Question

A mile-runner’s times for the mile are normally distributed with a mean of 4 min. 3 sec. (This would have to be expressed in decimal minutes -- 4.05 minutes), and a standard deviation of 2 seconds (0.0333333··· minutes (the three dots indicate a repeating decimal)). What is the probability that on a given run, the time will be 4 minutes or less?

Answer 1

Answer: 0.0668

Step-by-step explanation:

Given: Mean : $\mu=\text{4 min. 3 sec.=4.05 minutes}$

Standard deviation : $\sigma = \text{2 seconds=0.033333 minutes }$

The formula to calculate z-score is given by :_

$z=\dfrac{x-\mu}{\sigma}$

For x= 4 minutes  , we have

$z=\dfrac{4-4.05}{0.03333}\approx-1.5$

The P-value = $P(z\leq-1.5)=0.0668072\approx0.0668$

Hence, the probability that on a given run, the time will be 4 minutes or less = 0.0668