Question

A set of data has a normal distribution with a mean of 56 and a standard deviation of 9. Find the percent of data within the following interval. from 29 to 83 The percent of data within the given interval is ____%.

Answer 1

Solution

We are given

$\begin{gathered} \mu=56 \\ \sigma=9 \end{gathered}$

First, we will find the probability p(29 < x-bar < 83)

Note: z score formula

$z=\frac{\bar{x}-\mu}{\sigma}$

To find the probability

$\begin{gathered} p(29<\bar{x}<83)=p(\bar{x}<83)-p(\bar{x}<29) \\ p(29<\bar{x}<83)=p(z<\frac{83-56}{9})-p(z<\frac{29-56}{9}) \\ p(29<\bar{x}<83)=p(z<3)-p(z<-3) \\ p(29<\bar{x}<83)=0.99865-0.0013499 \\ p(29<\bar{x}<83)=0.9973001 \\ p(29<\bar{x}<83)=0.9973\text{ \lparen to four decimal places\rparen} \end{gathered}$

Therefore, the percentage within the given interval will be

$0.9973\times100=99.73\%$

The answer is

$\begin{equation*} 99.73\% \end{equation*}$