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=(x-\mu)/(\sigma) \\ z=(104-110)/(20) \\ z=(-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.