if it's Q3, then for part a), expand out the sum,
 + X_2 - \bar(X) + X_3-\bar(X) + ... + X_n - \bar(X) = X_1+X_2+X_3+...+X_n - n\bar(X) = X_1+X_2+X_3+...+X_n - n(n^{-1}\sum X_i) = 0)
as required
for part b) diff (X_i-c)^2 with respect to the variable X_i obviously becomes 2(X_i-c) now if we sum it from i = 1 to n we have 2(X_1-c) + 2(X_2-c) + ... + 2(X_n-c)
factor the 2 out 2[(X_1-c) + (X_2-c) + ... + (X_n-c)] now look at the inside bracket thats just the expression in part a) thats minimised when c = mean of X
so [(X_1-c) + (X_2-c) + ... + (X_n-c)] = 0 when c = mean of X thus 2(X_i-mean of X) = 0 which means at c = mean of X there is the turning point for the parabola
and u know since this is a positive parabola then that point must be the global minimum
for part c)
http://en.wikipedia.org/wiki/Bessel%27s_correctioni hope im talking about the right assignment lololol
...and uom offers this subject in first year? that's actually a pretty good subject from the looks of this assignment, very very very helpful in your latter careers in commerce, stats and commerce go hand in hand.
Home
Help
Search
Login
