Question

Scores for a common standardized college aptitude test are normally distributed with a mean of 499 and a standard deviation of 97. 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 563.7. P(X > 563.7) = 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 9 of the men are randomly selected, find the probability that their mean score is at least 563.7. P(M > 563.7) = Enter your answer as a number accurate to 4 decimal places. If the random sample of 9 men does result in a mean score of 563.7, 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 563.7. 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 563.7.

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Answer to a math question Scores for a common standardized college aptitude test are normally distributed with a mean of 499 and a standard deviation of 97. 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 563.7. P(X > 563.7) = 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 9 of the men are randomly selected, find the probability that their mean score is at least 563.7. P(M > 563.7) = Enter your answer as a number accurate to 4 decimal places. If the random sample of 9 men does result in a mean score of 563.7, 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 563.7. 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 563.7.

Expert avatar
Maude
4.7
105 Answers
Given: X \sim N(499, 97^2)
We need to find:
1. P(X > 563.7)
2. P(M > 563.7) , where M is the mean of a sample of size n = 9

Step 1: Calculate the z-score:
z = \frac{563.7 - 499}{97} = \frac{64.7}{97} \approx 0.6680

1. P(X > 563.7) = P\left(Z > \frac{563.7 - 499}{97}\right) = P(Z > 0.6680) \approx 0.2525

Step 2: For the sample mean with n = 9 :
\sigma_M = \frac{97}{\sqrt{9}} = \frac{97}{3}
z_M = \frac{563.7 - 499}{\frac{97}{3}} = \frac{64.7}{\frac{97}{3}} \approx 2

2. P(M > 563.7) = P\left(Z > \frac{563.7 - 499}{97/3}\right) = P\left(Z > 2\right) \approx 0.0228

Therefore,
1. P(X > 563.7) \approx \boxed{0.2525}
2. P(M > 563.7) \approx \boxed{0.0228}

Since the probability of obtaining a sample mean of 563.7 or higher by chance alone is 0.0228, it is unlikely by chance alone, supporting the claim that the course is effective.

Hence, the answer is:
Yes. The probability indicates that it is highly unlikely that by chance, a randomly selected group of students would get a mean as high as 563.7.

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