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Edexcel (IAL): Statistics 1, WST01 (27th January 2016) Watch

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    (Original post by Awaiskorai)
    I could be wrong too but I got different answer for P(Y>X) I got 1/4
    P(X=-2) = 1/4
    P(x= 1) = 1/6
    P(X=3) = 1/3
    P(X=4) = something
    P(X=something) = something.

    P(Y > X) = P(7-2x > x) = p(3x < 7) = p (X < 2.333333...) = P(x=-2) + P(X=1).


    Got exact same answers for q2. Though they tried to confuse us using thr boundaries and range
    Great!

    For Question 3 I got the same answers as you but the intercept of my equation was negative so although it was of same value it was negative
    A negative intercept doesn't make sense, though? You can't have a negative number of sales.

    For question 4 again, got the same answers,
    Great.


    For 5 Again the same answers but with different second dp 5.42 and and 22.56
    Did you look at the percentage points table or the normal normal distribution table?

    For 6 all the same but with calculations shown
    Great! Sounds like you've done fairly well.
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    (Original post by Awaiskorai)
    Either we have all messed up or are we actually getting a 90+ UMS? XD Not sure.
    I'm not sure what you're trying to say.
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    I am trying to say, as we have got almost the same answers, either we have fared great or everything is wrong and down the hill xP

    About the negative sales part I thought it could be negative because of a loss in sales :/ I would have certainly rechecked if I had time.
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    (Original post by Zacken)
    Disclaimer: If you disagree with any of my answers, I'm probably wrong and you're probably right

    1. (a) F(5) = 5/6
    (b) E(X) = 2
    (c) Var(X) = 15/2
    (d) (i) E(Y) = 3
    (ii) Var(Y) = 30
    (iii) P(Y > X) = 5/12

    2. (a) 30 marks is the lowest pass mark.
    (b) 46 marks is the lowest merit mark.
    (c) c=70, d=6
    (d) 68, 72, 79 are the three highest marks
    (e) Outlier at 5, range starts at 10, not sure why they mentioned 15, did I do something wrong?
    (f) 0.5 * 0.25 * 0.25 * 3 = 3/32

    I hate box plots, probably failed that question.

    3. (a)S_vs = 43.911 and S_ss = 40.44
    (b) pmcc = 0.929
    (c) pmcss approx 1 blah blah blah
    (d) s = 1.72 + 0.794 x
    (e) y = 1920 + 3.97x
    (f) coeff of x higher for textbooks blah blah blah

    4. (a) 0.3, 0.5, 0.7, 0.9 - those are the fails. Success on the last is 0.1
    (b) 0.9055 = 0.7 + 0.3 * 0.5 + 0.3 * 0.5 * 0.3 + 0.3 * 0.5 * 0.7 * 0.1
    (c) 1700/1811
    (d)Shall we make a petition? Hannah's sweet much. p + (1-p)(p-0.2) = 0.95
    (e) 1.1 - sqrt(3/50) approx 0.855

    5(a) 0.3446
    (b) 0.0583
    (c) mean = 22.57, standard dev = 5.43

    6(a) mean = 836, standard dev = 79.9
    (b) postiive skew, mean > median.
    (c) no change
    (d) standard dev decrease.
    For 2(f) I used the P(A|B) formula. 'A' being the probability of two of them getting merit, and 'B' being the probability of three of them passing. Do you think this would work..?
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    (Original post by Zacken)
    6(d) quantitatively? You get something like sqrt(5000 something), don't you? Lower value is correct, I said it decreases? Although my reason was:

    60^2 + 60^2 = 7200 &lt; 2 \times \frac{68 430}{10} or whatever the 68430 was. We should both get the marks.

    Thank god for the 2(e), why did they mention 15 marks, though?

    How'd you do 2(f) and what did you do that differs from mine, what did I mess up? :eek:

    Congratulations! This was a 6-question paper, those generally have lower grade boundaries.
    Yep, I think I got 73.9. I noticed that \[\sum (x-\bar{x})^{2} = S_{xx}, so I just found the new \sum x^{2} by finding \frac{old \sum x^{2} + (value_{1})^{2} + (value_{2})^{2}}{12}\]. After that, it was just the regular formula.

