The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles. What is the probability that a randomly selected set of tires will have a life of 36,000 to 46,000 miles? Round your answer to 4 decimal places.
Accepted Solution
A:
Answer: 0.6731Step-by-step explanation:Given : Mean : [tex]\mu = 40,000\text{ miles}[/tex]Standard deviation : [tex]\sigma = 5,000\text{ miles}[/tex]The formula to calculate the z-score :-[tex]z=\dfrac{x-\mu}{\sigma}[/tex]For x=36,000[tex]z=\dfrac{36000-40000}{5000}=-0.8[/tex]For x=46,000[tex]z=\dfrac{46000-40000}{5000}=1.2[/tex]The P-value : [tex]P(-0.8<z<1.2)=P(z<1.2)-P(z<-0.8)[/tex][tex]=0.8849303-0.2118554=0.6730749\approx0.6731[/tex]Hence, the probability that a randomly selected set of tires will have a life of 36,000 to 46,000 miles =0.6731