Question

The average teacher's salary in Connecticut is $57,337. Suppose that the distribution of salaries is normal with a standard deviation of $7500. If we sample 55 teachers' salaries, what is the probability that the sample mean is less than $55500?

Answer 1

SOLUTION

To solve this, we will apply the formula

$\text{Zscore = }\frac{x-\mu}{\frac{\sigma}{\sqrt[]{n}}}$

Where

$\begin{gathered} x=\text{ sample mean = 55500} \\ \mu=\text{ population mean = 57337} \\ \sigma=\text{ standard deviation = 7500} \\ n\text{ = number of samples = 55} \end{gathered}$

So from the formula

$\begin{gathered} \text{Zscore = }\frac{x-\mu}{\frac{\sigma}{\sqrt[]{n}}} \\ \\ \text{Zscore = }\frac{55500-57337}{\frac{7500}{\sqrt[]{55}}} \\ \\ \text{Zscore =}\frac{-1837}{\frac{7500}{\sqrt[]{55}}} \\ \\ \text{Zscore =-0.033026} \end{gathered}$

Now, we will calculate the Zscore for -0.033026 using a Zscore table or calculator

This becomes

[tex]P(xTherefore, the probabillity = 0.48683 or 48.68%