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    (Original post by veldt127)
    Ok, I managed to get all my answers stored in my calculator at the end, this is what I got. Not 100% sure on all of them, but we can compare.

    Question 1
    Spoiler:
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    a. Show that
    b. W - can't remember reasoning
    c. 0.994
    d. 0.0772 + 0.0142d
    e. i. 5.69
    ii. Reliable as the data was interpolated; it was within the range
    Question 2
    Spoiler:
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    a. 2p + q = 0.5
    b. q = 0.15
    c. p = 0.175
    d. 2.12
    e. 0.45
    f. i. 0.475
    ii. 0.375
    Question 3
    Spoiler:
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    a. 2.35
    b. Show that
    c. -133.9 - possibly -131.9
    d. -0.754
    e. Increase - I tried using the variables from the new week and recalculating r, and it was higher than the original. I said because the values were both higher than the means; I'm not sure at all on this one.
    Question 4
    Spoiler:
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    a. 0
    b. t = 0.03
    c. u = 0.22
    d. i. 0.45
    ii. 15/37
    e. 33 people have dinner (40*0.18) + (37*(15/37))
    Question 5
    Spoiler:
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    a. Width 0.5, Height 17
    b. 3.47 (median)
    c. i. Show that
    ii. Standard deviation = 0.680 - possibly 0.689
    d. P(W<3) = 0.2546
    e. No - the mean is not equal to the median suggesting asymmetrical skew18% of babies were less than 3kg, but her model predicted 25.46%
    f. i. Mean remains the same as the new value is the mean
    note: mean might decrease as it adds a value to the 3-3.5 group.
    ii. The standard deviation decreases as the variance decreases with a new value close to the mean
    Question 6
    Spoiler:
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    a. 0.0668
    b. 206 - The lowest 20% of times were the fastest 20% of runners.
    c. 0.36
    I swear 6b was like 176.336 or something? If the mean was 200 minutes how can the fastest 20% be P(X<206) like you got? Surely P(X<206) would be >0.5
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    (Original post by SANTR)
    So you divided each of the observations by 1 and then multiplied the 'same' probabilities with each corresponding 'new' observation i.e. 1/ each observation?
    Sounds correct because the p(X=x), but the P(R=1/X)
    therefore you take the reciprocal of each observation in the pdf and using knowledge of the expectation E(x), which is the sum of the P(x=x) multiplied by each observation x then you should get a value for E(S). Which btw was P(X=S)
    so it's about how u interpret the question and manipulate so x would be s and from there if you know what your doing you should end up with the right answer does anyone remember the table the x values not the P(X=x)?
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    Anyone know the marks for the other questions? I remember a few. It would be useful to add to the first post.

    Bee wiggle question
    Response variable (1 mark)
    Work out regression line (3 marks, may have been 4)

    Sarah rebecca prob distribution
    Make an equation with p and q (1 mark)
    Find p (2 mark)
    Hence find q (1 mark maybe 2??)
    Find var x (2 mark, maybe 3 cos u had to find E(x squared) aswell)
    Find E(R) (3 marks)
    Probabilities of them winning a game (4 marks) split into two parts

    Rainfall question
    Find Syy (1 mark lol)
    Work out R (2 marks)
    State effect on pmcc with reason (2 marks)

    Baby question
    Work out width/height of bar (3 marks)
    Linear intepolation for median (3 marks)
    Show estimate of mean is... (2 marks)
    Standard deviation (2 marks)
    P(x<3) (2 marks i think)
    Was shyan right to use normal distribution (3 marks)

    Hotel question
    Question about probability of booking a breakfast without room was (1 mark)
    Working out the value of t was (4 marks) and then
    Working out the value of U was (2 marks.)
    The conditional probabilities of the people eating dinner question (2 marks for each part so 4 marks overall)
    The part about estimating the amount of people from the coach who had dinner was (2 marks)

    Normal dist. question about marathon
    1st part (3 marks)
    Finding how fast he can run to be in top 20% (3 marks)
    Finding probability of X<u-30|x<u (3 marks)

    Only have 59/75 marks here written down, anyone else remember the other 16? They are missing from the bee question and the rainfall question, I think i got the other ones completed
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    (Original post by igotohaggerston)
    We are talking about Edexcel here
    Actually Edexcel have always been nice with stats grade boundaries, having touched 56 last year in the IAL paper and a few times in 2010-2011. So I wouldn't rule out 56 or 57 (doubting 55 though)
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    (Original post by Katieramsay)
    How did everyone get the probability for Sarah and Rebecca wining. I got 0.4625 and 0.4450 so clearly I've got it wrong but I don't understand how?? 🤓🤓
    Check back through the thread for a pic. I know what you did wrong: the question said that a computer selected a single value of X, then both Sarah and Rebecca are allocated a score based on this using X or 1/X.
    You treated it as one value of X picked for sarah, then a different value picked for Rebecca. A lot of calculation and cases later you got your answers.
    You might get marks for a special case misread, as your question is actually harder but testing the same maths.
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    (Original post by Roadman)
    I swear 6b was like 176.336 or something? If the mean was 200 minutes how can the fastest 20% be P(X<206) like you got? Surely P(X<206) would be >0.5
    Did anyone else get 6c as 0.64 not 0.36?

    I remember the question being the probability between mew + 30 and mew - 30.
    The probability W < mew + 30 was 0.82 so P(W<mew -30) and P(W> mew+30) = 0.18. So bit in middle was 1-0.36 = 0.64.

    I'm pretty sure I'm wrong considering everyone got 0.36. I guess I just read the question wrong as I was half asleep. Does anyone remember the exact wording of the question?

