# Solved: The tread wear (in thousands of kilometers) that car owners get with a certain kind of tire is a random variable whose probability density

#### ByDr. Raju Chaudhari

Feb 24, 2021

The tread wear (in thousands of kilometers) that car owners get with a certain kind of tire is a random variable whose probability density is given by

 $$f(x) = \frac{1}{30}e^{-x/30}, x>0$$

Find the probabilities that one of these tires will last

(a) at most 18,000 kilometers;
(b) anywhere from 27,000 to 36,000 kilometers;
(c) at least 48,000 kilometers

#### Solution

The distribution function of $x$ is

 \begin{aligned} F(x) &= 1- e^{-x/30}. \end{aligned}

(a) The probabilities that one of these tires will last at most 18,000 kilometers is $P(X\leq 18)$

 \begin{aligned} P(X \leq 18) &= P(X\leq 18)\\ & = F(18)\\ & = \big(1- e^{-18/30}\big) \\ & = 1- 0.5488116 \\ &= 0.4511884 \end{aligned}

(b) The probabilities that one of these tires will last anywhere from 27,000 to 36,000 kilometers is $P(27 < X < 36)$

 \begin{aligned} P(27 < X < 36) &= P(X < 36) - P(X < 27)\\ & = F(36) - F(27)\\ & = \big(1- e^{-36/30}\big) - \big(1- e^{-27/30}\big)\\ &= e^{-27/30} -e^{-36/30}\\ & = 0.4065697 - 0.3011942\\ &= 0.1053754 \end{aligned}

(c) The probabilities that one of these tires will last at least 48,000 kilometers is $P(X\geq 48)$.

 \begin{aligned} P(X\geq 48) &= 1- P(X< 48)\\ & = 1- F(48)\\ & = 1- \big(1- e^{-48/30}\big) \\ & = e^{-48/30} \\ & = 0.2018965 \end{aligned}