<PRE><B>Question 1050176</B>
 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 &lt; x-bar &lt; 30) = P(-0.5208 &lt; z &lt; +0.5208) = normalcdf(-0.5208&lt;0.5208)
= 0.3975
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Cheers,
Stan H.
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