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The admissions committees for most Masters of Business Administration (MBA) programs require the Graduate Management Admission Test (GMAT) as part of the application for new students. It has been shown that the scores on the GMAT are normally distributed with a mean of 542.3 and a standard deviation of 120.54. (Source: MBA)

The minimum GMAT score required for admission to the MACC program in the School of Accountancy at BYU is 500, but it is rare for students with scores less than 600 to be admitted. (Source: Marriott School of Business - BYU)

Part 1: What is the probability that a randomly selected student will score above 542.3 on the GMAT?

Answer :

Final answer:

The probability that a randomly selected student will score above 542.3 on the GMAT, given a normal distribution with a mean of 542.3, is exactly 50%.

Explanation:

The question asks about the probability that a randomly selected student will score above 542.3 on the GMAT, given that the scores are normally distributed with a mean of 542.3 and a standard deviation of 120.54. In a normal distribution, the mean represents the point at which half of the observations fall above and half fall below. Therefore, without doing any calculations, we can ascertain that the probability of a randomly selected student scoring above the mean (542.3) is exactly 50% or 0.5 since the distribution is symmetric around the mean.

This understanding is based on the properties of the normal distribution, where the mean serves as the central tendency, and the distribution is symmetrical. As a result, 50% of the observations lie above the mean, and 50% lie below it.

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Rewritten by : Barada

Final answer:

The probability that a randomly selected student will score above 542.3 on the GMAT is 0.5.

Explanation:

To find the probability that a randomly selected student will score above 542.3 on the GMAT, we need to calculate the z-score for that score and then use the z-score to find the probability using a standard normal distribution table. The formula for calculating the z-score is:

z = (x - μ) / σ

Plugging in the values, we get:

z = (542.3 - 542.3) / 120.54 = 0

The z-score of 0 tells us that the score of 542.3 is equal to the mean. Since we want the probability of scoring above the mean, we need to find the area to the right of the z-score. Looking up the z-score in the standard normal distribution table, we find that the probability is approximately 0.5.

Learn more about Probability here:

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