Question

Statistics Question: I have the answer, need help on the "how they got there" piece!

"If a random sample of size n is selected from the finite population that consists of the integers 1, 2,...N, show that :

The mean of (X-bar) is (N+1)/2

Answer 1

Now, I have the solution, but Im not here fishing for answers so much as trying to understand how they get from one end of the equation to the next. The solution is \[E[X^-] = \mu = \frac{ 1 }{ N } \sum_{k=1}^{N}k = \frac{ 1 }{ N }[\frac{ N(N+1) }{ 2 }] = \frac{ N+1 }{ 2 }\]

Answer 2

What I dont understand is that my book defines Xbar as just \[\sum_{i=1}^{n} \frac{ x_i }{ n }\]
So either I am mis-interpreting the problem, or I just am not getting how to translate the summation right. Anyone know where the N(N+1)/2 comes from?

Answer 3

recall the \[E[X] = \sum_{n=1}^{N}P[X=n]*n\]
If each sample is equally likely , then \[P[X=n] = 1/N\]
So \[E[X]=\sum_{n=1}^{N}(1/N)*n\]
\[E[X]=(1/N)\sum_{n=1}^{N}n\]
But 1 + 2 +3 = 3 + 2 + 1
1 + 2 + ... N = N + ... 2 + 1
Add the first term on the left to the first term on the right, the second term on the left to the second term on the right and so on...
(1+ N) + (2 + N-1) + ... (N + 1)
This give N*(N+1)
But this is twice the original sum, so divide it by 2
1 + 2 + ... + N = N*(N+1)/2
Therefore
\[E[X]=(1/N)\sum_{n=1}^{N}n\]
\[= (1/N)*N*(N+1)/2\]
\[=(N+1)/2\]

Answer 4

I cant pretend to know the reasoning for doing that, but I appreciate you explaining it, I'll play around with some similar problems to see if I can make it click. Cheers!