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 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 ? ‑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 ? Use Table A, or software to calculate your answer. (Enter your answer rounded to four decimal places.) probability: (b) You sample 36 students. What is the standard deviation of the sampling distribution of their average score ?¯ ? (Enter your answer rounded to two decimal places.) standard deviation: (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.) probability:

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 )

= 0.6772 - 0.3228 = 0.3544

probability =0.3544

b) n = 36

= = 495

= / n = 10.8/ 36 = 1.8

P(490< <500 )  

= P[(490-495) /1.8 < ( - ) / < (500-495) /1.8 )]

= P( -2.78< Z < 2.78)

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

= 0.9973 - 0.0027 = 0.9946

probability = 0.9946