Question

The life of a certain type of automobile tire is normally distributed with mean 34,000 miles and standard deviation 4000 miles.

Answer 1

Answer:

$\mu = 34000$

$\sigma = 4000$

(a) What is the probability that such a tire lasts over 40,000 miles?

We are supposed to Find P(x>40000)

$z=\frac{x-\mu}{\sigma}$

at x =40000

$z=\frac{40000-34000}{4000}$

$z=1.5$

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)

$z=\frac{x-\mu}{\sigma}$

at x =30000

$z=\frac{30000-34000}{4000}$

$z=-1$

refer the z table

P(x<30000)=P(z<-1)=0.1587

$z=\frac{x-\mu}{\sigma}$

at x =35000

$z=\frac{35000-34000}{4000}$

$z=0.25$

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

$P(x>40000 | x>30000) = \frac{P(x>40000)}{P(x>40000)}$

$=\frac{1-P(x$

$=\frac{1-P(z$

$=\frac{1-P(z$

$=\frac{1-0.9332}{1-0.1587}$

$=\frac{0.0668}{0.8413}$

$=0.0794$

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