Answer to Question #339646 in Statistics and Probability for sezi

Question

The life of a certain type of automobile tire is normally distributed with mean 34,000 miles and standard deviation 4000 miles. a. What is the probability that such a tire lasts over 40,000 miles? b. What is the probability that it lasts between 30,000 and 35,000 miles? c. Given that it has survived 30,000 miles, what is the conditional probability that it survives another 10,000 miles? 

Answer 1

a)

$$P(X>40000)=1-P(Z\le\dfrac{40000-34000}{4000})$$

$$=1-P(Z\le1.5)=0.0668$$

b)

$$P(30000<X<35000)$$

$$=P(Z<\dfrac{35000-34000}{4000})$$

$$-P(Z\le\dfrac{30000-34000}{4000})$$

$$=P(Z<0.25)-P(Z\le-1)$$

$$\approx0.598706-0.158655\approx0.4401$$

c)

$$P(X\ge30000+10000|X>30000)$$

$$=\dfrac{P(X\ge40000\cap X>30000)}{P(X>30000)}$$

$$=\dfrac{1-P(Z<\dfrac{40000-34000}{4000})}{1-P(Z\le\dfrac{30000-34000}{4000})}$$

$$=\dfrac{0.0668072}{0.8413447}=0.0794$$