1.
A fair coin is tossed 12 times. Let X be the number of heads obtained in 12 tosses. Then X ~ Bin(12, 1/2).
Find the probability P(x) of obtaining 3 heads.
(Correct your answer to 5 decimal places.)
2.
Suppose that in a certain company, the probability of any member of staff taking a sick leave is 0.07.
If there are 13 clerks in the company, find the probability that exactly 4 of them take a sick leave.
(Correct your answer to 5 decimal places.)
3.
The probability distribution of X, the number of telephone calls received per hour in a certain house, can be approximated by a Poisson probability distribution, Po(1.23).
Find the probability that, in an hour, 4 is received.
(Correct your answer to 5 decimal places.)
Binomial and Poisson Distribut
2009-05-18 7:32 pm
回答 (4)
2009-05-22 6:37 am
✔ 最佳答案
1:X ~ Binomial (n = 12, p = 1 / 2)
pX(x) = nCx * px * (1 - p)n - x
= 12Cx / 212
pX(x = 3) = 12C3 / 212
= 5.37109%
2:
X ~ Binomial (n = 13, p = 0.07)
pX(x) = nCx * px * (1 - p)n - x
= 13Cx * 0.07x * 0.9313 - x
pX(x = 4) = 13C4 * 0.074 * 0.939
= 0.89340%
3:
X ~ Poisson (p = 1.23)
pX(x) = px * e-p / x!
= 1.23x * e-1.23 / x!
pX(x = 4) = 1.234 * e-1.23 / 4!
= 2.78758%
2009-05-21 22:39:30 補充：
manyu_yan is wrong for question 1
2009-05-20 7:26 pm
1. Pr(X=3) = 12C3 * 0.5^3 * (1-0.5)^(12-3)
=0.05371
2. Pr(X=4) = 13C4*0.07^4*(1-0.07)^(13-4)
=0.00893
3. Pr(X=4) = [exp(-1.23)*1.23^4] / 4!
= 0.02788
=0.05371
2. Pr(X=4) = 13C4*0.07^4*(1-0.07)^(13-4)
=0.00893
3. Pr(X=4) = [exp(-1.23)*1.23^4] / 4!
= 0.02788
2009-05-18 9:53 pm
the answer for problem 1 should be (12C3)(0.5)^3(0.5)^9
2009-05-18 8:06 pm
1. (12C3)(0.5)^3(0.5)^8=0.10742
2. (13C4)(0.07)^4(1-0.07)^9=0.00893
3. (e^(-1.23) x 1.23^4) /4! = 0.02788
2. (13C4)(0.07)^4(1-0.07)^9=0.00893
3. (e^(-1.23) x 1.23^4) /4! = 0.02788
收錄日期: 2021-04-23 18:19:10
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090518000051KK00477