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QUESTION

An electrical shop replenishes its supply of a specialized
   type of bulb at the beginning of each week. The demand in a
   week for this type of bulb has a Poisson distribution with mean
   3.

   Find the smallest number of bulbs which should be in stock at
   the beginning of the week to be at least 90\% confident of
   satisfying the demand in a week. If the number in stock at the
   beginning of the week is set at that number find the expected
   number of satisfied demands in the week.

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 ANSWER

Demand $\sim P(3)$. We can use either $\displaystyle
p(x)=\frac{\mathrm{e}^{-3}3^x}{x!}$ or tables to find $F(x)$.

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\begin{center}
\begin{tabular}{lr}
$p(0)=.050$&\hspace{.5in}$F(0)=.050$\\ $p(1)=.149$&$F(1)=.199$\\
$p(2)=.224$&$F(2)=.423$\\ $p(3)=.224$&$F(3)=.647$\\
$P(4)=.168$&$F(4)=.815$\\ &$F(5)=.916$
\end{tabular}
\end{center}
(Note that there are two modes, 2 and 3.)

Hence the least number to be at least 90\% certain is 5. So stock
5. If the demand is $0,1,2,3,4$ then the satisfied demand is
$0,1,2,3,4.$ If the demand is $\ge5$ then the satisfied demand is
5.

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$E(\mathrm{satisfied\ demand\
})\\=0p(0)+1p(1)+2p(2)+3p(3)+4p(4)+5(p(5)+p(6)+\ldots)\\
=0p(0)+1p(1)+2p(2)+3p(3)+4p(4)+5(1-p(4))\\
=.149+.448+.672+.672+5\times.185=2.866$



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