Question

Scores for a common standardized college aptitude test are normally distributed with a mean of 515 and a standard deviation of 113. Randomly selected men are given a Test Prepartion Course before taking this test. Assume, for sake of argument, that the test has no effect.

A. If 1 of the men is randomly selected, find the probability that his score is at least 584.2.
P(X > 584.2) =  Round to 4 decimal places.

NOTE: Answers obtained using exact z-scores or z-scores rounded to 2 decimal places are accepted.

B. If 6 of the men are randomly selected, find the probability that their mean score is at least 584.2.
P(¯¯¯XX¯ > 584.2) =  Round to 4 decimal places.

NOTE: Answers obtained using exact z-scores or z-scores rounded to 2 decimal places are accepted.

C. If the random sample of 6 men does result in a mean score of 584.2, is there strong evidence to support the claim that the course is actually effective?

• 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 584.2.
• No. The probability indicates that is is possible by chance alone to randomly select a group of students with a mean as high as 584.2.

Answer 1

Solution :

Given that ,

mean = = 515

standard deviation = = 113

a) P(x > 584.2) = 1 - P(x < 584.2 )

= 1 - P[(x - ) / < (584.2 - 515) /113 ]

= 1 - P(z < 0.61)

= 1 - 0.7291 = 0.2709

probability= 0.2709

b) n = 6

=   = 515

= / n = 113 / 6 = 46.1321

P( >584.2 ) = 1 - P( < 584.2)

= 1 - P[( - ) / < (584.2-515) /46.1321 ]

= 1 - P(z <1.50 )

= 1 - 0.9332 = 0.0668

probability = 0.0668

c) No. The probability indicates that is is possible by chance alone to randomly select a group of students with a mean as high as 584.2.