    Edit - I ****ed up this paper big time. How didnt i notice that question was conditional probability
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    (Original post by Katieramsay)
    How did everyone get the probability for Sarah and Rebecca wining. I got 0.4625 and 0.4450 so clearly I've got it wrong but I don't understand how?? 🤓🤓
    I got exactly that!
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    (Original post by 666kappa)
    Did anyone else get 6c as 0.64 not 0.36?

    I remember the question being the probability between mew + 30 and mew - 30.
    The probability W < mew + 30 was 0.82 so P(W<mew -30) and P(W> mew+30) = 0.18. So bit in middle was 1-0.36 = 0.64.

    I'm pretty sure I'm wrong considering everyone got 0.36. I guess I just read the question wrong as I was half asleep. Does anyone remember the exact wording of the question?
    The bit in the middle didn't have much to do with the question I'm afraid. You had to work out 18/50 as it was p(anb)/p(b) with a being chance of less than u - 30 and b being less than u.
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    (Original post by 666kappa)
    Did anyone else get 6c as 0.64 not 0.36?

    I remember the question being the probability between mew + 30 and mew - 30.
    The probability W < mew + 30 was 0.82 so P(W<mew -30) and P(W> mew+30) = 0.18. So bit in middle was 1-0.36 = 0.64.

    I'm pretty sure I'm wrong considering everyone got 0.36. I guess I just read the question wrong as I was half asleep. Does anyone remember the exact wording of the question?
    Not the middle bit, sorry.
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    (Original post by PrinceAli)
    The bit in the middle didn't have much to do with the question I'm afraid. You had to work out 18/50 as it was p(anb)/p(b) with a being chance of less than u - 30 and b being less than u.
    I worked out P(mew-30 < W < mew+30). Along with working out the 20% slowest time nots the fastest. RIP
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    (Original post by studentsixth)
    Actually Edexcel have always been nice with stats grade boundaries, having touched 56 last year in the IAL paper and a few times in 2010-2011. So I wouldn't rule out 56 or 57 (doubting 55 though)
    Yeah! I agree. The 2015 IAL was easier than today's paper so the grade boundaries could be as low as that


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    (Original post by studentsixth)
    I got exactly that!
    Yeah apparently we read the question wrong, but still hoping we got it right and everyone else is wrong 😂😂
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    (Original post by 666kappa)
    Did anyone else get 6c as 0.64 not 0.36?

    I remember the question being the probability between mew + 30 and mew - 30.
    The probability W < mew + 30 was 0.82 so P(W<mew -30) and P(W> mew+30) = 0.18. So bit in middle was 1-0.36 = 0.64.

    I'm pretty sure I'm wrong considering everyone got 0.36. I guess I just read the question wrong as I was half asleep. Does anyone remember the exact wording of the question?

    Edit - I ****ed up this paper big time. How didnt i notice that question was conditional probability
    You weren't looking for the bit in the middle. It was P(W< mew-30 I W< mew)
    as u said P(W< mew-30) is 0.18 due to symmetry then u divide that by P(W<mew) which is 0.5 which gives u 0.36
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    (Original post by Mr T Pities You)
    Check back through the thread for a pic. I know what you did wrong: the question said that a computer selected a single value of X, then both Sarah and Rebecca are allocated a score based on this using X or 1/X.
    You treated it as one value of X picked for sarah, then a different value picked for Rebecca. A lot of calculation and cases later you got your answers.
    You might get marks for a special case misread, as your question is actually harder but testing the same maths.
    Thanks do you have a picture??
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    (Original post by Katieramsay)
    Thanks do you have a picture??
    Only one I'm posting!
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    (And you can faintly see my wrong answer rubbed out - the same error you made! When will I learn to read the question properly?)
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    (Original post by azizadil1998)
    Mate there was a VERY similar question in one of the past papers, fortunately did it the night before hahah - You just had to subtract the value they gave from 1 to get 0.18, and "P(X<u)" is just the mean, and the mean represents 0.5 in a normal distribution model; therefore 0.18/0.5 = 0.36.

    You're so lucky no way! Now youve said it, makes perfect sense ty so much but ugh im kicking myself fml
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    (Original post by Roadman)
    I swear 6b was like 176.336 or something? If the mean was 200 minutes how can the fastest 20% be P(X<206) like you got? Surely P(X<206) would be >0.5
    The mean was 240.
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    (Original post by uk_shahj)
    Anyone know the marks for the other questions? I remember a few. It would be useful to add to the first post.
    [...]
    Only have 59/75 marks here written down, anyone else remember the other 16? They are missing from the bee question and the rainfall question, I think i got the other ones completed
    Someone else posted this on the first page:
    (Original post by Patrick Gekko)
    Q1: 2,1,2,4,2Q2: 1,3,2,2,3,4Q3: 3,1,2,2,2Q4: 1,4,2,4,2Q5: 3,2,3,3,3,3Q6: 3,3,3those were the marks for each part of the question.
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    (Original post by Roadman)
    You weren't looking for the bit in the middle. It was P(W< mew-30 I W< mew)
    as u said P(W< mew-30) is 0.18 due to symmetry then u divide that by P(W<mew) which is 0.5 which gives u 0.36
    Generally if P(Z>z) then you do 1-P(Z<z) as this is a special case only if it exceeds x~z(0,1). Also to standardise a probability use Z=X-u/o and if p(Z<-z) then you just find the p(Z<z) by symmetry of the normal distribution graph always draw the standard graph with mean 0 and sd 1 to compare with given and work from there.
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    The last time they asked the conditional probability question it was:
    1 mark for stating the p(X<u-30)/p(X whatever it was)
    1 mark for putting the correct numbers into it
    1 mark for the answer

    There was also a mark available for if you treated it like an independent event like in probability.
 
 
 
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