Question

assume that the salaries of elementary school teachers in the united states are normally distributed with a mean of $32,000 and a standard deviation of $3000. if 100 teachers are randomly selected, find the probability that their mean salary is greater than $32,500.
a. 0.0475
b. 0.9525
c. 0.3312
d. 0.1312

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$. Given $\sigma = 3000$ and $n = 100$, we have $\sigma_{\bar{x}}=\frac{3000}{\sqrt{100}}=\frac{3000}{10}=300$.

Step2: Calculate the z - score

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$. Here, $\bar{x}=32500$, $\mu = 32000$, and $\sigma_{\bar{x}} = 300$. So, $z=\frac{32500 - 32000}{300}=\frac{500}{300}\approx1.67$.

Step3: Find the probability

We want $P(\bar{X}>32500)$, which is equivalent to $P(Z > 1.67)$. Using the standard normal distribution table, $P(Z\leq1.67)=0.9525$. Then $P(Z > 1.67)=1 - P(Z\leq1.67)=1 - 0.9525 = 0.0475$.

Answer:

A. 0.0475