I have trouble solving the following statistics exercise:

Question

A company produces lightbulbs with an average lifetime of 1000 hours and standard deviation of 50 hours. Find the probability that in a sample of 100 bulbs there are at least two which stop working before 900 hours of lifetime.

When I was approaching this exercise I noticed it didn't say how the lifetime was distributed. But then I remembered that according to the CLT I we can approximate the sampling distribution to a normal distribution, when n (dimension of the sample) tends to infinity.

I know that the sample mean is equal to the population mean, I calculated the standard deviation for the sample $\bar{x}$ $$ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{50}{\sqrt{100}} = 5 $$

But after that I'm kinda stuck, I'm not sure about the at least part. I tried calculating $$ P =\ (\bar{X} <900) $$ standardising and using the z-scores table but I get $ P =\ (Z < -20) $, which can't be right.

Any ideas on how approach this exercise?

Answer 1

HINT…First find the probability of any one light bulb having a lifetime of less than $900$ hours, which you can calculate using the normal distribution.

The probability you get will be $p$, say.

Then use the Binomial distribution $X \rightarrow Bin(100, p)$ to calculate $p(X\geq 2)$

Answer 2

Let $X$ be the lifetime of a lightbulb. Then $X$ is normally distribution with mean 1000 hours and standard deviation of 50 hours.

$$ P(X<900) = P\left(\frac{X-1000}{50} < \frac{900-1000}{50} \right) = P(Z<-2) $$