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 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 10.4 . Suppose that, unknown to you, the mean score of those taking the MCAT on your campus is 500 . 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 495 and505 ? Use Table A, or software to calculate your answer. (Enter your answer rounded to four decimal places.) probability: (b) You sample 25 students. What is the standard deviation of the sampling distribution of their average score x¯ ? (Enter your answer rounded to two decimal places.) standard deviation: (c) What is the probability that the mean score of your sample is between 495 and 505 ? (Enter your answer rounded to four decimal places.) probability:

Answer 1

Solution :

Given that ,

mean = = 500

standard deviation = = 10.4

a) P(495 < x < 505) = P[(495 - 500)/ 10.4) < (x - ) /  < (505 - 500) / 10.4) ]

= P(-0.48 < z < 0.48)

= P(z < 0.48) - P(z < -0.48)

Using z table,

= 0.6844 - 0.3156

= 0.3688

b) n = 25

= = 500

= / n = 10.4 / 25 = 2.08

P(495 < < 505)  

= P[(495 - 500) / 2.08 < ( - ) / < (505 - 500) / 2.08)]

= P(-2.40 < Z < 2.40)

= P(Z < 2.40) - P(Z < -2.40)

Using z table,  

= 0.9918 - 0.0082  

= 0.9836