    It was just to trip us up. I realised the moment I read it. For 2(f) what I did was 3 x (0.75)2 x (0.25), since 0.75 students pass and 0.25 students get a merit. I might be wrong though.

    The normal distribution question was really nice. Plenty of people at my centre couldn't do it from the general reaction. It's definitely a boundary dropper! (which makes it nicer for me 'cause I could do it )

    See you on the M2 thread.

    EDIT: Apologising for the terrible latex
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    How many marks are we going to lose for negative intercept? Is Ecf there?
    And I looked at the normal distribution table.
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    (Original post by aymanzayedmannan)
    Yep, I think I got 73.9. I noticed that \[\sum (x-\bar{x})^{2} = S_{xx}, so I just found the new \sum x^{2} by finding \frac{old \sum x^{2} + (value_{1})^{2} + (value_{2})^{2}}{12}\]. After that, it was just the regular formula.

    It was just to trip us up. I realised the moment I read it. For 2(f) what I did was 3 x (0.75)2 x (0.25), since 0.75 students pass and 0.25 students get a merit. I might be wrong though.


    See you on the M2 thread.

    EDIT: Apologising for the terrible latex
    That's certainly wrong because two students that passed were on merit so (.25*.25 *. 75) and there were three such possibilities so, 3(.25*.25*.75)
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    (Original post by Awaiskorai)
    That's certainly wrong because two students that passed were on merit so (.25*.25 *. 75) and there were three such possibilities so, 3(.25*.25*.75)
    Did you get a final answer of 9/64? I got that.
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    (Original post by Awaiskorai)
    I am trying to say, as we have got almost the same answers, either we have fared great or everything is wrong and down the hill xP
    I doubt we all got the same wrong answer.

    About the negative sales part I thought it could be negative because of a loss in sales :/ I would have certainly rechecked if I had time.
    That doesn't make sense, oh well - you shan't lose too many marks. Maybe 1 or 2.

    (Original post by elhm18)
    For 2(f) I used the P(A|B) formula. 'A' being the probability of two of them getting merit, and 'B' being the probability of three of them passing. Do you think this would work..?
    I'll try your method and get back to you in a few minutes.

    (Original post by Awaiskorai)
    And I looked at the normal distribution table.
    You were meant to look at the percentage points table which give more exact values for z.

    (Original post by aymanzayedmannan)
    Yep, I think I got 73.9. I noticed that \[\sum (x-\bar{x})^{2} = S_{xx}, so I just found the new \sum x^{2} by finding \frac{old \sum x^{2} + (value_{1})^{2} + (value_{2})^{2}}{12}\]. After that, it was just the regular formula.
    Sounds right. I did that as well but I was wary of using nothing but calculations so I explained it the other way just to be safe.

    It was just to trip us up. I realised the moment I read it. For 2(f) what I did was 3 x (0.75)2 x (0.25), since 0.75 students pass and 0.25 students get a merit. I might be wrong though.
    Phew! I'm not very good with box plots, so I stayed confused.

    I think you mean 3 x (0.25)^2 x (0.75) - I considered doing that as well, but I figured that the 0.75 should be a 0.5 because otherwise it includes people getting merits. In either case, neither of us should lose more than 2 marks.

    The normal distribution question was really nice. Plenty of people at my centre couldn't do it from the general reaction. It's definitely a boundary dropper! (which makes it nicer for me 'cause I could do it )
    It was very nice! The mark distribution was nice as well - 3, 5, 6. Looks like you're on track for full UMS.

    See you on the M2 thread.
    Now that isn't going to go well...

    EDIT: Apologising for the terrible latex
    Don't apologise!
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    (Original post by Zacken)
    Disclaimer: If you disagree with any of my answers, I'm probably wrong and you're probably right

    1. (a) F(5) = 5/6
    (b) E(X) = 2
    (c) Var(X) = 15/2
    (d) (i) E(Y) = 3
    (ii) Var(Y) = 30
    (iii) P(Y > X) = 5/12

    2. (a) 30 marks is the lowest pass mark.
    (b) 46 marks is the lowest merit mark.
    (c) c=70, d=6
    (d) 68, 72, 79 are the three highest marks
    (e) Outlier at 5, range starts at 10, not sure why they mentioned 15, did I do something wrong?
    (f) 0.5 * 0.25 * 0.25 * 3 = 3/32

    I hate box plots, probably failed that question.

