Question

The life of a certain type of automobile tire is normally distributed with mean $34,000$miles and standard deviation $4,000$miles.

(a) What is the probability that such a tire lasts more than $40,000$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 the tire survives another $10,000$miles?

Short answer

a) $0.06681$is the probability last more than $40,000$miles

b) $0.44$is the probability between $30,000$and $35,000$miles

c)$0.0794$is the conditional probability of another$10,000$miles

Step-by-step solution

Step:1 Find the probability of last more than 40,000miles.(Part a)

a) Create a random value of $X$.

Representing the lifespan of a chosen randomly automotive tire.

That is given to us $X~\mathcal{N}\left(34000,{4000}^2\right)$

$\begin{array}{r}P(X>40000)=1-P(X\le 40000)\\ \end{array}$

$\begin{array}{r}=1-P\left(\frac{X-34000}{4000}\le \frac{40000-34000}{4000}\right)\\ \end{array}$

$=1-\mathrm{\Phi }\left(1.5\right)$

$=1-0.93319$

$=0.06681$

Step:2 Find the probability between 30,000and 35,000. (Part b)

b) The given information is $X~\mathcal{N}\left(34000,{4000}^2\right)$

$P(30000\le X\le 35000)=P\left(\frac{30000-34000}{4000}\le \frac{X-34000}{4000}\le \frac{35000-34000}{4000}\right)$

localid="1649488038646" role="math" $=P\left(-1\le \frac{X-34000}{4000}\le 0.25\right)$

$=\mathrm{\Phi }\left(0.25\right)-\mathrm{\Phi }(-1)$

role="math" localid="1649488073471" $=0.59871-0.15866$

$=0.44$

Step:3  Find the conditional probability. (Part c)

c). Given $X~\mathcal{N}\left(34000,{4000}^2\right)$

$P(X\ge 40000\mid X\ge 30000)=\frac{P(X\ge 40000,X\ge 30000)}{P(X\ge 30000)}$

$=\frac{P(X\ge 40000)}{P(X\ge 30000)}$

The preceding expression is equivalent to

$\frac{P(X\ge 40000)}{P(X\ge 30000)}=\frac{1-P\left(\frac{X-34000}{4000}\le \frac{4000034000}{4000}\right)}{1-P\left(\frac{X34000}{4000}\le \frac{30000-34000}{4000}\right)}$

$=\frac{1-\mathrm{\Phi }\left(1.5\right)}{1-\mathrm{\Phi }(-1)}$

$=\frac{1-0.93319}{1-0.15866}$

$=0.0794$