find the value a such that the probability that a tire lasts more than a miles is approximately 0.8

Question

The lifetime of a certain tire has normal distribution with $\mu = 50,000$, and $\sigma = 5,000$.

(a) Find the probability that the tire will last between 48,000 and 56,000 miles.

Let $L$ = Lifetime of tire ~ $N(50000,5000)$. Let $Z = \frac{L-50000}{5000}, Z$~$(0,1)$. We want $P(\frac{-2000}{5000}<Z<\frac{6000}{5000})$ = $\Phi(\frac{6}{5}) - \Phi(-\frac{2}{5}) = .8849 - .3446 = .5403$.

(b) Find the value $a$ such that the probability that a tire last more than $a$ miles is approximately 0.8.

Here is where I am getting stuck, how should I set up the probability so that I can find $a$? Below is what I have so far...

$P(Z<a) = .8 =$ $P(\frac{L-50000}{5000}<\frac{a-50000}{5000})$

However, I dont know where to proceed from here. Thanks in advance for any help.

Answer 1

Hint. You are looking for a number $a$ such that $$ P(L>a)=0.8 $$ or equivalently, by the change of variable $L \to Z$, $$ P(Z>\frac{a-50000}{5000})=0.8 $$ or $$ P(Z\leq\frac{50000-a}{5000})=0.8 $$ Using a table your $Z$-table you can find that $$ P(Z\leq \color{red}{0.84}) \approx 0.8. $$ Then

$$ \frac{50000-a}{5000}=\color{red}{0.84} $$

and you may conclude easily.