Difficult Poisson distribution Question - S2 Edexcel

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The Joker
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Question:

The probability that a match will not strike is 0.009. Calculate the probability that in a box of 100 matches:

(a) they all strike satsfactorily
(b) at least 2 do not strike
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The Joker
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bump I tried everything but the calculator always gives me an error because 100! cannot be done.
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natty_d
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Shouldn't you be using Binomial? But the numbers are so big that you use Poisson as an approximation - therefore you use np and your lambda value for Poisson so multiply 0.009 by 100 and then use this number and use the possion equation.
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The Joker
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Can you guide me through this please, repped.
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zomgleh
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(Original post by The Joker)
Can you guide me through this please, repped.
for huge numbers use the- nCr button in ur calculator.
ex- 100nCr100.

btw what are the answers?
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zomgleh
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(Original post by natty_d)
Shouldn't you be using Binomial? But the numbers are so big that you use Poisson as an approximation - therefore you use np and your lambda value for Poisson so multiply 0.009 by 100 and then use this number and use the possion equation.
hmmm i don't think the poisson is the right distribution to use because the mean and the variance don't give about the same values?

or is that only a condition if you make it one? i'm not sure.
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The Joker
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The correct answers are:

(a) 0.4066
(b) 0.2275

How it got to there I don't know.
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natty_d
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(Original post by zomgleh)
hmmm i don't think the poisson is the right distribution to use because the mean and the variance don't give about the same values?

or is that only a condition if you make it one? i'm not sure.
Yes, you're right - but if you work it out mean being np = 0.009 x 100 = 0.9 and variance = 0.9 (1-0.009)= 0.8919. They are so similar that you can use the Poisson distribution. :borat:
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zomgleh
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(Original post by The Joker)
The correct answers are:

(a) 0.4066
(b) 0.2275

How it got to there I don't know.
i used the binomial formula for the 1st one- (0.991)^100 and got 0.4049, can't think of any other way.
you can't use a normal coz p isn't close to 0.5.
can't use poisson coz mean and variance aren't about the same.
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zomgleh
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(Original post by zomgleh)
i used the binomial formula for the 1st one- (0.991)^100 and got 0.4049, can't think of any other way.
you can't use a normal coz p isn't close to 0.5.
can't use poisson coz mean and variance aren't about the same.
okay my bad, you can use a poisson.
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zomgleh
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(Original post by natty_d)
Yes, you're right - but if you work it out mean being np = 0.009 x 100 = 0.9 and variance = 0.9 (1-0.009)= 0.8919. They are so similar that you can use the Poisson distribution. :borat:
oops, yes sorry I miscalculated.
but how would you divide by 100 factorial?
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The Joker
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(Original post by zomgleh)
i used the binomial formula for the 1st one- (0.991)^100 and got 0.4049, can't think of any other way.
you can't use a normal coz p isn't close to 0.5.
can't use poisson coz mean and variance aren't about the same.
Okay thank you for your help, can you help me with this last question.

Question
: From previous experience of word-processing her work, a sociology student is known to make an average of 1.5 errors per complete page. The numbers of errors on different pages are independent.

(a) Find the probability that the essay will contain exactly 10 errors.

P(X = 10) so e-1.5 x 1.5^10 over 10!= 0.0000000035.

The correct answer is actually 0.0413.
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natty_d
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(Original post by The Joker)
The correct answers are:

(a) 0.4066
(b) 0.2275

How it got to there I don't know.
Right - you don't need poisson at all.

Use Binomial, I think you must be calculating something wrong.

For question 2: you want at least 2 so the probability of they all strike satisfactorily + the probability of one not striking correctly taken away from 1.

So for P of 0 = from previous question you get 0.4049

then P of 1 = 100C1 x 0.009^1 x 0.991^ 99 = 0.3677

then 1 - (0.4049 + 0.3677) = 0.2774 :awesome:

I've obviously made a small rounding error but there you are.
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natty_d
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(Original post by The Joker)
Okay thank you for your help, can you help me with this last question.

Question
: From previous experience of word-processing her work, a sociology student is known to make an average of 1.5 errors per complete page. The numbers of errors on different pages are independent.

(a) Find the probability that the essay will contain exactly 10 errors.

P(X = 10) so e-1.5 x 1.5^10 over 10!= 0.0000000035.

The correct answer is actually 0.0413.
Does it say that there's a certain amount of pages in her essay or something?
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Krebs
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the S2 exam is in 9 days.

(Y)


And the sociology question should mention how many pages there are in the essay.

You then multiply the average rate (1.5) by that factor.
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The Joker
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(Original post by natty_d)
Does it say that there's a certain amount of pages in her essay or something?
Sorry it has 4 pages my mistake.
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zomgleh
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can someone help me with the hypothesis testing using 2-tailed tests please?

thing is- I don't understand when you use p(x>/=) or p(x</=) ?
here's a link-
http://tinypic.com/view.php?pic=dpg7b7&s=6

why couldn't they have used-

p(x>/=) 35 instead??
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natty_d
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(Original post by The Joker)
Sorry it has 4 pages my mistake.
Then multiply 4 by 1.5 so you get the correct rate and then use the answer for your poisson distribution.
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The Joker
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(Original post by natty_d)
Then multiply 4 by 1.5 so you get the correct rate and then use the answer for your poisson distribution.
Yeah I realised that eventually thanks.
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Krebs
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(Original post by zomgleh)
can someone help me with the hypothesis testing using 2-tailed tests please?

thing is- I don't understand when you use p(x>/=) or p(x</=) ?
here's a link-
http://tinypic.com/view.php?pic=dpg7b7&s=6

why couldn't they have used-

p(x>/=) 35 instead??

When you read the text it says that the 35 out of the hundred people were sold something.

35 < 45 (45% of 100) , so you go for that tail.

But yes, the question does seem a bit strange. I would have thought you should do both tails as H1 is P =/= 0.45. Is it from the S2 book?
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