AQA A2 Statistics 2 MS2B Unofficial Mark Scheme

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the extra sales were poisson distribute with a mean of 2, a was the probability of selling 4, which is 3 extra sales, so (e^-2)(2^3)/3! = a

oh my god I don't believe it, thanks for this.

By extra demand I thought it meant having more than 5 in the week!!!!!!!!!!

Very disappointed :frown:

Reply 21

alex27996

i think i got 65/75. would this be enough for an A do you think?

Reply 22

Josh771

Seems good :smile:

I messed up on the probability that 1 of 4 was negative, for some reason I did all 4.... still would get working marks for finding the probability of 1 being negative even if its not worth much! But the rest was really good! Im glad the last question's part a) was a show that, I did the whole question before going to do that which saved a load of time as I didn't understand it at first! Pretty nice paper though compared to what I had imagined, especially as some parts felt like they were cut straight out of last years (i.e two missing probabilities a and b, show the variance, rectangle with variance of 3!!)

Reply 23

ibanezmatt13

On finding a and b in the last question, at the time I didn't know what I was doing, but I blagged it and I'm wondering if this will get the marks.

I looked in the poisson tables and realised that 1 - P(X<=3) = 0.143, though I didn't know the significance of this

So my answer to part i) was:
======================

P(X <= 3) = 0.8571

a = 1 - 0.8571 = 0.143

and then sum of probabilities is 1, so found b based on that.

Any marks?

UPDATE: This is a load of rubbish I think. What a way to loose 8 marks...

(edited 8 years ago)

Reply 24

HeyC97

How did you do 2)b)ii) I put prob or being negative = 1/3 so prob of being positive = 2/3 and then 1/3 x2/3^3? But I'm sensing this may have been wrong ..?

Reply 25

ibanezmatt13

Original post by HeyC97

How did you do 2)b)ii) I put prob or being negative = 1/3 so prob of being positive = 2/3 and then 1/3 x2/3^3? But I'm sensing this may have been wrong ..?

Very close. I think it asked for the probability of X being negative once in 4 times, so you'd have had to multiply that by 4C1 = 4 as there are 4 ways of that occurring.

Reply 26

ec1209

Does anyone know how you were supposed to calculate the profit for stocking 4 magazines in question 7? I don't even care about my grade anymore I just want to understand it! :smile:

Reply 27

Josh771

Original post by ec1209

Does anyone know how you were supposed to calculate the profit for stocking 4 magazines in question 7? I don't even care about my grade anymore I just want to understand it! :smile:

I did it by saying the probability of 4 selling was the same as 4 and 5 selling when selling 5 magazines (i think it was 0.343 or something). Then find E(X), do 1 x E(X) -0.5(4-E(x)) just as with the previous part

Reply 28

anise_mc

Original post by alex27996

i think i got 65/75. would this be enough for an A do you think?

i'd say so, as it was 65 for an A in june 2014 i think and those boundaries were very high :smile:

Reply 29

ibanezmatt13

Original post by anise_mc

i'd say so, as it was 65 for an A in june 2014 i think and those boundaries were very high :smile:

Do you think the boundary for the A* this year will be lower than last year?

I personally think the paper was very nice, but I struggled with the last question which lost me most on offer for the question.

Reply 30

jjsnyder

Here is the method for 7c and 7d as a few people asked. Could not remember exact values but it has the method.

Posted from TSR Mobile

Reply 31

jjsnyder

ImageUploadedByStudent Room1434123236.168973.jpg

This is the method for 7c and 7d :smile:

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Reply 32

Cal_Fitz

For Q 2 bii, wouldn't it be 2/7 as the random values can be -2,-1,0,1,2,3 and 4?

Reply 33

jjsnyder

Original post by Cal_Fitz

For Q 2 bii, wouldn't it be 2/7 as the random values can be -2,-1,0,1,2,3 and 4?

No, as its continuous it doesn't take set values so it has a range of 4-(-2) which is 6. So 2/6

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Reply 34

Cal_Fitz

Ah damn, I did that then crossed it out! Still think I got around 64 which should still hopefully be an A!

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Reply 35

PrimeLime

Original post by jjsnyder

No, as its continuous it doesn't take set values so it has a range of 4-(-2) which is 6. So 2/6

Posted from TSR Mobile

4C1 x (1/3) x (2/3)^3 for that question right?

Reply 36

Oraeng

For the chi-squared 5)a) I combined the 17 or 18 and 19 or more columns... was this right?

Reply 37

jjsnyder

Original post by Oraeng

For the chi-squared 5)a) I combined the 17 or 18 and 19 or more columns... was this right?

Yes this was correct :smile:

Posted from TSR Mobile

Reply 38

jjsnyder

Original post by PrimeLime

4C1 x (1/3) x (2/3)^3 for that question right?

Yes, I have put this on the mark scheme :smile:

Posted from TSR Mobile

Reply 39

Madmatician

Guys I made a really stupid mistake on Q1b

Instead of doing P(X=2), I did P(X≤2).
I got X~Po(4.4) and in my method I had (e^-4.4 x 4.4^2)/2!

How many marks will I have lost out of 3?