A film supplier produces 10 rolls of a specifically sensitized film each year. If the film is not sold within a year it must be discarded. Past experience indicates that D, the small demand for the film, is a Poisson-distributed random variable with parameter 8. If a profit of 7 dollars is made on every roll which is sold, while a loss of 2 dollars is incurred on every roll which must be discarded, compute the expected profit which the supplier may realize on the 10 rolls which he produces.
Well, if the profit random variable, P, is a function of D, the variable representing the demand for films, we have:
$P = 7D + 3 (10-D) = 10 D-30$
So the expected value should be
$E[P] = E[10D - 30] = 10E[D] - 30 = 10\times8 - 30 = 50 $
Since 8 is the Poisson parameter, and the expected value of a Poisson distribution is it's parameter.
But apparently, my approach to the problem is wrong. The answer is 45,74. What am I forgetting?
Edit: There's a mistake I did while copying the question from the book. The loss should be of 3 dollars, not 2. Also, It should be $P = 7D - 3(10-D)$, but the result is still in accordance with my reasoning.
There is some confusion in what you write. Even if you ignore the sign error in your computation, you appear to switch the loss of $2$ per unsold roll to a loss of $3$.
But the big error you make is that you ignore the fact that the fellow can't sell more than $10$ even if there is demand for more, as he's only got $10$ to sell. Thus the answer is $$E[P]=\sum_{i=0}^9 P(D=i)\times (7i-3(10-i))+P(D≥10)\times 70$$
(trusting that you meant a loss of $3$ per unsold roll...that's what you need to get to the official answer)
And the rest follows by direct computation.