Question

A new Firestone tire is guaranteed to last for 40,000 miles. The actual mean life of the tires is 47,000 miles with a standard deviation of 4,000 miles.

a) What percent of the tires will last for at least 40,000 miles?
b) What percent of the tires will not last for at least 40,000 miles?
c) What is the probability that a tire will last for more than 50,000 miles?
d) The Firestone Company wants to advertise how long some of their tires last. They decide to state how long the top 3% of their tires will last. How many miles will the top 3% of their tires last?

Answer 1

a) Let's find z - score:
z = (x-mu)/sigma = (40,000-47,000)/4,000 = - 1.75
p-value for z = 1.75 is 0.0401, for estimating the percent of the tires which last for at least 40,000 miles we have to consider right side of the distribution:
P(x > 40,000) = (1-P(x < 40,000))*100% = 95.99 %


b) the percent of the tires which are not last for at least 40,000 miles would be equal to P (x < 40,000) = 0.0401*100% = 4.01 %;
c) for more than 50,000 miles
z = (50,000-40,000)/4,000 = 2.5
the probability would be
P(x>50,000) = 00.62 %