The life expectancy of a certain tire is a normal distribution with the mean = 40,000 miles, standard deviation is 1,000 miles. Find the probability that a tire will last more than 39,000 miles.

Answer:[tex]P = 84\%[/tex]Step-by-step explanation:The average is: [tex]\mu = 40,000\ miles[/tex]The standard deviation is: [tex]\sigma = 1,000\ miles[/tex]We want the probability that a tire lasts more than 39,000 miles. This is: [tex]P (X>39,000)[/tex]Now we must transform these values to those of a standard normal distribution to facilitate calculation by using the probability tables. [tex]P (X> 39,000)\\\\P (X-\mu> 39,000-40,000)\\\\P (\frac{X-\mu}{\sigma}> \frac{39,000 -40,000}{1,000})\\\\P (Z> -1)[/tex]This is: [tex]P(Z> -1) = P(Z<1)[/tex] ---------- (For the symmetry of the standard normal distribution) When you search for the normal standard table, you get the following value: [tex]P(Z <1) = 0.8413\\\\P(Z> -1) = 0.8413[/tex]