Question

Salaries for teachers in a particular elementary school district are normally distributed with a mean of $41,000 and a standard deviation of $7,200. We randomly survey ten teachers from that district. (Round your answers to the nearest dollar.)
(a) Find the 90th percentile for an individual teacher's salary.
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(b) Find the 90th percentile for the average teacher's salary.
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Answer 1

Using the normal distribution and the central limit theorem, it is found that:

a) The 90th percentile for an individual teacher's salary is $50,216.

b) The 90th percentile for the average teacher's salary is $43,914.

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 $s = \frac{\sigma}{\sqrt{n}}$.

In this problem:

• Mean of $41,000, thus $\mu = 41000$.
• Standard deviation of $7,200, thus $\sigma = 7200$.

Item a:

The 90th percentile is X when Z has a p-value of 0.9, so X when Z = 1.28. Thus:

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

$1.28 = \frac{X - 41000}{7200}$

$X - 41000 = 1.28(7200)$

$X = 50216$

The 90th percentile for an individual teacher's salary is $50,216.

Item b:

Average for a sample of 10, thus $n = 10, s = \frac{7200}{\sqrt{10}}$.

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

By the Central Limit Theorem

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

$1.28 = \frac{X - 41000}{\frac{7200}{\sqrt{10}}}$

$X - 41000 = 1.28\frac{7200}{\sqrt{10}}$

$X = 43914$

The 90th percentile for the average teacher's salary is $43,914.

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