 30 is approximately 0.2198.
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Final answer:
An MCAT score is a random, quantitative variable. The probabilities for mean scores and individual scores between 26 and 30 can be found using z-scores and the normal distribution.
Explanation:
An MCAT score is a random variable because it is a numerical outcome of a random phenomenon, with different possible outcomes for different students. It is a quantitative variable because it is numerical and arithmetic operations can be performed on it.
For the probability of the mean score between 26 and 30 with a sample of 25 students, we can use the central limit theorem and z-scores to calculate it. The standard error (SE) is the standard deviation divided by the square root of the sample size, so SE=6.4/sqrt(25)=1.28. We find the z-scores for 26 and 30, which are (26-25)/1.28 and (30-25)/1.28 respectively. We then check these z-scores on the z-table (or use a calculator) to find the probability.
When a single student is selected, the standard deviation remains 6.4. Again, calculate the z-scores for 26 and 30 and find the probability using the z-table or a calculator. Both cases involve an understanding of the normal distribution and z-scores.
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