Question

Scores for a common standardized college aptitude test are normally distributed with a mean of 480 and a standard deviation of 106. 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 553.4.
P(X > 553.4) =
Enter your answer as a number accurate to 4 decimal places.

If 12 of the men are randomly selected, find the probability that their mean score is at least 553.4.
P(M > 553.4) =
Enter your answer as a number accurate to 4 decimal places.

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 12 men does result in a mean score of 553.4, is there strong evidence to support the claim that the course is actually effective?

• No. The probability indicates that is is possible by chance alone to randomly select a group of students with a mean as high as 553.4.
• 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 553.4.

Answer 1

1)

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Given mean = 480

S.d = 106

1)

P(x>553.4)

Z = (553.4 - 480)/106 = 0.69

From z table, P(z>0.69) = 0.2451

2)

In case of sample

Z = (x-mean)/(s.d/√n)

So

Z = (553.4 - 480)/(106/√12) = 2.4

From z table, P(z>2.4) = 0.0082

2)

Yes

As 0.0082 is less than 0.05

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 553.4.