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?
Accepted Solution
A:
Answer: 0.0668Step-by-step explanation:Given: Mean : [tex]\mu=\text{4 min. 3 sec.=4.05 minutes}[/tex]Standard deviation : [tex]\sigma = \text{2 seconds=0.033333 minutes }[/tex]The formula to calculate z-score is given by :_[tex]z=\dfrac{x-\mu}{\sigma}[/tex]For x= 4 minutes , we have[tex]z=\dfrac{4-4.05}{0.03333}\approx-1.5[/tex]The P-value = [tex]P(z\leq-1.5)=0.0668072\approx0.0668[/tex]Hence, the probability that on a given run, the time will be 4 minutes or less = 0.0668