Question

Scores for a common standardized college aptitude test are normally distributed with a mean of 515 and a standard deviation of 107. 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 18 of the men are randomly selected, find the probability that their mean score is at least 557.9.

P( ¯x > 557.9) =

Answer 1

To calculate the probability that the mean score of 18 randomly selected men is at least 557.9, we need to convert the mean score to a z-score and find the corresponding area under the standard normal curve.The probability P(¯x > 557.9) is approximately 0.0459 or 4.59%.

Step-by-step explanation:

To calculate the probability that the mean score of 18 randomly selected men is at least 557.9, we need to convert the mean score to a z-score and find the corresponding area under the standard normal curve.

First, we calculate the standard error of the mean (SEM) using the formula SEM = standard deviation / square root of sample size. In this case, SEM = 107 / sqrt(18) ≈ 25.3.

Next, we convert the desired mean score of 557.9 to a z-score using the formula z = (x - mean) / SEM. The z-score is approximately (557.9 - 515) / 25.3 ≈ 1.697.

Using a standard normal distribution table or a calculator, we find the area to the right of the z-score (1.697) is approximately 0.0459. Therefore, the probability P(¯x > 557.9) is approximately 0.0459 or 4.59%.