Can someone please let me know if I have this question correct:
Ron drives a transport truck across North America. The average number of
kilometers a driving tire will last is 83000 km with a standard deviation of
12000 km. Ron likes to replace his tires before the lifespan is likely to be
passed. If he wishes to replace tires so that there is at most a 20% likelihood
that the tires are passed their lifespan, at what distance should he replace the
tires? Report your answer to the nearest hundred km.
I drew a normal distribution curve like so to represent what I'm trying to find:
I figured the area for $Z_1$ is $70 \%$ because from the left to the middle is $50\%$ and than another $20\%$ on the right side, so $0.50 + 0.20 = 0.70$ or $70\%$. The inverse of the area of $0.70$ is $0.5244$, thus:
$$Z_1 = 0.5244$$ $$\mu = 83000$$ $$\sigma = 12000$$
Then using
$$Z_1 = \frac{x - \mu}{\sigma} $$
I did:
$$0.5244 = \frac{x - 83000}{12000}$$ $$0.5244 \times 12000 = x - 83000$$ $$6292.81 + 83000 = x$$ $$x = 89292.81 \approx 89300$$
Therefore, Ron would change his tires after driving about 89300km.
Is this correct?
The picture is not correct. We want a probability of $20\%$ in the left tail.
If Ron replaces at $83000$, then with probability $0.5$ the tire is by then "unsafe." If later, as in your answer, then the probability the tire is unsafe by the time it is replaced is $70\%$.