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Business Weekly conducted a survey of graduates from 30 top MBA programs.

On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is $187,000.
Assume the standard deviation is $45,000.
Suppose you take a simple random sample of 36 graduates.
Round to 4 decimal places.
What is the distribution of X?
For a single randomly selected graduate, find the probability that her salary is between $186,650 and $195,500.
For the simple random sample of 36 graduates, find the probability that the average salary is between $186,650 and $195,500.

1 Answer

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Final answer:

The distribution of a single MBA graduate's salary is normally distributed with a mean of $187,000 and a standard deviation of $45,000. For a sample of 36 graduates, the distribution of the mean salary is normally distributed with the same mean but a smaller standard deviation of $7,500. The probabilities are calculated using Z-scores and the standard normal distribution.

Step-by-step explanation:

The question involves understanding the distribution of salaries among MBA graduates and conducting statistical analysis using the concepts of normal distribution and sampling distributions. Given the mean annual salary of $187,000 and a standard deviation of $45,000 for MBA graduates, we can model individual salaries using a normal distribution. However, when considering the average salary of a sample of 36 graduates, the distribution of the sample mean salary will also be normally distributed, according to the Central Limit Theorem, with the same mean but a smaller standard deviation equal to the original standard deviation divided by the square root of the sample size (the standard error).



For a single randomly selected graduate, we calculate the probability that her salary falls between $186,650 and $195,500 using the Z-score and consulting the standard normal distribution table (or using a calculator with normal distribution functions).

For the sample of 36 graduates, we calculate the probability that the average salary is between $186,650 and $195,500. Since the sample size is 36, the standard deviation of the sample mean (standard error) will be $45,000 / √36, which simplifies to $7,500. We then find the Z-scores for $186,650 and $195,500 and again consult the standard normal distribution to find the probability.

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