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Apologies for the lack of a descriptive title but I don't know well how to describe this problem. I received it during an online pre-interview test earlier this year for an algo trading role. I was unable to solve it from a probabilistic standpoint and have still been unsuccessful, only doing so by coding a quick MC simulation. Perhaps this was their goal given it was an algo-based position, but I can't help but feeling there is a simpler method. I don't know the exact figures from my question, but found an identical format online.

Two laptop manufacturing companies A and B are fierce rivals. A's laptops have a mean life of 4.9 years with a standard deviation of 1.2 years, while B's laptops have a mean life of 4 years with a standard deviation of 1 years. What is the probability that a random sample of 25 of A's laptops will have a mean life that is at least 1 year longer than that of a random sample of 49 of B's laptops?

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    $\begingroup$ This is a straightforward application of the central limit theorem and should be solvable by any undergraduate student of probability theory or statistics. No simulation is needed; in fact, all that is required is a standard normal distribution table and a hand calculator. If one has memorized certain empirical rules, even the former is not required. $\endgroup$ – heropup Aug 24 '18 at 7:44
  • $\begingroup$ heropup, perhaps you or someone else can provide additional guidance. I read about and watched a couple of lectures on CLT but despite a now cursory understanding, haven't found content covering applications behind multiple independent random variables as we have in my question above and am still unable to solve it. Most applications involve some distribution and its parameters with some n, and an x (i.e. probability x<=xbar) but I haven't found applications beyond this. $\endgroup$ – NaT3z Aug 24 '18 at 9:46
  • $\begingroup$ You need to find the expected values and variances of (a) the sample mean of the A laptops, (b) the sample mean of the B laptops, (c) the difference in the sample means $(\bar{A}- \bar{B})$, and (d) that difference minus $1$. You might approach (a) and (b) via the sample sums if that is easier for you $\endgroup$ – Henry Aug 24 '18 at 11:58
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First of all we have to know how the means of the samples are distributed. Let´s say you have a sample of $n$ independent normally distributed random variables with the parameters $\mu$ and $\sigma^2$. Then the mean of the sample is distributed as $ \mathcal N\left(\mu, \frac{\sigma^2}{n} \right)$. Therefore the mean of the sample of A laptops is distributed as $\overline X_{25}\sim \mathcal N\left(4.9, \frac{1.2^2}{25} \right)$. And the the sample of B laptops is distributed as $\overline Y_{49}\sim \mathcal N\left(4, \frac{1^2}{49} \right)$

Then it is asked for $\overline X_{25}-\overline Y_{49}>1$ or equivalently $\overline Y_{49}-\overline X_{25}<-1$

The difference of two normally and independent distributed random variables are normally distributed as well. The new mean is the difference of the mean of the two random variables.The new variance is the sum of the variances.

$D=\overline Y_{49}-\overline X_{25}\sim \mathcal N\left(4-4.9, \frac{1.2^2}{25}+ \frac{1^2}{49} \right)$. For the standard deviation we have approximately $0.2793$

Finally you calculate the value of $P(D<-1)=\Phi\left(\frac{-1-(-0.9)}{0.2793} \right)$, where $\Phi\left(z \right)$ is the cdf of the standard normal distribution.

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    $\begingroup$ Thank you! I made the ridiculously ignorant error of summing the standard deviations rather than the variances which is a frustratingly silly mistake given the nature of standard deviation of a sample... $\endgroup$ – NaT3z Aug 25 '18 at 0:18
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    $\begingroup$ @NaT3z You´re welcome. Don´t worry, this is a common mistake. I´ve made it too. But after some of these frustratingly events I never sum any standard deviations again. $\endgroup$ – callculus Aug 25 '18 at 8:35

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