Answer:


(a) What is the probability that such a tire lasts over 40,000 miles?
We are supposed to Find P(x>40000)

at x =40000


Refer the z table
P(z<1.5)=0.9332
P(x>4000)=1-P(z<1.5)=1-0.9332=0.0668
The probability that such a tire lasts over 40,000 miles is 0.0668
b)What is the probability that it lasts between 30,000 and 35,000 miles?
P(30000<x<35000)

at x =30000


refer the z table
P(x<30000)=P(z<-1)=0.1587

at x =35000


refer the z table
P(x<35000)=P(z<0.25)=0.5987
P(30000<x<35000) =P(x<35000)-P(x<30000)=P(z<0.25)-P(z<-1)=0.5987-0.1587=0.44
So,the probability that it lasts between 30,000 and 35,000 miles is 0.44
c) Given that it has survived 30,000 miles, what is the conditional probability that the tire survives another 10,000 miles?
P(x>40000 | x>30000)
Using formula given above







Given that it has survived 30,000 miles, The conditional probability that the tire survives another 10,000 miles is 0.0794