Q): Suppose that you wish to sample $12$ observations randomly from a uniform distribution on the interval $(0,2)$. An approximate value of the probability that the average of your sample will be less than $0.5$.

what i did was i found the $\mathbb E(x)$ to be $1$ and variance to be $1/3$ and then i wasnt sure where to go with this so i decided to do the $z$-formula of $(0.5-1)/\sqrt{1/3}/\sqrt{12}$ to get me $z = -3$ then i got $p= 0.001350$ from that , i was just wondering if that was correct.

  • $\begingroup$ Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. $\endgroup$ – Brian Mar 28 at 12:11
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    $\begingroup$ Your approximation to an answer is correct. $\endgroup$ – NCh Mar 28 at 13:32
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    $\begingroup$ Simulation in R: a = replicate(10^6, mean(runif(12, 0, 2))); mean(a < .5) returns 0.000979, accurate to about 3 places and agrees with $\approx 0.001$ from good normal approx. $\endgroup$ – BruceET Mar 28 at 15:57
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    $\begingroup$ @BruceET -but how many simulations would you need to distinguish between $0.001350$ and $0.001007$? $\endgroup$ – Henry Mar 28 at 16:55
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    $\begingroup$ @Henry. I was just doing a 'reality check' on a tail probability so far out. With a million iterations it's $.001 \pm 0.0006.$ Maybe $10^7$ or $10^8$ would settle it, But the point is that the normal aprx should be good enough--even with an avg of only 12. $\endgroup$ – BruceET Mar 28 at 18:25

The expectation of a single sample is $1$, with variance $\frac13$, so the average of twelve samples has an expectation of $1$ and variance of $\frac{1}{36}$, saying (as you found) that for this average $0.5$ would be $3$ standard deviations below the mean. For a standard normal distribution, you are correct that the probability of values less than $3$ standard deviations below the mean is about $0.00135$

The difficult question is whether a normal approximation is a good approximation in this case: in particular normal approximations can sometimes perform relatively poorly in the tails

Your question actually involves a Bates distribution, but it is easier to handle translating it to the Irwing-Hall distribution: you want the probability that the sum of $12$ uniform i.i.d. random variables on $[0,1]$ is less than $3$. This is $$\frac{1}{12!}\left({12 \choose 0}(3-0)^{12} - {12 \choose 1}(3-1)^{12} + {12 \choose 2}(3-2)^{12}- {12 \choose 3}(3-3)^{12}\right) \approx 0.001007$$

So the exact answer is rather less than the normal approximation. Whether the normal approximation is good enough, I leave for you to judge


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