Here is the question:
A shopkeeper hires vacuum cleaners to the public at £5 per day. The mean daily demand is 2.6. If only 3 vacuum cleaners are available for hire calculate the mean of the daily income.
The answers say £10.70 however I can't seem to get to this value.
My thoughts:
Since there are only 4 possible values left (0, 1, 2, 3) I would take the probability of each occuring using regular poisson distribution with a mean of 2.6 and multiply each of those possibilities by the value that created them, then add this together to get what I believed would be the new average demand, however it does not give us the expected demand of 2.14 ($\frac{10.7}{2.6}$) instead returning 1.34
e.g. $P(X=0)\times0+P(X=1)\times1+P(X=2)\times2+P(X=3)\times3$
My reasoning for the above is because if you dont stop and continue with it approaching $P(X=\infty)*\infty$ it approaches 2.6 our current demand.
I realised this is probably due to the fact that the new data set actually follows a new average that is not 2.6, but I don't know how to take on this question. Even taking the cumulative probabilities of 0, 1, 2, and 3 and multiplying by 2.6 returns an incorrect value.
So how do we get the answer of £10.70?
Let the daily demand $X$ be a Poisson random variable with mean $\lambda = 2.6$. Then the daily income random variable is $Y = 5(X \wedge 3) = 5\min(X,3)$. We want to find $\operatorname{E}[Y]$, the expected value of the daily income, so $$\begin{align*} \operatorname{E}[Y] &= \operatorname{E}[5 \min(X,3)] = 5 \sum_{x=0}^\infty \min(x,3) \Pr[X = x] \\ &= 5 \Biggl(\sum_{x=0}^2 (x-3) \Pr[X = x] + \sum_{x=0}^\infty 3 \Pr[X = x] \Biggr) \\ &= 5(-3\Pr[X = 0] - 2\Pr[X = 1] - \Pr[X = 2] + 3) \\ &= 15 - 5e^{-\lambda}\left(3 + 2\lambda + \frac{\lambda^2}{2}\right) \\ &\approx 10.6996.\end{align*}$$ The explanation is as follows: if $X \ge 3$, then there can be at most a daily income of $15$, since the demand exceeds the supply of $3$ vacuum cleaners. So another way to write this is $$\begin{align*} \operatorname{E}[Y] &= 5(0 \Pr[X = 0] + 1 \Pr[X = 1] + 2 \Pr[X = 2] + 3 \Pr[X \ge 3]) \\ &= 5(\Pr[X = 1] + 2\Pr[X = 2] + 3(1 - \Pr[X \le 2])) \\ &= 5(3 - 3\Pr[X = 0] - 2 \Pr[X = 1] - \Pr[X = 2]), \end{align*}$$ which is exactly the same expression as above.