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stas 2 mei poisson distribution
stuck on part of a question,
1, Weak spots occur at random in the manufacture of a certain cable at an average rate of 1 per 100 metres. If x represents the number of weak spots in 100 metres of cable, write down the distribution of x.
x ~ B(100, 0.01)
x ~ P(1)
Lengths of this cable are wound on to drums. Each drum carries 50 metres of cable. Find the probability that a drum will have 3 or more weak spots.
λ = 0.5
P(x>3) = 1 - P(x<2)
= 0.0144 (from tables)
A contractor buys 5 such drums. Find the probability that two have just one weak spot each and the other three have none.
This is the part im stuck on, if anyone could help and write out what to do sort of like i have above that would be great
1, Weak spots occur at random in the manufacture of a certain cable at an average rate of 1 per 100 metres. If x represents the number of weak spots in 100 metres of cable, write down the distribution of x.
x ~ B(100, 0.01)
x ~ P(1)
Lengths of this cable are wound on to drums. Each drum carries 50 metres of cable. Find the probability that a drum will have 3 or more weak spots.
λ = 0.5
P(x>3) = 1 - P(x<2)
= 0.0144 (from tables)
A contractor buys 5 such drums. Find the probability that two have just one weak spot each and the other three have none.
This is the part im stuck on, if anyone could help and write out what to do sort of like i have above that would be great
A similar question to above, which i am equally stuck on!
A soiologist claims that only 3% of all suitably qualified students from inner city shool go to university. Use his Vlaim and the Poisson approixmation to the binomial distribution to estimate the probability that in a randomly chosen group of 200 such students
i) exactly 5 go to university
ii) more than 5 go to university
^ these I'm okay with, its the next part..
iii) Another group of 100 students is also chosen. Find the probability that exactly 5 of each group go to university.
A soiologist claims that only 3% of all suitably qualified students from inner city shool go to university. Use his Vlaim and the Poisson approixmation to the binomial distribution to estimate the probability that in a randomly chosen group of 200 such students
i) exactly 5 go to university
ii) more than 5 go to university
^ these I'm okay with, its the next part..
iii) Another group of 100 students is also chosen. Find the probability that exactly 5 of each group go to university.
dreamer86
A contractor buys 5 such drums. Find the probability that two have just one weak spot each and the other three have none.
There are 5C2 = 10 ways in which that can happen:
11000 (first two drums have one weak spot each, last three have none)
10100
10010
10001
01100
01010
01001
00110
00101
00011
Each of these 10 possibilities has probability
[e^(-0.5)*0.5]^2 [e^(-0.5)]^3 = 0.25e^(-2.5)
So the answer is 10*0.25e^(-2.5) = 2.5e^(-2.5) = 0.205.
iii) Another group of 100 students is also chosen. Find the probability that exactly 5 of each group go to university.
P(first group has exactly 5 who go to uni)*P(second group has exactly 5 who go to uni)
= [e^(-6) 6^5/5!]*[e^(-3) 3^5/5!]
= 0.0162
Thankyou, you helped me yesterday!
with the second one i realised that you have to multiply the two together, as they are not the same because different number of students if that makes sense, but im okay on that now
with the first question, I understand the 5C2 part and that
[e^(-0.5)*0.5]^2 [e^(-0.5)]^3 = 0.25e^(-2.5)
the first bracket is ^2 because 2 drums have a weak spot and the second bracket is ^3 as 3 drums dont, and there are 5 altogether, but how do you know what goes in the actual brackets?
im sorry to be a pain, you have helped me loads though
with the second one i realised that you have to multiply the two together, as they are not the same because different number of students if that makes sense, but im okay on that now
with the first question, I understand the 5C2 part and that
[e^(-0.5)*0.5]^2 [e^(-0.5)]^3 = 0.25e^(-2.5)
the first bracket is ^2 because 2 drums have a weak spot and the second bracket is ^3 as 3 drums dont, and there are 5 altogether, but how do you know what goes in the actual brackets?
im sorry to be a pain, you have helped me loads though
dreamer86
with the second one i realised that you have to multiply the two together, as they are not the same because different number of students if that makes sense, but im okay on that now
Sorry, I misread the question: I thought the groups were of equal size.
The answer is
P(first group has exactly 5 who go to uni)*P(second group has exactly 5 who go to uni)
= [e^(-6) 6^5/5!]*[e^(-3) 3^5/5!]
= 0.0162
with the first question, I understand the 5C2 part and that
[e^(-0.5)*0.5]^2 [e^(-0.5)]^3 = 0.25e^(-2.5)
the first bracket is ^2 because 2 drums have a weak spot and the second bracket is ^3 as 3 drums dont, and there are 5 altogether, but how do you know what goes in the actual brackets?
[e^(-0.5)*0.5]^2 [e^(-0.5)]^3 = 0.25e^(-2.5)
the first bracket is ^2 because 2 drums have a weak spot and the second bracket is ^3 as 3 drums dont, and there are 5 altogether, but how do you know what goes in the actual brackets?
The number of weak spots on a drum has the Poisson(0.5) distribution.
P(a drum has exactly one weak spot) = e^(-0.5)*0.5
P(a drum has no weak spots) = e^(-0.5)
Jonny W
Sorry, I misread the question: I thought the groups were of equal size.
The answer is
P(first group has exactly 5 who go to uni)*P(second group has exactly 5 who go to uni)
= [e^(-6) 6^5/5!]*[e^(-3) 3^5/5!]
= 0.0162
The number of weak spots on a drum has the Poisson(0.5) distribution.
P(a drum has exactly one weak spot) = e^(-0.5)*0.5
P(a drum has no weak spots) = e^(-0.5)
The answer is
P(first group has exactly 5 who go to uni)*P(second group has exactly 5 who go to uni)
= [e^(-6) 6^5/5!]*[e^(-3) 3^5/5!]
= 0.0162
The number of weak spots on a drum has the Poisson(0.5) distribution.
P(a drum has exactly one weak spot) = e^(-0.5)*0.5
P(a drum has no weak spots) = e^(-0.5)
i totally understand now, thankyou very much
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