Question 8.5: An advertising board is illuminated by several hundred bulbs......

(a) It seems appropriate to model the number of fused bulbs with a Poisson distribution. We assume, however, that the probabilities of fused bulbs on two consecutive days are independent of each other; i.e. they only depend on λ but not on the time t.

(b) The arithmetic mean is

\bar{x} =\frac{1}{30} \left(0 + 1 · 8 + 2 · 8+· · ·+5 · 1\right)=\frac{52}{30}=1.7333

which means that, on an average, 1.73 bulbs are fused per day. The variance is

s^{2}= \frac{1}{30} \left(0 + 1^{2} · 8 + 2^{2}· 8+· · ·+5^{2} · 1\right)− 1.7333^{2}

= \frac{142}{30} − 3.0044 = 1.72889.

We see that mean and variance are similar, which is an indication that the choice of a Poisson distribution is appropriate since we assume E(X) = λ and Var(X) = λ.

(c) The following table lists the proportions (i.e. relative frequencies f_{j} ) together with the probabilities P(X = x) from a Po(1.73)-distribution. As a reference, we also list the probabilities from a Po(2)-distribution since it is not practically possible that 1.73 bulbs stop working and it may hence be an option to round the mean.

One can see that observed proportions and expected probabilities are close together which indicates again that the choice of a Poisson distribution was appropriate. Chapter 9 gives more details on how to estimate parameters, such as λ, from data if it is unknown.

(d) Using λ = 1.73, we calculate

P(X > 5) = 1 − P(X ≤ 5) = 1 − \sum\limits_{i=0}^{5}{\frac{\lambda ^{i}}{i!} exp\left(-\lambda \right) }

=1-exp\left(−1.73\right)\left(\frac{1.73^{0}}{0!} +\frac{1.73^{1}}{1!}+ \ldots +\frac{1.73^{5}}{5!}\right)

= 1 − 0.99 = 0.01.

Thus, the bulbs are replaced on only 1 % of the days.

(e) If X follows a Poisson distribution then, given Theorem 8.2.1, Y follows an exponential distribution with λ = 1.73.

(f) The expectation of an exponentially distributed variable is

E(Y) = \frac{1}{\lambda}=\frac{1}{1.73}=0.578.

This means that, on average, it takes more than half a day until one of the bulbs gets fused.