Question 327158: I have two questions that I am in dire need of answers to. I have been stumped for 3 days:
A math teacher gives two different tests to measure students' aptitude for math. Scores on the first test are normally distributed with a mean of 20 and a standard deviation of 5.3. Scores on the second test are normally distributed with a mean of 68 and a standard deviation of 11.6. Assume that the two tests use different scales to measure the same aptitude. If a student scores 27 on the first test, what would be his equivalent score on the second test? (That is, find the score that would put him in the same percentile.)
a) 75.4 b) 81.3 c) 83.3 d) 84.1
The law firm of Dewey, Cheetham, and Howe has a pool of candidates wishing for internship in the law firm. The law firm has decided to test each candidate on his or her ability to win a nuisance lawsuit. Candidates who score in the top 33% will be given an internship. The law firm has issued similar tests in the past and knows that the average score is 80 (out of 100), the standard deviation is 10, and that test scores are normally distributed. What is the minimum score that a candidate must earn in order to place in the top 33%?
a) 81.25 b) 84.40 c) 86.30 d) 88.60
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! 3 days??
a score of 27 is 7 points above the mean , or 1.321 (7/5.3) standard deviations
the same percentile on the 2nd test would be ___ 68 + (1.321 * 11.6) or 83.3
from a z-score table , the top 33% is greater than .44 standard deviations above the mean
80 + (.44 * 10) = 84.4
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