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# Poisson distribution question watch

1. A garage uses a spare part at an average rate of 5 per week. Assuming that usage of this part follows a Poisson distribution, find the probability that:

v) exactly 5 are used in each of two successive weeks.

what does it mean by two successive weeks?

the mean is 5
for two weeks, it is 10.

P(x=5) for this
2. (Original post by 0utdoorz)
A garage uses a spare part at an average rate of 5 per week. Assuming that usage of this part follows a Poisson distribution, find the probability that:

v) exactly 5 are used in each of two successive weeks.

what does it mean by two successive weeks?

the mean is 5
for two weeks, it is 10.

P(x=5) for this
Well in two weeks you would expect an average of 10, so X ~ Po(10) and P(X=5).

By "successive weeks" it means that you assume that the average will remain the same each week.
3. (Original post by As_Dust_Dances_)
Well in two weeks you would expect an average of 10, so X ~ Po(10) and P(X=5).

By "successive weeks" it means that you assume that the average will remain the same each week.
but I'm trying lambda = 10
p(x=5) =P(x is less than and equal to 5) - P(x is less than and equal to 4)

in poisson tables:
p(x=5) = 0.0671
and p(x=4) = 0.0293

0.0671-0.0293 = 0.378

but the answer is meant to be '0.0308'
4. (Original post by 0utdoorz)
but I'm trying lambda = 10
p(x=5) =P(x is less than and equal to 5) - P(x is less than and equal to 4)

in poisson tables:
p(x=5) = 0.0671
and p(x=4) = 0.0293

0.0671-0.0293 = 0.378

but the answer is meant to be '0.0308'
for p(x=5) just use this formula:-

and you should get 0.0378
5. (Original post by As_Dust_Dances_)
for p(x=5) just use this formula:-

and you should get 0.0378

oh thanks! is that for two successive weeks??

also how would I do the next part of the question:
"If stocks are replenished weekly, determine the number of spare parts which should be in stock at the beginning of each week to ensure that on average the stock will be insufficient on no more than one week in a 52-week year"
6. (Original post by 0utdoorz)
but I'm trying lambda = 10
p(x=5) =P(x is less than and equal to 5) - P(x is less than and equal to 4)

in poisson tables:
p(x=5) = 0.0671
and p(x=4) = 0.0293

0.0671-0.0293 = 0.378

but the answer is meant to be '0.0308'
Look I don't know what to say apart from that some people on here are absolute jokers. The probability is indeed 0.0308. Think of this question:

What is the probability of getting two heads in successive tosses:

(1/2) x (1/2) =1/4

now compare:

What is the probability of getting 5 whatevers in two successive weeks?

The mistake in the other's logic is that you can't have 4 in one week and 6 in the next etc. with their mean of 10.
7. (Original post by metaltron)
Look I don't know what to say apart from that some people on here are absolute jokers. The probability is indeed 0.0308. Think of this question:

What is the probability of getting two heads in successive tosses:

(1/2) x (1/2) =1/4

now compare:

What is the probability of getting 5 whatevers in two successive weeks?

The mistake in the other's logic is that you can't have 4 in one week and 6 in the next etc. with their mean of 10.
so do I do x=2.5?
8. (Original post by metaltron)
Look I don't know what to say apart from that some people on here are absolute jokers. The probability is indeed 0.0308. Think of this question:

What is the probability of getting two heads in successive tosses:

(1/2) x (1/2) =1/4

now compare:

What is the probability of getting 5 whatevers in two successive weeks?

The mistake in the other's logic is that you can't have 4 in one week and 6 in the next etc. with their mean of 10.
^ Yeah this is correct, ignore me. I'm thinking of something else. I assumed it meant in a two week period, in which case it would be 0.0378. And ok, people do tend to make mistakes!
9. (Original post by 0utdoorz)
so do I do x=2.5?
Nope put P(X=5) into the formula with lambda = 5

then square your answer and you should get 0.0308
10. (Original post by 0utdoorz)
so do I do x=2.5?
Say each week is independent of the other, which they will be. You want the probability that you will need 5 spare parts in a week. This is:

The probability of needing 5 spare parts in the second week is the same, of course. Now you have to fill in the value of lambda. Remember this is for one week.

Then use a similar reasoning to where you multiply (1/2) by (1/2) to get the probability of getting two consecutive heads.
11. (Original post by As_Dust_Dances_)
^ Yeah this is correct, ignore me. I'm thinking of something else. I assumed it meant in a two week period, in which case it would be 0.0378. And ok, people do tend to make mistakes!
Sorry, I though there was around 3 other people giving the wrong answer. But since it was just you, I can accept that you made a mistake
12. thanks!!

also how would I do the next part of the question:
"If stocks are replenished weekly, determine the number of spare parts which should be in stock at the beginning of each week to ensure that on average the stock will be insufficient on no more than one week in a 52-week year"
13. (Original post by 0utdoorz)
thanks!!

also how would I do the next part of the question:
"If stocks are replenished weekly, determine the number of spare parts which should be in stock at the beginning of each week to ensure that on average the stock will be insufficient on no more than one week in a 52-week year"
If you think about it you want there to be so much stock that, if s is the number of spare part at the start of each week:

14. (Original post by metaltron)
If you think about it you want there to be so much stock that, if s is the number of spare part at the start of each week:

= 1-p(x less than or equal to 5)=1/52
1-(1/52)= 0.9808

in tables, for lambda (5)

when x=9: 0.9682
when x=10: 0.9863
when x=11: 0.9945

so is it 9?
15. (Original post by 0utdoorz)

= 1-p(x less than or equal to 5)=1/52
1-(1/52)= 0.9808

in tables, for lambda (5)

when x=9: 0.9682
when x=10: 0.9863
when x=11: 0.9945

so is it 9?
From the values you have quoted, you are looking for the first time is above 0.9808.

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