Question

suppose that the time required to complete a 1040R tax form is normally distributed with a mean of 110 minutes and a standard deviation of 20 minutes. What proportion of 1040R tax forms will be completed in less than 104 minutes?

Answer 1

For the normal distribution, given,

$\begin{gathered} \mu=110 \\ \sigma=20 \end{gathered}$

Let use the z-score formula to find the zscore corresponding to the value x = 104.

$\begin{gathered} z=\frac{x-\mu}{\sigma} \\ z=\frac{104-110}{20} \\ z=\frac{-6}{20} \\ z=-0.3 \end{gathered}$

Thus, we need

$P(z<-0.3)$

Using a normal distribution table/calculator, we can figure out the answer.

$P(z<-0.3)=0.38209$

Answer

About 38.215% of 1040R tax forms will be completed in less than 104 minutes.