更新1:
(b) Find the probability of having at most 15 days of bad transmission in the next 100 days.
probability?
2015-12-02 3:22 pm
Failure in the wifi service occurs at a mean rate of 3 instances per day, and the transmission on a day is consider “bad” if there are at least 5 failures that day.Find the probability of having exactly 2 days of bad transmission in the coming 5 days.
回答 (2)
2015-12-02 6:19 pm
✔ 最佳答案
(a)Suppose X failures per day
P(x) ≡ P( X = x ) = 3^x * e^(-3) / x !
P(bad)
= P( X ≧ 5 )
= 1 - [ P(0) + P(1) + P(2) + P(3) + P(4) ]
= 1 - e^(-3)*[ 3^0/0! + 3^1/1! + 3^2/2! + 3^3/3! + 3^4/4! ]
= 1 - 16.375*e^(-3)
≒ 0.185
C(5,2)*0.185^2*( 1 - 0.185 )^3 ≒ 0.185 ..... Ans
(b)
p = 0.185
q = 1 - p = 0.815
Using normal approximation to binomial probability
μ = np = 100*0.185 = 18.5
σ = √( npq ) = √( 18.5*0.815 ) ≒ 3.88
P( D ≧ 15 )
= P( Z > (15-18.5)/3.88 ) , where Z ~ N(0,1)
≒ P( Z > 0.90 )
= 1 - P( Z < 0.90 )
= 1 - 0.8159
= 0.1841 ..... Ans
2015-12-09 5:49 am
mean=3
(a)
P(X>5 or X=5)
=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)
=1-(3^0)(e^-3)/0!-(3^1)(e^-3)/1!-(3^2)(e^-3)/2!-(3^3)(e^-3)/3!-(3^4)(e^-3)/4!
=0.184736755
P(X=2)
=(5C2)(1-0.184736755)^(5-2) (0.184736755)^2
=0.184926952
(b)
P(X<15 or =15)
=100C0 (1-0.184736755)^100 (0.184736755)+100C1 (1-0.184736755)^99 (0.184736755)+100C2 (1-0.184736755)^98 (0.184736755)^2+...+100C15 (1-0.184736755)^85 (0.184736755)^15
=0.225172728
(a)
P(X>5 or X=5)
=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)
=1-(3^0)(e^-3)/0!-(3^1)(e^-3)/1!-(3^2)(e^-3)/2!-(3^3)(e^-3)/3!-(3^4)(e^-3)/4!
=0.184736755
P(X=2)
=(5C2)(1-0.184736755)^(5-2) (0.184736755)^2
=0.184926952
(b)
P(X<15 or =15)
=100C0 (1-0.184736755)^100 (0.184736755)+100C1 (1-0.184736755)^99 (0.184736755)+100C2 (1-0.184736755)^98 (0.184736755)^2+...+100C15 (1-0.184736755)^85 (0.184736755)^15
=0.225172728
收錄日期: 2021-04-18 14:08:06
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151202072234AAwfOEi