Question

a companys annual income tax data for 2021 shows a mean annual employee income of $55,000, with a standard deviation of annual income of $12,000. what percentage of employees have an annual income between $55,000 and $79,000? 13.5% 34% 2.5% 47.5%

Explanation:

Step1: Calculate z - scores

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. Given $\mu = 55000$, $\sigma=12000$. For $x = 55000$, $z_1=\frac{55000 - 55000}{12000}=0$. For $x = 79000$, $z_2=\frac{79000 - 55000}{12000}=\frac{24000}{12000}=2$.

Step2: Use the standard normal distribution

The area under the standard - normal curve between $z = 0$ and $z = 2$ can be found using the properties of the standard normal distribution. The area to the left of $z = 0$ is $0.5$, and the area to the left of $z = 2$ is approximately $0.9772$. The area between $z = 0$ and $z = 2$ is $0.9772-0.5 = 0.4772\approx47.5\%$.

Answer:

47.5%