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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?

Elijah
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2 Answers2

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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 $p(X\geq 2)$

David Quinn
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  • Thank you, so i solve P(x < 900) right? Which then becomes P(Z < (900-1000)/sample standard dev.), right? But then I get P(Z < -20), which is equal to 0... – Elijah Jan 11 '23 at 10:36
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    You use the standard deviation $50$ not the standard deviation of the mean, so you should have $p(z<-2)$ not $p(z<-20)$ – David Quinn Jan 11 '23 at 16:20
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    You’re not looking for the probability the mean of a sample of $100$ is less than $900$ which is obviously vanishingly small… – David Quinn Jan 11 '23 at 16:23
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    @Elijah It's important here to note which quantity we're using the standard deviation of. The standard deviation derived from the CLT gives us information on the distribution of the mean of our sample, but if we just want the probability of any individual bulb dying out early, then that relies on the distribution of the lifetimes itself, so we use the original standard deviation. What this answer is trying to guide you toward is that if we have that probability for a single bulb, then the bit having to do with the sample of 100 bulbs is really setting up Bernoulli trials. Does that help? – Stephen Donovan Jan 11 '23 at 20:00
  • Yeah, a lot! Thank you – Elijah Jan 12 '23 at 11:12
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Let $X$ be the lifetime of a lightbulb. Then $X$ is normally distribution with mean 1000 hours and standard deviation of 50 hours.

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

Someone
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  • But I don't understand why we don't put the standard deviation of the sample, which is 5, at the denominator... – Elijah Jan 11 '23 at 12:11