Almost all medical schools in the United States require students to take the Med

Question

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate the mean score 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 6.3 . Suppose that (unknown to you) the mean score of those taking the MCAT on your campus is 24 .

In answering the following, usez-scores rounded to two decimal places.

If you choose one student at random, what is the probability (±0.0001; that is round to a 4th decimal place) that the student's score is between 20 and 30?

You sample 27 students. What is the standard deviation (±0.01, that is round to 2 decimal places) of sampling distribution of their average score xx?

What is the probability (±0.0001; that is round to a 4th decimal place) that the mean score of your sample is between 20 and 30?

Explanation / Answer

Mean ( u ) =24

Standard Deviation ( sd )=6.3

Normal Distribution = Z= X- u / sd ~ N(0,1)   

a.

To find P(a < = Z < = b) = F(b) - F(a)

P(X < 20) = (20-24)/6.4

= -4/6.3 = -0.6349

= P ( Z <-0.6349) From Standard Normal Table

= 0.2628

P(X < 30) = (30-24)/6.3

= 6/6.3 = 0.9523

= P ( Z <0.9523) From Standard Normal Table

= 0.8295

P(20 < X < 30) = 0.8295-0.2628 = 0.5667   

b.

Number ( n ) = 27

Standard Deviation ( sd )= 6.3/ Sqrt(n) = 1.3966

c.

To find P(a <= Z <=b) = F(b) - F(a)

P(X < 20) = (20-24)/6.3/ Sqrt ( 27 )

= -2.0867

= P ( Z <-2.0867) From Standard Normal Table

= 0.0185

P(X < 30) = (30-24)/6.3/ Sqrt ( 27 )

= 6.3/5.1961 = 4.9488

= P ( Z <4.9488) From Standard Normal Table

= 1

P(20 < X < 30) = 1-0.0185 = 0.9815

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