Let "X" be a binomial random variable based on "n" trials with success probability "p."
If the binomial probability histogram is not too skewed, "X" has approximately a normal distribution with "\\mu=np" and "\\sigma=\\sqrt{p(1-p)\/n}."
In practice, the approximation is adequate provided that both "np\\geq10" and "n(1-p)\\geq10," since there is then enough symmetry in the underlying binomial distribution.
Check normal approximation to binomial
"n=1000, p=0.8"
"np=1000(0.8)=800\\geq10,"
"n(1-p)=1000(1-0.8)=200\\geq10"
Hence we can use normal approximation to binomial. Then
C> 1.264%
Comments
Leave a comment