Question

NOT A QUESTION....

Answer 1

@lgbasallote

Answer 2

proofs?

Answer 3

Most of the statstics book do not have that... so i did it

Answer 4

cool. takes great smarts to do those =))

Answer 5

Thanx

Answer 6

@eliassaab can u plz check is my maths correct....

Answer 7

Notice first that
\[
k \binom n k = n \binom {n - 1} {k - 1}
\]

\[
E(X)= \sum_{k \mathop = 0}^n k \binom n k p^k q^{n - k}=\\
\sum_{k \mathop = 0}^n n \binom {n - 1} {k - 1} p^k q^{n - k}=\\
n p \sum_{k \mathop = 1}^n \binom {n - 1} {k - 1} p^{k - 1} q^{\left({n - 1}\right) - \left({k - 1}\right)}
\\
\text{ Changing the variables } m=n-1,\, j=k-1\\
n p \sum_{j \mathop = 0}^m \binom m j p^j q^{m - j}=\\np (p+q)^m=np (1)^m=np
\]

Answer 8

how is