mean_and_variance_of_binomial_distribution
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| Both sides previous revisionPrevious revisionNext revision | Previous revision | ||
| mean_and_variance_of_binomial_distribution [2024/10/09 10:27] – [For variance] hkimscil | mean_and_variance_of_binomial_distribution [2024/10/14 17:06] (current) – [For variance] hkimscil | ||
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| Line 32: | Line 32: | ||
| ====== For Mean ====== | ====== For Mean ====== | ||
| \begin{eqnarray*} | \begin{eqnarray*} | ||
| - | E(X) & = & \sum_{x}x p(x) \\ | + | E(X) & = & \sum_{x}x p(x) \;\;\; \because \; p(x) = {{n}\choose{x}} p^x (1-p)^{n-x} |
| & = & \sum_{x=0}^{n} x {{n} \choose {x}} p^x(1-p)^{n-x} | & = & \sum_{x=0}^{n} x {{n} \choose {x}} p^x(1-p)^{n-x} | ||
| & = & \sum_{x=0}^{n} x \frac{n!}{x!(n-x)!} p^x(1-p)^{n-x} | & = & \sum_{x=0}^{n} x \frac{n!}{x!(n-x)!} p^x(1-p)^{n-x} | ||
| Line 92: | Line 92: | ||
| \text {we know that the underline part is} \\ | \text {we know that the underline part is} \\ | ||
| + | (p+(1-p))^m \\ | ||
| + | \text {and, we also know that it is 1} \\ | ||
| (p+(1-p))^m = 1^m \\ | (p+(1-p))^m = 1^m \\ | ||
| & = n(n-1)p^2 (p + (1-p))^m \\ | & = n(n-1)p^2 (p + (1-p))^m \\ | ||
| Line 100: | Line 102: | ||
| E[X(X - 1)] & = n(n-1)p^2 \\ | E[X(X - 1)] & = n(n-1)p^2 \\ | ||
| E[X^2 - X] & = n(n-1)p^2 \\ | E[X^2 - X] & = n(n-1)p^2 \\ | ||
| - | E[X^2]- E[X] & = n(n-1)p^2 \\ | + | E[X^2]- E[X] & = n(n-1)p^2 |
| - | E[X^2]- np & = n(n-1)p^2 \\ | + | E[X^2]- np & = n(n-1)p^2 |
| E[X^2]& = n(n-1)p^2 + np \\ | E[X^2]& = n(n-1)p^2 + np \\ | ||
| \\ | \\ | ||
mean_and_variance_of_binomial_distribution.1728437223.txt.gz · Last modified: 2024/10/09 10:27 by hkimscil