uniform distribution with interval (0,2) and sample 12

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.

• 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. – Brian Mar 28 at 12:11
• Your approximation to an answer is correct. – NCh Mar 28 at 13:32
• 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. – BruceET Mar 28 at 15:57
• @BruceET -but how many simulations would you need to distinguish between $0.001350$ and $0.001007$? – Henry Mar 28 at 16:55
• @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. – 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