Probability 28

👨🏻‍🏫 學術台數學

還不明白的薑

2009-07-14 7:51 pm

The number X of customers per day requiring a copy of a particular newspaper at a newsagent's shop is a discrete random variable having distribution

P(X = k) = pq^(k-30) , k = 30,31,32,...

where p = 0.05 , q = 1-p .

(a) Assuming that on each day the newsagent has an unlimited supply of copies of the newspaper, calculate the mean number of copies sold per day.

(b) If the newsagent has only N copies of the newspaper available for sale each day, find the smallest value of N for there to be a probability of at least 0.7 that the daily demand will be met.

(c) The newsagent makes a profit of $1.2 on every copy of the newspaper thath he sells, and loss $1.8 on every copy left unsold at the end of the day. If the newsagent has N copies available for sale each day and the number of customers requiring the paper is x, write down expressions for the daily profit in the cases when x<N and x>=N, respectively. By considering the additional daily profit in each of these cases if the newspaper had (N+1) copies available each day, find the value of N which will maximise the newsagent's mean daily profit from the sale of this newspaper.

更新1:

Answers (a) 49 (b) 53 (c) 40

回答 (2)

用戶9112

2009-07-15 7:52 am

✔ 最佳答案

(a)
Pre-calculation:
Σ(r=0 to ∞) xr = 1 / (1-x), for |x|<1
Differetiate wrt x,
Σ(r=0 to ∞) rxr-1 = 1 / (1-x)2, for |x|<1
Multiply by x each side,
Σ(r=0 to ∞) rxr = x / (1-x)2, for |x|<1
Sub x = q
Σ(r=0 to ∞) rqr = q / p2
The mean number of copies sold per day = Σ(k=30 to ∞)kP(k)
= Σ(k=30 to ∞)kpqk-30
Let r = k-30; k = r+30
Σ(k=30 to ∞)kpqk-30
=Σ(r=0 to ∞)(30+r)pqr
=30Σ(r=0 to ∞)pqr + pΣ(r=0 to ∞)rqr
=30 + q/p
=49
(b)Σ(k=30 to N)pqk-30 >= 0.7
(p + pq + pq2 +...+ pqN-30)>=0.7
p(1 - qN-29)(1 - q) >= 0.7
0.3 > qN-29
log(0.3) > (N-29)log(0.95)
N-29 > 23.47
N > 52.47
N must be at least 53.

(c) When x<N, Profit = 1.2x - 1.8(N-x) = 3x - 1.8N
When x>=N, Profit = 1.2N
Mean profit for N copies P(N)
= Σ(k=30 to N-1)pqk-30(3k-1.8N) + Σ(k=N to ∞)pqk-30(1.2N)
= 3Σ(k=30 to N-1)kpqk-30 - 1.8NΣ(k=30 to N-1)pqk-30 + 1.2N[1 - Σ(k=30 to N-1)pqk-30]
= 1.2N - 3NΣ(k=30 to N-1)pqk-30 + 3Σ(k=30 to N-1)kpqk-30
P(N+1) = 1.2(N+1) - 3(N+1)Σ(k=30 to N)pqk-30 + 3Σ(k=30 to N)kpqk-30
= 1.2N + 1.2 - 3(N+1)Σ(k=30 to N-1)pqk-30 - 3(N+1)pqN-30 + 3Σ(k=30 to N-1)kpqk-30 + 3NpqN-30
Incremental profit if increase from N to N+1 copies
P(N+1) - P(N) = 1.2 - 3Σ(k=30 to N-1)pqk-30 - 3(N+1)pqN-30 + 3NpqN-30
= 1.2 - 3Σ(k=30 to N-1)pqk-30 - 3pqN-30
= 1.2 - 3(p + pq + pq2 +... + pqN-30)
= 1.2 - 3p(1 - qN-29)/(1-q)
= 1.2 - 3(1 - qN-29)
When N=39, P(N+1) - P(N) is roughly zero, so the profit reaches a maximum at around 39-40

👍 0 👎 0

六呎將軍

2009-07-14 10:11 pm

Any ans for reference ?

👍 0 👎 0

收錄日期: 2021-04-23 23:18:49
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090714000051KK00594

檢視 Wayback Machine 備份