Question

independent practice
the average annual salary for accountants is $73,560 with standard deviation $35,802.
determine the probability that a randomly selected accountant has a salary greater than $100,000.
.3419
.00976
.23010
.11342

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 100000\), \(\mu=73560\), and \(\sigma = 35802\).

$$z=\frac{100000 - 73560}{35802}=\frac{26440}{35802}\approx0.74$$

Step2: Find the probability

We want \(P(X>100000)\), which is equivalent to \(P(Z > 0.74)\).
Since \(P(Z>z)=1 - P(Z\leq z)\), and from the standard normal table \(P(Z\leq0.74)=0.7704\)

$$P(Z > 0.74)=1 - 0.7704=0.2296\approx0.2301$$

Answer:

\(0.23010\)