Question

Scores for a common standardized college aptitude test are normally distributed with a mean of 506 and a standard deviation of 102. Randomly selected men are given a Prepartion Course before taking this test. Assume, for sake of argument, that the Preparation Course has no effect on people's test scores. If 1 of the men is randomly selected, find the probability that his score is at least 544.2. P(X > 544.2) = Enter your answer as a number accurate to 4 decimal places. If 14 of the men are randomly selected, find the probability that their mean score is at least 544.2. P(x-bar > 544.2) = Enter your answer as a number accurate to 4 decimal places. If the random sample of 14 men does result in a mean score of 544.2, is there strong evidence to support a claim that the Preapartion Course is actually effective? (Use the criteria that "unusual" events have a probability of less than 5%.)

No. The probability indicates that is is possible by chance alone to randomly select a group of students with a mean as high as 544.2 if the Preparation Course has no effect. Yes. The probability indicates that is is highly unlikely that by chance, a randomly selected group of students would get a mean as high as 544.2 if the Preparation Course has no effect.

Answer 1

µ = 506, σ = 102

a)

P(X > 544.2) =

= P( (X-µ)/σ > (544.2-506)/102)

= P(z > 0.3745)

= 1 - P(z < 0.3745)

Using excel function:

= 1 - NORM.S.DIST(0.3745, 1)

= 0.354

b)

P(X̅ > 544.2) =

= P( (X̅-μ)/(σ/√n) > (544.2-506)/(102/√14) )

= P(z > 1.4013)

= 1 - P(z < 1.4013)

Using excel function:

= 1 - NORM.S.DIST(1.4013, 1)

= 0.0806

c) Yes. The probability indicates that is is highly unlikely that by chance, a randomly selected group of students would get a mean as high as 544.2 if the Preparation Course has no effect.