Question

What is the probability that a random sample of 100 accounting graduates will provide an average(X¯) that is within $902 of the population mean (µ)?

Answer 1

Using the normal distribution and the central limit theorem, it is found that there is a 0.6328 = 63.28% probability that a random sample of 100 accounting graduates will provide an average that is within $902 of the population mean.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation given by $s = \frac{\sigma}{\sqrt{n}}$

Hence, the probability of sample having mean within M of the population mean is the p-value of $Z = \frac{M}{\frac{\sigma}{\sqrt{n}}}$ subtracted by the p-value of $Z = -\frac{M}{\frac{\sigma}{\sqrt{n}}}$.

In this problem, we suppose $\sigma = 10000$, and thus, with a sample of 100, we have that $s = \frac{10000}{\sqrt{100}} = 1000$.

• Within $902, hence $M = 902$.

$Z = \frac{M}{\frac{\sigma}{\sqrt{n}}}$

$Z = \frac{902}{1000}$

$Z = 0.902$

$Z = 0.902$ has a p-value of 0.8164.

$Z = \frac{M}{\frac{\sigma}{\sqrt{n}}}$

$Z = -\frac{902}{1000}$

$Z = -0.902$

$Z = -0.902$ has a p-value of 0.1836.

0.8164 - 0.1836 = 0.6328.

0.6328 = 63.28% probability that a random sample of 100 accounting graduates will provide an average that is within $902 of the population mean.

A similar problem is given at brainly.com/question/24663213