SOLUTION: Of the over 1.6 million graduating high school students who took the American College Test (ACT) in 2012, the mean score was 21.1 and the standard deviation was 5.3. Assuming that

Question 910054: Of the over 1.6 million graduating high school students who took the American College Test (ACT) in 2012, the mean score was 21.1 and the standard deviation was 5.3. Assuming that the scores are distributed normally;
a. Find P(X≤28)
b. Find P(18≤X≤21)
c. Find the score such that P(X>x) = .66
d. Assume that the University of Kentucky will only accept applicants that scored higher than a 21 on the ACT. What is the probability that a student taking the exam in 2012 would not qualify for admission?

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Of the over 1.6 million graduating high school students who took the American College Test (ACT) in 2012, the mean score was 21.1 and the standard deviation was 5.3. Assuming that the scores are distributed normally;
a. Find P(X≤28)
z(28) = (28-21.1)/5.3 = 1.3019
P(x <= 28) = P(z <= 1.3019) = normalcdf(-100,1.3019) = 0.9035
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b. Find P(18≤X≤21)
Find the z-value for 18 and for 21
Find the probability z lies between those z-values
Ans:: 0.2132
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c. Find the score such that P(X>x) = .66
Find the z-value with a left-tail of 0.66
invNorm(0.66) = 0.4125
Find the corresponding score::
x = z*s + u = 0.4125*5.3+21.1 = 23.29
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d. Assume that the University of Kentucky will only accept applicants that scored higher than a 21 on the ACT. What is the probability that a student taking the exam in 2012 would not qualify for admission?
z(21) = (21-21.1)/5.3 = -0.1/5.3 = -0.0189
P(score < 21) = P(z < -0.0189) = normalcdf(-100,-0.0189) = 0.4925
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Cheers,
Stan H.
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