Sampling distributions

Activity 5.1   Continuing Activities 3.8 and 3.11,show that the mean and variance of the binomial distribution in (which is the same as the sum of $n$ independent Bernoulli trials) has mean $n\pi$ and variance $n\pi(1-\pi)$.
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Answer 5.1   From the formulae, the proportion of successes, which is the mean of the $n$ Bernoulli trials has mean $\pi$ and variance $\pi(1-\pi)/n$. So the sum of $n$ Bernoulli trials has mean $n\pi$ and variance $n\pi(1-\pi)$.
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Activity 5.2   From the results in 5.1, by fixing $\lambda=n\pi$, and allowing $n$ to increase, and $\pi$ to get small, discover that the mean and variance of a Poisson distribution are both $\lambda$.
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Answer 5.2   The mean is $n\pi=\lambda$. The variance is $n\pi(1-\pi)=\lambda(1-\pi)$. This tends to $\lambda$ as $\pi$ tends to 0.
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