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S2 problem question

It is part C in which I am struggling with - I have looked on the tables for p(x<= 10) for the last value in which it is greater than 0.9, I found this value, but then the final answer on the markscheme correlates to a different value?

Thank you :smile:
Reply 1
Original post by iMacJack
It is part C in which I am struggling with - I have looked on the tables for p(x<= 10) for the last value in which it is greater than 0.9, I found this value, but then the final answer on the markscheme correlates to a different value?

Thank you :smile:


Look at the percentage points table, i.e: the table right below the normal distribution one in your formula booklet, you'll see that it gives a z value of 1.2816.
Reply 2
Original post by Zacken
Look at the percentage points table, i.e: the table right below the normal distribution one in your formula booklet, you'll see that it gives a z value of 1.2816.


I don't really see how this is helping me :frown:
Reply 3
Original post by iMacJack
I don't really see how this is helping me :frown:


Oh, sorry. I assumed it was a normal distribution question without looking at the previous parts.

P(C>10)<0.1    P(C10)>0.9\mathbb{P}(C > 10) < 0.1 \iff \mathbb{P}(C \leq 10) > 0.9. Now look in the x=10 row of your poisson table.
(edited 7 years ago)
Reply 4
Original post by Zacken
Oh, sorry. I assumed it was a normal distribution question without looking at the previous parts.

P(C>10)<0.1    P(C10)>0.9\mathbb{P}(C > 10) < 0.1 \iff \mathbb{P}(C \leq 10) > 0.9.


According to the markscheme, apparently not :frown:
Reply 5
Original post by iMacJack
According to the markscheme, apparently not :frown:


Does this help:
Reply 6
Original post by Zacken
Does this help:


Yeah - they were the two values I found, but then in the markscheme the final answer is lambda = 7/4?
Reply 7
Original post by iMacJack
Yeah - they were the two values I found, but then in the markscheme the final answer is lambda = 7/4?


Yes, because Charlotte does it over 4 hours. So, really what you've found is CPo(μ)C \sim Po(\mu) where μ=7\mu = 7.

But the random variable XPo(λ)X \sim Po(\lambda) is over 1 hour. So CPo(4λ)C \sim Po(4\lambda), that is: μ=4λ\mu = 4\lambda.
Reply 8
Original post by Zacken
Yes, because Charlotte does it over 4 hours. So, really what you've found is CPo(μ)C \sim Po(\mu) where μ=7\mu = 7.

But the random variable XPo(λ)X \sim Po(\lambda) is over 1 hour. So CPo(4λ)C \sim Po(4\lambda), that is: μ=4λ\mu = 4\lambda.


Oh yeah... should've realised that :redface:

Thank you :biggrin:
Reply 9
Original post by iMacJack
Oh yeah... should've realised that :redface:

Thank you :biggrin:


No problem.

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