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AQA M1SB 8th of june 2016 unofficial markscheme watch

  • View Poll Results: [AQA MS1B 8th of June 2016] What raw mark do you think the A grade will be ?
    66
    20.00%
    65
    5.00%
    64
    10.00%
    63
    8.33%
    62
    13.33%
    61
    11.67%
    60
    15.00%
    59
    5.00%
    58
    11.67%

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    (Original post by -jordan-)
    It was 8/40 because one of them was equal to 200 euros.
    yes one was equal and the question said 'less than 20' so it was 7, but tbh i recon the mark scheme will be nice
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    (Original post by clairebear101)
    yes one was equal and the question said 'less than 20' so it was 7, but tbh i recon the mark scheme will be nice
    Maybe I'm going mad but I seem to remember counting all of the items before 167.67 and there were 8.
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    (Original post by clairebear101)
    this is what i put too, i messed up the CI tho, put the wrong values in, does this mean i loose all 4 marks? and will i get ECF for the claim, as normally markschemes say M1 for comapring the CI with the given value? so should only loose 3 marks max here??
    Confidence interval questions always have these marks:
    1 for the upper limit
    1 for the lower limit
    1 for stating the standard error (sd/rootn)
    1 for stating how many sd the interval is above and below the mean (ie + or - 2.58sd)

    You should get 1 mark for a correct claim and another for an explanation. You might only drop a couple for the wrong numbers imo.
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    (Original post by Sam Webb)
    Confidence interval questions always have these marks:
    1 for the upper limit
    1 for the lower limit
    1 for stating the standard error (sd/rootn)
    1 for stating how many sd the interval is above and below the mean (ie + or - 2.58sd)

    You should get 1 mark for a correct claim and another for an explanation. You might only drop a couple for the wrong numbers imo.
    but my claim is incorrect as a did the CI wrong will i get EFC for using it in the claim?

    I worked out the standard error then used it in the formula, by replacing the s.d in the formula with it, so essentially i used it twice! which i shouldnt have done but i didnt really unstand the formula but yeah :/ i think i got most of the rest of the paper correct
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    (Original post by Cascadess)
    Because it was not from their sample, there was no decreasing denominators I'm pretty sure. You where just using the results in the table on an infinite population of people at the dentists.

    So I did 164/500 * 164/500 * 121/500 * 121/500 * 4!

    Which is the same as (164/500)^2 * (121/500)^2 * 4!

    Because we are not taking our selecting from our sample, I don't think the sample size goes down.

    (Btw i have no idea if those numbers are correct, i'm just going off you here )
    Yes! I agree with squaring the probability of each condition but why do you multiply by 4!

    There is only one possible combination, right? Please correct me :ashamed2:
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    (Original post by Shaks.smxth)
    Yes! I agree with squaring the probability of each condition but why do you multiply by 4!

    There is only one possible combination, right? Please correct me :ashamed2:
    No, there are actually 6 possible combinations and should be x6. Draw a probability tree and it becomes clear. E.g. AABB ABAB, and so on...
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    (Original post by -jordan-)
    Maybe I'm going mad but I seem to remember counting all of the items before 167.67 and there were 8.
    it was 7, not including that value
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    (Original post by Shaks.smxth)
    Yes! I agree with squaring the probability of each condition but why do you multiply by 4!

    There is only one possible combination, right? Please correct me :ashamed2:
    Don't worry Shakd.smxth I think that's the main thing people are arguing/confused about on this part. I multiplied by 4! because I thought that if you had A(1) and A(2) as the two different people of the first condition and B(1) and B(2) as the two different people of the second condition then there are four different events, which can happen in any order, and the way to find out the amount of ways things can be arranged is by factorial the quantity of items, which would be 4!

    But -jordan- Has told me that infact we should be multiplying by 4!/(2!*2!) which is 6.

    So i believe the full, correct, calculation is:

    (164/500)^2 * (121/500)^2 * 6

    We both got it wrong :/
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    (Original post by clairebear101)
    but my claim is incorrect as a did the CI wrong will i get EFC for using it in the claim?

