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S2 - Binomial and Poisson Distribtuions
Hi,
I'm a bit stuck on part d of this question, as I don't know what to do, and so would really appreciate someone explaining how to do it.
The number, X, of breakdowns per week of the lifts in a large block of flats has Poisson distribution with a mean 0.25. Find, to 3dp, the probability that on an particular week:
a) there will be at least one breakdown : 0.221
b) there will be at most two breakdowns: 0.998
c) Show that, to 3dp, the probability that during a 12 week period there will be no lift breakdowns is 0.050.
The residents in the flats have a maintenance contract with Liftserve. The contract is for 20 blocks of 12 weeks. For every block of 12 weeks with no breakdowns the residents pay Liftserve £500. If there is atleast 1 breakdown in a 12 week block then Liftserve will mend the lift free of charge and the residents pay nothing for the block of 12 weeks.
d) Find the probability that over the period of the contract the residents pay no more than £1000.
Answer: 0.925
Also, for the following question, would you use a Binomial or a Poisson distribution and why? (as I'm not sure which one to use)
Stormville is hit by a hurricane on averagee once every 30 days during the hurricane season. Find the probability that a) in a 30 day period it is hit at least twice, b) in a 60 day period it is hit at least 4 times, c) in three consecutive 20-day periods it is hurricane-free in just two of them.
Thank you very much!
I'm a bit stuck on part d of this question, as I don't know what to do, and so would really appreciate someone explaining how to do it.
The number, X, of breakdowns per week of the lifts in a large block of flats has Poisson distribution with a mean 0.25. Find, to 3dp, the probability that on an particular week:
a) there will be at least one breakdown : 0.221
b) there will be at most two breakdowns: 0.998
c) Show that, to 3dp, the probability that during a 12 week period there will be no lift breakdowns is 0.050.
The residents in the flats have a maintenance contract with Liftserve. The contract is for 20 blocks of 12 weeks. For every block of 12 weeks with no breakdowns the residents pay Liftserve £500. If there is atleast 1 breakdown in a 12 week block then Liftserve will mend the lift free of charge and the residents pay nothing for the block of 12 weeks.
d) Find the probability that over the period of the contract the residents pay no more than £1000.
Answer: 0.925
Also, for the following question, would you use a Binomial or a Poisson distribution and why? (as I'm not sure which one to use)
Stormville is hit by a hurricane on averagee once every 30 days during the hurricane season. Find the probability that a) in a 30 day period it is hit at least twice, b) in a 60 day period it is hit at least 4 times, c) in three consecutive 20-day periods it is hurricane-free in just two of them.
Thank you very much!
d) That is the same as there being breakdowns in at most 2 blocks of 12 weeks.
The probability of there being at least one breakdown is 0.221. You want the sum of P(no breakdowns) + P(breakdowns in one block of 12 weeks only) + P(breakdowns in 2 blocks of 12 weeks only).
I'd use Bin(20, 0.221) and calculate it using nCr.
For the second bit:
Use a poisson distribution if it gives you an average rate of occurence, but binomial if you have to work out combinations of True/False occurences. Basically, if you know the probability of an outcome of a specific event, use binomial.
The probability of there being at least one breakdown is 0.221. You want the sum of P(no breakdowns) + P(breakdowns in one block of 12 weeks only) + P(breakdowns in 2 blocks of 12 weeks only).
I'd use Bin(20, 0.221) and calculate it using nCr.
For the second bit:
Use a poisson distribution if it gives you an average rate of occurence, but binomial if you have to work out combinations of True/False occurences. Basically, if you know the probability of an outcome of a specific event, use binomial.
Reply 2
So would the second bit be a binomial question, because you've got a fixed number of trials (days)?
Reply 4
For the first question I did
x~B(20,0.2211...)
so
P(X<=2)=P(x=0)+P(x=2)+P(x=2)
=(20C0)(0.2211...)^0(0.7788...)^20 + (20C1)(0.2211...)^1(0.778...)^19 + (20C2)(0.2211...)^2(0.7788...)^18
=0.00673794716 + 0.038274964 + 0.103275096
= 0.148288007
But the answer in the book is 0.925? Where am I going wrong?
x~B(20,0.2211...)
so
P(X<=2)=P(x=0)+P(x=2)+P(x=2)
=(20C0)(0.2211...)^0(0.7788...)^20 + (20C1)(0.2211...)^1(0.778...)^19 + (20C2)(0.2211...)^2(0.7788...)^18
=0.00673794716 + 0.038274964 + 0.103275096
= 0.148288007
But the answer in the book is 0.925? Where am I going wrong?
First, I've just noticed a mistake. The Binomial distribution I should have told you to use was Bin(20,0.050) or whatever the full answer to part c) is. Your working is sound though.
Second, the answer 0.925 seems very high as a probability that a 95% certain event won't happen at least 10% of the time. Are you sure the figures are correct?
Second, the answer 0.925 seems very high as a probability that a 95% certain event won't happen at least 10% of the time. Are you sure the figures are correct?
Reply 6
Using Bin(20,0.050) gives me very small numbers on my calculator. Does it seem to work for you?
I checked the book again and it is 0.925. I don't think most of these questions are very realistic, but maybe they've made an error when writing the answers?
I checked the book again and it is 0.925. I don't think most of these questions are very realistic, but maybe they've made an error when writing the answers?
Reply 8
Oh right. Thanks very much for helping me with the question!
Original post by Trangulor
Nope, I get a very tiny answer. That should be expected though, the probability of an almost certain outcome appearing only one in ten trials should be very low, definitely not 0.925. Its very possible that there has been an error in printing, be it the question or the answer
Please take no offence. There is no misprint.
You were right about it being X~B(20, 0.05) but the you may have put into your calculator (20C2)(0.05^2)(0.95^18). This give you a small answer, but it's only working out P(X=2) not P(X<=2).
For that, you can use the statistical table as n=20 p=0.05 is in there, giving an answer 0.9245.
Alternatively, you can find out P(X=0) +P(X=1)+P(X=2).
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