Question

(12.17) The Normal distribution with mean ? ? = 24.9 and standard deviation ? ? = 6.4 is a good description of the total score on the Medical College Admission Test (MCAT). This is a continuous probability model for the score of a randomly chosen student.

Use Table A to find the probability of the event "the student chosen has a score of 35 or higher". P P ("the student chosen has a score of 35 or higher") ( ± ± 0.0001) =

Answer 1

Let X be the score of the student. Given ?= 24.9 and  ? = 6.4 , this implies that

X ~N(24.9, 6.4^2) => X~ N(24.9, 40.96).

We have to determine the probability of the event the student chosen has a score of 35 or higher. So we have to find

P(X>=35) = P((X-24.9/(40.96^.5))>= (35-24.9)/(40.96^.5)) (Z= (X-?)/ ?) ,( ? =40.96^.5)

= P(Z>10.1/6.4)

=P(Z>1.58)

= 1- P(Z< 1.58)

Using normal probability table P(Z<1.58) = .9427

Therefore P(X>=35) = 1-P(Z<1.58) = 1-.94295 = .05705

Hence the probability that the student chosen has a score of 35 or higher is equal to 0.05705