Statistic probability

A travel agency is selling some first-class air-tickets. Let Y denote the number of first-class air-tickets sold per week. According to the past experience

the probability disrtibution of Y is as follows:Y 0 1 2 3 4P(Y) 1/15 2/15 7/15 4/15 1/15Each first-class air-ticket costs $15

000. The travel agency makes a profit of $5

000 for each ticket sold and the airline agrees to buy back all the unsold tickets from the travel agency with price S for each unsold ticket

whereS = $6000 x (Y 1)^(1/2)In order to maximize the weekly profit

the travel agency buys n first-class air-tickets from the airline.(a) Find n

the number of first-class air-tickets that the travel agency buys from the airline.(b)Find the mean and the standard diviation of the number of tickets that the airline has to buy back if the weekly profit of the travel agency is maximized.第一次接觸以概率去找最大值的題目 (找最大值不是要用微分嗎?)腦內一片空白望各位指教
(a) when n=0

the expected profit is 0when n=1

the expected profit is (14/15)(5000) (1/15)(6000-15000)=4066.67when n=2

the expected profit is (12/15)(2)(5000) (2/15)(6000√2-15000) (1/15)(2)(6000-15000)=5931.37when n=3

the expected profit is (5/15)(3)(5000) (7/15)(6000√3-15000) (2/15)(2)(6000√2-15000) (1/15)(3)(6000-15000)=-687.52when n=4

the expected profit is (1/15)(4)(5000) (4/15)(6000√4-15000) (7/15)(2)(6000√3-15000) (2/15)(3)(6000√2-15000) (1/15)(4)(6000-15000)=-9080.25So

the travel agency should buy 2 first-class air-tickets in order to maximize her weekly profit.(b) For n=2

Let X be the number of tickets that the airline has to buy back Then X 0 1 2P(X) 12/15 2/15 1/15So E(X)=4/15E(X^2)=2/15 4/15=6/15=2/5Var(X)=2/5-4/15=2/15 and standard deviation=0.3651

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3651
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