Probability of lamp after $10500$ h when already reached $9000$ h

Question

An energy-saving lamp lights up an average of $10000$ hours before it fails, with a standard deviation of $800$ hours. Gaussian distribution.

What is the probability of a lamp lighting up after $10500$ hours still when it has already reached $ 9000 $ hours?

Equation for standard distribution:

$z = \frac{x-\mu}{\sigma}$

Probability of reaching $9000$ hours $=0.1056$

Probability of reaching $10500$ hours $=0.2659$

I tried things like subtracting the difference of the probabilities from 1, subtracting quotients, subtracting product. I didn't came to any solution.

Answer 1

If $X$ is the random variable that represents lifetime of the bulb,

$P(X \gt 9000) \ne 0.1056$. Rather $P(X \gt 9000) = 1 - 0.1056$. The integral will be,

$ \displaystyle P(X \gt 9000) = \int_{9000}^{\infty} \frac {1}{800 \sqrt{2 \pi}} e^{- (x - 10000)^2 / 1280000} ~dx$ $\approx 0.89435$

$P(X \gt 10500) \approx 0.266$ as you obtained.

So, $ \displaystyle P(X \gt 10500 | X \gt 9000) \approx \frac{0.266}{0.89435}$

Answer 2

Just use standard probability first, if $X$ is the life time of the lamp:

$$P(X \ge 10500\mid X\ge 9000) = \frac{P(X \ge 10500)}{P(X \ge 9000}$$

a special case of $P(A\mid B)=\frac{P(A \cap B)}{P(B)}$ etc. You already computed both parts of the fraction, I believe? They cannot be correct though, why?