P(X < A) = P(Z < (A - mean)/standard deviation)
mean = 500
standard deviation = 10.6
(a) P(X > 520) = 1 - P(X < 520)
= 1 - P(Z < (520 - 500)/10.6)
= 1 - P(Z < 1.89)
= 1 - 0.9706 (from standard normal distribution table)
= 0.0296
Percentage over 520 = 2.96 %
(B) P(515 < X < 525) = P(X < 525) - P(X < 515)
= P(Z < (525 - 500)/10.6) - P(Z < (515 - 500)/10.6)
= P(Z < 2.36) - P(Z < 1.42)
= 0.9909 - 0.9222
= 0.0687
Percentage between 515 and 525 = 6.87 %
Almost all medical schools in the United States require students to take the Medical College Admission...
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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 10.8. Suppose that, unknown to you, the mean score of those taking the MCAT on your campus is 500. In...
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