    I worked out the standard error then used it in the formula, by replacing the s.d in the formula with it, so essentially i used it twice! which i shouldnt have done but i didnt really unstand the formula but yeah :/ i think i got most of the rest of the paper correct
    I think you'll miss a mark for stating wrong claim incorrect. But as long as you timed your confidence interval answer's limits by 1.2 and used them in your explanation I reckon you'd probably still get the EFC mark.
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    (Original post by Cascadess)
    Don't worry Shakd.smxth I think that's the main thing people are arguing/confused about on this part. I multiplied by 4! because I thought that if you had A(1) and A(2) as the two different people of the first condition and B(1) and B(2) as the two different people of the second condition then there are four different events, which can happen in any order, and the way to find out the amount of ways things can be arranged is by factorial the quantity of items, which would be 4!

    But -jordan- Has told me that infact we should be multiplying by 4!/(2!*2!) which is 6.

    So i believe the full, correct, calculation is:

    (164/500)^2 * (121/500)^2 * 6

    We both got it wrong :/
    All three of us it seems... it's the most discriminating question for sure.
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    (Original post by Sam Webb)
    I think you'll miss a mark for stating wrong claim incorrect. But as long as you timed your confidence interval answer's limits by 1.2 and used them in your explanation I reckon you'd probably still get the EFC mark.
    Okay thanks, i hope so, only 4 or so marks lost anyway, since rest of paper was okay and grade boundaries are around 60 for an A, should be okay
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    (Original post by Language student)
    I used the Normal Stat Mode on the calculator and enter the following: Lower : 505
    Upper: 99999999999999
    s.d: 3.5
    mean: 508.5

    Which I get the probability of 0.841, then 0.841^6 which equals to 0.354 to 3sfg

    But i thought the question said when the bottles had volume greater than 505

    So you'd get P(z>508.5-505/3.5)
    => P(z>1) = 1 - p(z<1) = 1-0.8143 = 0.1587
    i got a crazy decimal when i did power of 6 to that but i thought it was meant to be can someone explain
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    (Original post by clairebear101)
    Okay thanks, i hope so, only 4 or so marks lost anyway, since rest of paper was okay and grade boundaries are around 60 for an A, should be okay
    Ya you'll be fine
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    (Original post by -jordan-)
    All three of us it seems... it's the most discriminating question for sure.
    For sure! I just drew out the tree, and you are very correct! I won't lie this is very frustrating xD
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    (Original post by manutd1324)
    But i thought the question said when the bottles had volume greater than 505

    So you'd get P(z>508.5-505/3.5)
    => P(z>1) = 1 - p(z<1) = 1-0.8143 = 0.1587
    i got a crazy decimal when i did power of 6 to that but i thought it was meant to be can someone explain
    It's Value - Mean / Standard Deviation. You got it the wrong way round. Should have got P(Z>-1)
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    (Original post by -jordan-)
    No, there are actually 6 possible combinations and should be x6. Draw a probability tree and it becomes clear. E.g. AABB ABAB, and so on...
    I stand corrected
    Thank you x
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    (Original post by Cascadess)
    For sure! I just drew out the tree, and you are very correct! I won't lie this is very frustrating xD
    Oh well I think we'll all have scored some amount of marks, 2 or 3 maybe. Not sure how many I'll have lost but I guess it's an error they're anticipating people to make and will be reasonable with it.
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    (Original post by Thequickspark)
    A few main questions.

    What did you say that the assumption was for part c?

    What did you get for the probability that it wouldn't be a particular value on the normal distribution (normally you say 100% if it's the mean, but in this case I did something else)

    Who else hopes the grade boundaries are lower than a donkey's testes?
    The assumption I put was that the bottles contents are all independent of one another.

    The probability of something the liquid content being exactly 1 number is 0. Therefore the probability of it not being exactly that number is 100%.

    And I'm hoping grade boundaries are on par with the testes of a standard donkey. But most people said they found it easy so I guess I'll expect high grade boundaries so I can't be disappointed.
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    (Original post by -jordan-)
    It's Value - Mean / Standard Deviation. You got it the wrong way round. Should have got P(Z>-1)
    Ahh cheers mate
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    (Original post by shaks.smxth)
    yes! I agree with squaring the probability of each condition but why do you multiply by 4!

    There is only one possible combination, right? Please correct me :ashamed2:
    yooooo thats exactly what i did, is that right or wrong?!?!?!?!
 
 
 
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