The class average on a recent test was 60. What is the probability that a random

Question

The class average on a recent test was 60. What is the probability that a randomly selected student from the class scored 72 or higher?

I have been given two answers:

Answer 1: P(X >= x) <= EX / x where EX = 60 and x = 72 for P <= 0.833.

Answer 2: P(x >= 72) = 1 - P(x < 72); since of 100 avg is 50 but mean is 60 so standard deviation is 10; P(x >= 72) = 1 - P(x < 72) = 1 - P((x - 60/10) < (72 - 60)/10) = 1 - ?(z < 1.2) from standard normal distribution table ?(1.2) = 0.8849; 1 - 0.8849 = 0.1151

Please identify (or provide) the correct answer and explain why the correct one is correct relative to the incorrect answer(s).

Please provide:
(i) the formulas used,
(ii) the steps followed to solve the problem, and
(iii) the answer.

NOTE: There is no missing information.

Explanation / Answer

Assuming the minimum possible score is 0 (a reasonable assumption), the first answer is correct. Mean score = P(scoring less than 72)*mean score(those scoring less than 72) + P(scoring 72 or more)*mean score(those scoring 72 or more) P(scoring less than 72) = 1 - P(scoring 72 or more), so we may rewrite this as Mean score = (1-P(scoring 72 or more))mean score(those scoring less than 72) + P(scoring 72 or more)*mean score(those scoring 72 or more) As the total mean score is fixed at 60, the lower the mean score of those scoring less than 72 and those scoring 72 or more, the greater the probability of scoring 72 or more. If the mean score of those scoring less than 72 is 0 (the lowest that it can be), and the mean score of those scoring 72 or more is 72(the lowest that it can be), then (1-P(scoring 72 or more))0 + P(scoring 72 or more)72 = 60 Therefore, P(scoring 72 or more)72 = 60, and P(scoring 72 or more) = 60/72 = 5/6 This is the maximum probability. If everyone scored exactly 60, then the mean score is 60, and P(scoring 72 or more) = 0 Thus, 0

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