Scores for a common standardized college aptitude test are
normally distributed with a mean of 491 and a standard deviation of
102. 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 555.6.
P(X > 555.6) =
Enter your answer as a number accurate to 4 decimal places.
If 9 of the men are randomly selected, find the probability that
their mean score is at least 555.6.
P(M > 555.6) =
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 9 men does result in a mean score of 555.6, is
there strong evidence to support the claim that the course is
actually effective?
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 = 491
S.d = 102
A)
P(x<555.6)
Z = (555.6-491)/102 = 0.63
From z table, P(z<0.63) = 0.7357
B)
For a sample z = (x-mean)/(s.d/√n)
N =9
Z = (555.6-491)/(102/√9)
Z = 1.9
From z table, P(z<1.9) = 0.9713
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