Question

provide an appropriate response. use the standard normal table to find the probability.
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 a teacher is selected at random, find the
probability that he or she makes more than $36,000.
a. 0.1056
b. 0.4040
c. 0.0918
d. 0.9082

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 36000\), \(\mu=32000\), and \(\sigma = 3000\).

$$ z=\frac{36000 - 32000}{3000}=\frac{4000}{3000}\approx1.33 $$

Step2: Find the probability

We want to find \(P(X>36000)\), which is equivalent to \(P(Z > 1.33)\).
Since \(P(Z>z)=1 - P(Z\leq z)\), and from the standard normal table \(P(Z\leq1.33)=0.9082\).

$$ P(Z>1.33)=1 - 0.9082=0.0918 $$

Answer:

C. 0.0918