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 the companys employees have an annual income between $43,000 and $67,000? 99.7% 95.4% 68% 34%

Explanation:

Step1: Recall the empirical rule for normal distribution

In a normal - distribution, approximately 68% of the data lies within 1 standard deviation of the mean, 95.4% lies within 2 standard deviations of the mean, and 99.7% lies within 3 standard deviations of the mean.
Let the mean $\mu = 55000$ and the standard deviation $\sigma=12000$.
We want to find the number of standard - deviations for the values 43000 and 67000.
For $x_1 = 43000$, the z - score is $z_1=\frac{x_1-\mu}{\sigma}=\frac{43000 - 55000}{12000}=\frac{- 12000}{12000}=-1$.
For $x_2 = 67000$, the z - score is $z_2=\frac{x_2-\mu}{\sigma}=\frac{67000 - 55000}{12000}=\frac{12000}{12000}=1$.

Step2: Apply the empirical rule

Since the values 43000 and 67000 are 1 standard deviation below and above the mean respectively, by the empirical rule of the normal distribution, approximately 68% of the data lies between these two values.

Answer:

68%