by Atul
Last Updated July 12, 2019 07:20 AM

A Binomial distribution Bp,n , where p≠0 , has the same mean and standard deviation, namely μ=σ .

Find the mean of Bp,n+1 .

$$\mu_1=np=\sqrt{np(1-p)}=\sigma_1$$ $$\implies np=1-p$$ $$\mu_2=(n+1)p=np+p=(1-p)+p=1$$

From $np=\mu=\sigma=\sqrt{np(1-p)}$ it follows that $np=0$ or $np=1-p$.

The first option is excluded on base of $p\neq0$.

Then: $$(n+1)p=np+p=1-p+p=1$$

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