Homework SourseThe admissions committees for most masters of business admin
| The admissions committees for most masters of business administration (MBA) programs require the Graduate Management Admission Test (GMAT) as part of the application for new students. It has been shown that the scores on the GMAT are normally distributed with a mean of 542.3 and a standard deviation of 120.54. (Source: MBA.com) The minimum GMAT score required for admission to the MAcc program in the School of Accountancy at BYU is 500, but it is rare for students with scores less than 600 to be admitted. (Source: Marriot school of Business-BYU) Use this information to answer questions 25 through 34. | |
| 23. | What is the probability that a randomly selected student will score above 542.3 on the GMAT? (Round to the nearest tenth.) |
| 24. | What is the probability that a randomly selected student will score below 600 on the GMAT? (Round to three decimal places) |
| 25. | What is the probability that a randomly selected student will score 600 or above on the GMAT? (Round your answer to three decimal places) |
| 26. | Find the GMAT score that corresponds to the 10th percentile. (Round to the nearest whole number) |
| 27. | Find the GMAT score that corresponds to the 90th percentile. (Round to the nearest whole number) |
| 28. | Find the first quartile of the distribution of GMAT scores. (Round to the nearest whole number) |
| 29. | Find the third quartile of the distribution of GMAT scores. (Round to the nearest whole number) |
| 30. | Consider a simple random sample of n=15 students taking the GMAT. What is the mean of the distribution of sample mean scores for all such samples? (Round to the nearest tenth, one decimal place) |
| 31. | What is the standard deviation of the distribution of sample mean scores for all such samples? (Round to three decimal places) |
| 32. | What is the probability that the sample mean GMAT score for a SRS of n=15 students will be greater than 600? (Round to three decimal places) |
| The admissions committees for most masters of business administration (MBA) programs require the Graduate Management Admission Test (GMAT) as part of the application for new students. It has been shown that the scores on the GMAT are normally distributed with a mean of 542.3 and a standard deviation of 120.54. (Source: MBA.com) The minimum GMAT score required for admission to the MAcc program in the School of Accountancy at BYU is 500, but it is rare for students with scores less than 600 to be admitted. (Source: Marriot school of Business-BYU) Use this information to answer questions 25 through 34. | |
Solution
(23) P(X>542.3) = P((X-mean)/s >(542.3-542.3)/120.54)
=P(Z>0)= 0.5 (from standard normal table)
--------------------------------------------------------------------------------------------------------------------
(24) P(X<600) = P(Z<(600-542.3)/120.54)
=P(Z<0.48) =0.684 (from standard normal table)
--------------------------------------------------------------------------------------------------------------------
(25)P(X>600) = P(Z>(600-542.3)/120.54)
=P(Z>0.48) =0.316(from standard normal table)
--------------------------------------------------------------------------------------------------------------------
(26) P(X<c)=0.1
--> P(Z<(c-542.3)/120.54) = 0.1
--> (c-542.3)/120.54 = -1.28 (from standard normal table)
So c= 542.3 -1.28*120.54 =388
--------------------------------------------------------------------------------------------------------------------
(27)P(X<c)=0.9
--> P(Z<(c-542.3)/120.54) = 0.9
--> (c-542.3)/120.54 = 1.28 (from standard normal table)
So c= 542.3 +1.28*120.54 =697
--------------------------------------------------------------------------------------------------------------------
(28)P(X<c)=0.25
--> P(Z<(c-542.3)/120.54) = 0.25
--> (c-542.3)/120.54 = -0.67 (from standard normal table)
So c= 542.3 -0.67*120.54 =462
--------------------------------------------------------------------------------------------------------------------
(29)P(X<c)=0.75
--> P(Z<(c-542.3)/120.54) = 0.75
--> (c-542.3)/120.54 = 0.67 (from standard normal table)
So c= 542.3 +0.67*120.54 =623
--------------------------------------------------------------------------------------------------------------------
(30)mean = 542.3 (stay the same)
--------------------------------------------------------------------------------------------------------------------
(31) standard deviation =s/vn =120.54/sqrt(15)
=31.123
--------------------------------------------------------------------------------------------------------------------
(32) P(xbar>600) = P(Z>(600-542.3)/31.123)
=P(Z>1.85) = 0.032 (from standard normal table)
