Poisson Distribution [急]

用戶25071

2010-09-28 12:24 pm

有冇可以幫我呀..?
1. The number of hits on a website is recorded.Over a long period of time the
mean number if hits per hour is 6.9.

i) explain briefly why the number of hits per unit time could be modelled by a
Poission distribution,

ii) Assuming a Poission model, calculate the probability that the websites has
(A) exactly 6 hits in an hour,
(B) exactly 3 hits in each of two successive 30-minute intervals

2. A manufacturer produces computer monitor screens in which the picture is
composed of 780 000 pixels. On average 1 in 500 000 of these pixels is faulty.
Let X represent the number of faulty pixels in one monitor screen.

i) State the condition required for X to be binomially distributed.

ii) Hence find the probability that a randomly selected monitor screen has
(A) exactly one faulty pixel,
(B) at least two faulty pixels.

iii) A retailer orders a batch of 5 monitor screens from the manufacturer. (The batch may be regarded as a random sample)

(A) Find the probability that exactly one of the five monitor screens has at
least two faulty pixels.

(B) Find the probability that there are at most 10 faulty pixels in total in the
batch,

iv) The manufacturer wishes to improve quality so that 90% of the monitor screens have no faulty pixels at all. To what value must the manufacturer reduce the probability of a piexl being faulty in order to achieve this?

回答 (1)

六呎將軍

2010-09-28 6:50 pm

✔ 最佳答案

1 i) Since the record is made over a long period of time, the average no. of hits over a period of time is almost constant and hence the case can be regarded as Poisson distribution.

ii A) With λ = 6.9 for a 1-hour period:

P(6 hits per hour) = 6.96 x e-6.9/6! = 0.1511

ii B) With λ = 3.45 for a 0.5-hour period:

P(3 hits per 1/2 hour) = 3.453 x e-3.45/3! = 0.2173

So to record exactly 3 hits in each of two successive 30-minute intervals, probability = 0.2173 x 0.2173 = 0.0472

2 i) 4 conditions are required for X to be binomially distributed:

1) There are n trials, where n is an integer.
2) Each trial results in a success or a failure.
3) The probability of a success, p, is constant from trial to trial.
4) The trials are independent.

ii A) Since the no. of trials is very large, this can be the limiting case of binomial distribution, i.e. Poisson distribution with λ = 1.56

P(1 faulty) = 1.561 x e-1.56/1! = 0.3278

B) P(2 faulty) = 1.562 x e-1.56/2! = 0.2557

iii A) 5C1 x 0.2557 x (1 - 0.2557)4 = 0.3924

B) For taking a random sample of 5 monitors, we have the distribution of faulty pixels being Poisson with λ = 5 x 1.56 = 7.8

Hence the required probability is given by:

7.80 x e-7.8/0! + 7.81 x e-7.8/1! + 7.82 x e-7.8/2! + ... + 7.810 x e-7.8/10! = 0.8352

iv) So we have to find out the new λ so that:

λ0 x e-λ/0! >= 0.9

e-λ >= 0.9

λ <= 0.1054

So at most only 1 in 780000/0.1054 = 7403153 of faulty pixels is allowed.

2010-09-29 09:23:50 補充：
e^(-λ) >= 0.9

Taking natural log on both sides:

-λ >= ln 0.9

λ <= - ln 0.9 = 0.1054

參考: Myself

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