Question

Scores for a common standardized college aptitude test are normally distributed with a mean of 502 and a standard deviation of 108. Randomly selected men are given a Test Preparation Course before taking this test. Assume, for sake of argument that the preparation course has no effect. If 1 of the men is randomly selected, find the probability that his score is at least 570.7. P(X> 570.7) = 0.3639 Enter your answer as a number accurate to 4 decimal places. If 8 of the men are randomly selected, find the probability that their mean score is at least 570.7. P(M> 570.7) = 0.7992 Enter your answer as a number accurate to 4 decimal places. s Assume that any probability less than 5% is sufficient evidence to conclude that the preparation course does help men do better. If the random sample of 8 men does result in a mean score of 570.7, is there strong evidence to support the claim that the course is actually effective? O No. The probability indicates that it is possible by chance alone to randomly select a group of students with a mean as high as 570.7. Yes. The probability indicates that it is highly ?) unlikely that by chance, a randomly selected group of students would get a mean as high as 570.7. > Next Question

Answer 1

Sol:

P(X > 570.7) = P( (X - mean ) / SD > (570.7 - mean) / SD )

= P(Z > (570.7 - 502) / 108)

= P( Z > 0.636)

= 1 - P( Z < 0.636)

= 1 - 0.7357

P(X > 570.7) = 0.2643

P(X > 570.7) = P( (X - mean ) / (SD / n^0.5) > (570.7 - mean) / (SD / n^0.5) )

= P(Z > (570.7 - 502) / (108 / 8^0.5))

= P( Z > 1.799)

= 1 - P( Z < 1.799)

= 1 - 0.9633

P(X > 570.7) = 0.0367

Since Probability = 0.0367 < 0.05 ( 5% ) , we can conclude that the preparation course does help men do better and there is strong evidence to support the claim that the course is actually effective.

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