Question 1050176: 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 6.8. Suppose that (unknown to you) the mean score of those taking the MCAT on your campus is 25.0.
A)You sample 25 students. What is the standard deviation of sampling distribution of their average score x? (Enter your answer to two decimal places.)
B)What is the probability that the mean score of your sample is between 20 and 30? (Round your answer to four decimal places.)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The scores follow a Normal distribution, and from published information you know that the standard deviation is 6.8. Suppose that (unknown to you) the mean score of those taking the MCAT on your campus is 25.0.
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Ans: 6.8/sqrt(25) = 6.8/5 = 48/5 = 9.6
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B)What is the probability that the mean score of your sample is between 20 and 30? (Round your answer to four decimal placesz.
z(20) = (20-25)/9.6 = -0.5208
z(30) = (30-25)/9.6 = +0.5208
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P(20 < x-bar < 30) = P(-0.5208 < z < +0.5208) = normalcdf(-0.5208<0.5208)
= 0.3975
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Cheers,
Stan H.
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