    3. (a)S_vs = 43.911 and S_ss = 40.44
    (b) pmcc = 0.929
    (c) pmcss approx 1 blah blah blah
    (d) s = 1.72 + 0.794 x
    (e) y = 1920 + 3.97x
    (f) coeff of x higher for textbooks blah blah blah

    4. (a) 0.3, 0.5, 0.7, 0.9 - those are the fails. Success on the last is 0.1
    (b) 0.9055 = 0.7 + 0.3 * 0.5 + 0.3 * 0.5 * 0.3 + 0.3 * 0.5 * 0.7 * 0.1
    (c) 1700/1811
    (d)Shall we make a petition? Hannah's sweet much. p + (1-p)(p-0.2) = 0.95
    (e) 1.1 - sqrt(3/50) approx 0.855

    5(a) 0.3446
    (b) 0.0583
    (c) mean = 22.57, standard dev = 5.43

    6(a) mean = 836, standard dev = 79.9
    (b) postiive skew, mean > median.
    (c) no change
    (d) standard dev decrease.
    All right except 2(f) if I am not wrong...
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    (Original post by Awaiskorai)
    That's certainly wrong because two students that passed were on merit so (.25*.25 *. 75) and there were three such possibilities so, 3(.25*.25*.75)
    But 0.75 is the probability of getting a merit or passing. 0.5 is the probability of passing only.
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    (Original post by Fuwad98)
    All right except 2(f) if I am not wrong...
    Lots of people seem to be saying that my 2(f) is wrong, it probably is.

    I'm not convinced why you do 3 * 0.25 * 0.25 * 0.75 though, surely the 0.75 should be (0.75 - 0.25) = 0.5 instead.
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    (Original post by Zacken)
    Lots of people seem to be saying that my 2(f) is wrong, it probably is.

    I'm not convinced why you do 3 * 0.25 * 0.25 * 0.75 though, surely the 0.75 should be (0.75 - 0.25) = 0.5 instead.
    I think getting a merit instantly means you'd be passing. They intersect. And 75% do pass, even if it includes the ones who got a merit.

    That's what I think, at least.
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    (Original post by elhm18)
    I think getting a merit instantly means you'd be passing. They intersect. And 75% do pass, even if it includes the ones who got merit.

    That's what I think, at least.
    But the question specifically said only 2 get a merit and 1 gets a pass.
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    (Original post by Zacken)
    But the question specifically said only 2 get a merit and 1 gets a pass.
    Did it? I remember it saying "given three of the students pass, what is the probability of two getting a merit?" or something along the lines. I'm not entirely sure, but that's what I remember.
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    (Original post by elhm18)
    Did it? I remember it saying "given three of the students pass, what is the probability of two getting a merit?" or something along the lines. I'm not entirely sure, but that's what I remember.
    I'm very sure it used the world "only" specifically, had it not - I'd have done the same thing as you.

    Edit: anyways, I'm not fussed. Either one would get 1/2 marks and that's fine so I'm going to stop arguing now. The boundaries should be fairly low anyway, with that normal dist. quesiton.
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    (Original post by Zacken)
    I'm very sure it used the world "only" specifically, had it not - I'd have done the same thing as you.
    Hm, maybe so. I might have read wrong, but I still feel as though they mentioned, "given three pass"

    I would have not used the P(A|B) formula otherwise.

    We'll get to know when the time comes, haha. But you're probably right.
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    (Original post by elhm18)
    Hm, maybe so. I might have read wrong, but I still feel as though they mentioned, "given three pass"

    I would have not used the P(A|B) formula otherwise.

    We'll get to know when the time comes, haha.
    Sooner, somebody will put up model answers within the next few days.
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    (Original post by aymanzayedmannan)
    Did you get a final answer of 9/64? I got that.
    Not sure if I told you but if you notice, the formula for Sxx is simply the formula for variance multiplied by 'n'. So I just divided the value of Sxx we were given with 10. And I got 79.9 too. :P

    variance = Name:  Screen Shot 2016-01-27 at 2.37.47 PM 1.png
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    (Original post by Zacken)
    Sooner, somebody will put up model answers within the next few days.
    Yeah, that's what I'm hoping for.
 
 
 
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