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{\bf Question}

Sudden dips in a supply of electrical current in occur in
accordance with the assumptions of a Poisson process with a rate
of 3 dips per hour.  What is the probability that the first dip in
the afternoon occurs later than 1.00 p.m.?

If a device fails when it has experienced the cumulative effect of
27 dips in the current, what is the probability density function
of the life of the device? Find approximately the probability that
such a device will last for longer than 12 hours.



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{\bf Answer}

Dips $\sim$ Poisson(3)

$P$(1st afternoon dip occurs after 1p.m.) $= P(W_1 &gt; 1 {\rm hour})
= e^{-3} = 0.0498$ (=$P$(no events))

The life of a device = waiting time until 27th dip, and so has
p.d.f.

$\ds \frac{3e^{-3t}(3t)^{26}}{26!} \hspace{.3in} t \geq 0$ $\ \ \
(\Gamma(3,27))$ 

Now $\ds \mu = \frac{27}{3} = 9 \, {\rm and} \, \sigma^2 =
\frac{27}{9} = 3$

So the lifetime $\sim N(9,3)$

$\ds P(L&gt;12) = P\left(Z &gt; \frac{12-9}{\sqrt 3} \right) = P(Z&gt;\sqrt
3) \approx P(Z &gt; 1.73) \approx 0.4182$ from tables.




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