Question

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 495.

In answering the questions, use z‑scores rounded to two decimal places.

(a) If you choose one student at random, what is the probability that the student's score is between 490 and 500? (Enter your answer rounded to four decimal places.)

(b) You sample 36 students. What is the standard deviation of the sampling distribution of their average score x¯ ? (Enter your answer rounded to two decimal places.)

(c) What is the probability that the mean score of your sample is between 490 and 500? (Enter your answer rounded to four decimal places.)

Answer 1

Solution :

Given that ,

mean = = 495

standard deviation = = 10.8

a) P(490 < x < 500) = P[(490 - 495)/ 10.8 ) < (x - ) /  < (500 - 495) / 10.8 ) ]

= P(-0.46 < z < 0.46)

= P(z < 0.46) - P(z < -0.46)

Using z table,

= 0.6772 - 0.3228

= 0.3544

b) n = 36

= = 495

= / n = 10.8 / 36 = 1.80

c) P(490 < < 500)  

= P[(490 - 495) /1.80 < ( - ) / < (500 - 495) / 1.80)]

= P(-2.78 < Z < 2.78 )

= P(Z < 2.78) - P(Z < -2.78 )

Using z table,  

= 0.9973 - 0.0027

= 0.9946