For a certain section of pine forest, the number of diseased trees per acre, Y, has a poisson distribution with mean of 10. The diseased trees are sprayed with an insecticide at a cost 3 dollars per tree, plus a fixed overhead cost for equipment rental of $50. Letting C denote the total spraying cost for a randomly selected acre, find the expected value and standard deviation for C. Within what interval would you expect C to lie with probability at least .75?
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1$\begingroup$ I downvoted because research must be done to ask a good question. Please see: Why is “Can someone help me?” not an actual question? $\endgroup$– EJoshuaS - Stand with UkraineCommented Mar 6, 2018 at 22:01
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$\begingroup$ Also downvoted because you do not show what you have tried or where you are stuck. But I have given you some clues how to start. This is not a site only for 'experts', but it is also not a homework-answer site. $\endgroup$– BruceETCommented Mar 6, 2018 at 23:28
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1 Answer
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You show no work of your own and reveal no context, so I'm flying almost blind as to what kind of help you need. Perhaps this is a start:
You have $C = 3Y + 50,$ where $Y \sim \mathsf{Pois}(\lambda = 10)$, so that $E(Y) = 10$ and $Var(Y) = 10$.
Then, $\mu_C = E(C) = E(3Y + 50) = 3E(Y) + 50 = 80$ dollars.
Also $Var(C) = Var(3Y + 50) = 9Var(Y).$ Then find $\sigma_c = SD(C) = \sqrt{Var(C)}$ dollars.
[You should be able to find relationships for E and Var such as these in your textbook. Please try to find them. Can you give reasons for each of my steps above?]
Assuming that $C$ is nearly normal, about 95% of the probability in the distribution of $C$ lies in the interval $\mu_C \pm 1.96\sigma_C,$ where the number 1.96 is derived (possibly from printed tables or software) using the CDF of the standard normal distribution (2.5% of the area is cut from upper and lower 'tails' to leave 95% in the 'middle').
[What does your textbook say about the normal approximation to Poisson probabilities? Even if the approximation is poor, guesses based on it may help you do do exact computations. Have you ever heard of the Empirical Rule (which is based on a normal approximation)?]
It seem you need to cut 12.5% of the probability from each tail of that distribution to Answer your question. (Maybe something like 1.1 or 1.2 instead of 1.96; I'll leave that part to you.) [Do you know how to use printed normal tables. Can you figure out where 1.96 comes from?]