# Uniformly distributed probability error

I'm having trouble with the following question:

The error for the production of a machine is uniformly distributed over $$[-0.7, 0.5]$$ unit. Assuming that there are $$10$$ machines working at the same time, approximate the probability that the final total production differ from the exact total production by more than $$1$$ unit?

I've attempted this multiple times and failed to get the correct answer. See the image below for my solution reaching 0.033625.

Any advice on where I'm going wrong and what I should be doing instead would be greatly appreciated. Thank you.

• Despite the praise below for providing details, we frown upon the use of images, unless vital (e.g., geometry). Please do your best to use text and mathjax in questions. Images can disappear, links rot over time. This site is an archive, as well as a Q&A site. Your post above would be rather easy to type, and equations will be formatted with on or two $-signs on each end. Just the effort, even if not perfect, will be well received. Oct 13 at 21:46 ## 2 Answers The requested probability that the error in production is greater than 1 means that $$\mathbb{P}[|\Sigma X|>1]=\mathbb{P}[|Z|>1.826]=2\times 0.034.$$ ...also an error of -2 even differs from exact production by more than 1 unit. • Correct method, but there is confusion about the standard deviation of the sum. Should be$\sigma_T=\sqrt{1,2},\$ Oct 2 at 0:52

Thanks for showing your work in detail; you are on the right track, with minor errors. @tommik (+1) has the correct analytic method, also using the fact that the sum of $$n=10$$ IID uniform random variables is very nearly normal. [By 'exact total', I think you mean $$E(T)=-1.]$$

By simulation of a million sums of ten, in R we have the following:

set.seed(2021)
t = replicate(10^6, sum(runif(10, -.7,.5)))
mean(abs(t -(-1)) > 1)
[1] 0.366562    # aprx prob of tails
mean(t)
[1] -0.9987682  # aprx -1
sd(t)
[1] 1.095333    # aprx 1.095445


Normal Approximation:

1 - diff( pnorm(c(-2,0), -1, sqrt(1.2)))
[1] 0.3613104


In the figure below, the desired probability is the the sum of the two areas outside the vertical red lines.

hdr= "UNIF(-.7, .5): Sums of Sample of 10 with Aprx. Normal Density"
hist(t, prob=T, br=40, col="skyblue2", main=hdr)
abline(v = c(-2, 0), col="red")