Question

The times to complete an obstacle course is normally distributed with mean 73 seconds and standard deviation 9seconds. What is the probability using the Empirical Rule that a rana mly selected finishing time is less than 100seconds?• Provide the final answer as a percent rounded to two decimal places.

Answer 1

From empirical probability,

`\begin{gathered} Z-\text{score = }(X-\mu)/(\sigma) \\ =(100-73)/(9) \\ =3 \\ \text{That means 100=}\mu+3\sigma \end{gathered}`

The percentage of the distribution that lies between 0 to 100 is

`\begin{gathered} The\text{ percentage probability of distribution that lie between }\mu+3\sigma\text{ and }\mu-3\sigma\text{ plus } \\ \text{percentage of distribution that lie between 0 and }\mu-3\sigma \\ \end{gathered}`
`\begin{gathered} =\text{ 99.7 + }(100-99.7)/(2) \\ =99.7\text{ + 0.15} \\ =99.85\text{ \% (2 decimal places)} \end{gathered}`