Question

Scores for a common standardized college aptitude test are normally distributed with a mean of 489 and a standard deviation of 113. 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 561.1. P(X>561.1)= Enter your answer as a number accurate to 4 decimal places. If 13 of the men are randomly selected, find the probability that their mean score is at least 561.1. P(x− bar >561.1)= Enter your answer as a number accurate to 4 decimal places.

Answer 1

The probability that a randomly selected man's score is at least 561.1 is approximately 0.2619. The probability that the mean score of 13 randomly selected men is at least 561.1 is approximately 0.0137.

Step-by-step explanation:

To find the probability that the score of a randomly selected man is at least 561.1, we need to calculate the z-score and then find the area to the right of that z-score on a standard normal distribution table.

1. Calculate the z-score: z = (561.1 - 489) / 113 ≈ 0.6416
2. Find the area to the right of the z-score: P(X > 561.1) = 1 - P(Z < 0.6416)

Using a standard normal distribution table, we find P(Z < 0.6416) ≈ 0.7381. Therefore, P(X > 561.1) ≈ 1 - 0.7381 ≈ 0.2619.

To find the probability that the mean score of 13 randomly selected men is at least 561.1, we need to calculate the z-score and find the area to the right of that z-score on a standard normal distribution table.

1. Calculate the z-score: z = (561.1 - 489) / (113 / sqrt(13)) ≈ 2.2224
2. Find the area to the right of the z-score: P(x-bar > 561.1) = 1 - P(Z < 2.2224)

Using a standard normal distribution table, we find P(Z < 2.2224) ≈ 0.9863. Therefore, P(x-bar > 561.1) ≈ 1 - 0.9863 ≈ 0.0137.