Question

# Scores for a common standardized college aptitude test are normally distributed with a mean of 483...

Scores for a common standardized college aptitude test are normally distributed with a mean of 483 and a standard deviation of 101. Randomly selected men are given a Test Prepartion Course before taking this test. Assume, for sake of argument, that the test has no effect.

If 1 of the men is randomly selected, find the probability that his score is at least 550.8.
P(X > 550.8) =
Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

If 8 of the men are randomly selected, find the probability that their mean score is at least 550.8.
P(M > 550.8) =
Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

If the random sample of 8 men does result in a mean score of 550.8, is there strong evidence to support the claim that the course is actually effective?

Sol)

Given

Mean = 483

S.D = 101

a)

Probability that the selected individual score is 550.8

P( X > 550.8 ) = P( Z > ( X - MEAN ) / S.D )

= P( Z > ( 550.8 - 483 ) / 101)

= P( Z > 67.8 / 101 )

= P( Z > 0.6712871)

= 0.2510

b)

n = 8

Probability that mean score is atleast 550.8

STANDARD ERROR ( S.E)

S.E = S.D / ✓ n

= 101 / ✓ 8

= 35.7088924

P( X > 550.8 ) = P( Z > ( X - MEAN ) / S.E )

= P( Z > ( 550.8 - 483 ) / 35.7088924)

= P( Z > 67.8/35.7088924)

= P( Z > 1.898686)

= 0.0288

Probability is less than 0.05 therfore

No there is no strong evidence to support the claim that the course is actually effective

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