# probability( central limit theory)

A seed manufacturer sells seeds in packets of 50. Assume that each seed germinates with probability .99 independently of all the others.The manufacturer promises to replace, at no cost to the buyer, any packet with 3 or more seeds that do not germinate. a) Use the poisson to estimate the probability a packet must be replaced. b) Use the normal to estimate the probability that the manufacturer has to replace more than 70 of the last 4000 packets sold.

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The number of bad seeds in a packet has binomial distribution with $p=0.01$ and $n=50$. So $p$ is small, $n$ fairly large, with $np$ neither large nor small. Under these conditions, the binomial is well approximated by the Poisson with parameter $\lambda=np$. In our case, $\lambda=0.5$.

The probability there are $2$ or fewer defective is therefore approximately $$e^{-\lambda}+e^{-\lambda}\frac{\lambda}{1!}+e^{-\lambda}\frac{\lambda^2}{2!}.\tag{1}$$

The probability that a packet has $3$ or more bad seeds is therefore has to be replaced is $1$ minus the number obtained in $(1)$. I get $0.0143877$, but my arithmetic is unreliable.

For the second, I would suggest using a normal approximation. We have an event (packet has to be replaced) that has probability $p\approx 0.0143877$. We want the probability that in $4000$ trials, the number that has to be replaced is $71$ or more. The number of packets that has to be replaced has approximately normal distribution, with mean $4000p$, and variance $4000p(1-p)$. Calculate the mean and standard deviation. Now you have a routine normal distribution calculation.

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\begin{align} m&=.99*50\\ p(x)&=\frac{e^{-m}m^x}{x!}\\ 1-R_p&=p(47)+p(48)+p(49)+p(50)\\ \end{align} $R_p$ is the required probability .

Can you continue the same line of thought for the second part?

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