2
$\begingroup$

We are given that numbers are selected at random from the interval $(0,1)$. If $100$ numbers are selected what is the probability that the average of the numbers is less than $0.5$?

We let $X_i$ be the $i$-th number selected, assuming $X_i\sim U(0,1)$, we need to find $$P\left(\dfrac{X_1 + X_2+ \dots +X_{100}}{100} < 0.5\right)$$ i.e $P(X_1 + X_2+ .... +X_{100} < 50)$.

Assuming $X_i$'s as independent, how can we find the given probability? I don't think $\sum_{i=1}^{n} X_i$ will come out to be a uniform distribution too. Can anyone help ?

$\endgroup$
  • 4
    $\begingroup$ If the numbers $X_i$ are independent and uniform on $(0,1)$ then so are the numbers $1-X_i$, hence the desired probability is $\frac12$ by symmetry. $\endgroup$ – Did Nov 22 '16 at 19:33
2
$\begingroup$

Comment: @Did's answer is simple and does not involve any approximation.

However, if you are in a beginning probability course and just covered the Central Limit Theorem, there is a chance that you are intended to use that. Each $X_i$ has $E(X_i) = 1/2$ and $Var(X_i) = 1/12.$

Therefore $S = \sum_{i=1}^{100} X_i$ has $E(S) = 50,$ $Var(S) = 100/12,$ and $SD(S) = 2.886751.$ By the CLT, $S$ is approximately normal. So you could standardize $S$ and use the (symmetrical!) normal distribution to get the answer.

This method has the advantage that it could also be used to find $P(S < a)$, for numbers $a$ other than 50.

A simple simulation (in R statistical software) makes it possible to illustrate that the distribution of $S$ is nearly normal.

m = 10^4;  n = 100;  x = runif(m*n)
DTA = matrix(x, nrow=m) # each row a sample of 100
s = rowSums(DTA) # vector of sums of the m samples
mean(s < 50)
## 0.499  # aprx P(S < 50),  proportion of s-values below 50

hist(s, prob=T, br=20, col="wheat", ylim=c(0,.15), 
   main="Simulated Distribution of Sums of 100 UNIF(0,1) Observations")
 curve(dnorm(x, 50, sqrt(100/12)), lwd=2, col="blue", add=T)
 abline(v=50, lwd=2, col="red")

enter image description here

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.