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    Not sure for 2(f) I got E(R) the same, but there was one of which they drew which had the probability of 0.2/0.3 Therefore P(1/X>X)+P(X>1/X) = 0.8/0.7. I think I got 0.625 and 0.175 not sure though
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    (Original post by veldt127)
    Yeah, I think you've worded that right. It was a weird question



    Ah I see, I'll get on that
    For some reason I didn't get the right answer, but showed the working out.
    How many marks would I drop?
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    (Original post by SANTR)
    Just out of interest how did you reach the answer for 2(e). What did you have to do to the given Probability Distribution table?
    for part e can you mention that the data was very slightly negatlively skewed and median and mean were close together therefore it can be modelled by a normal?
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    I saved my answers like you and got every question the same as you, I'm pretty sure they are correct! 😀


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    (Original post by neyjunior)
    for part e can you mention that the data was very slightly negatlively skewed and median and mean were close together therefore it can be modelled by a normal?
    Possibly, they were very close. The mark scheme might go either way; I haven't seen this kind of question enough times to comment
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    (Original post by azizadil1998)
    Hey everyone, I got 0.6 for 4.a (the hotel one), is it right?? I looked at the answers above it said 0 but I just guessed anyways

    what was the question again i can't remember?
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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
    Spoiler:
    Show
    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:
    Show
    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:
    Show
    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:
    Show
    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:
    Show
    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
    ii. The standard deviation decreases as the variance decreases with a new value close to the mean
    Question 6
    Spoiler:
    Show
    a. 0.0668
    b. 206 - The lowest 20% of times were the fastest 20% of runners.
    c. 0.36
    How did you get 0.03 for t in the Venn diagrams one? I got 0.17 and for u I got 0.08
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    (Original post by Einstein101)
    How did you get 0.03 for t in the Venn diagrams one? I got 0.17 and for u I got 0.08
    If you use the equation for independent events:

    P(A ∩ B) = P(A) * P(B)

    You get t = 0.03 and u = 0.22.
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    (Original post by Einstein101)
    How did you get 0.03 for t in the Venn diagrams one? I got 0.17 and for u I got 0.08
    You were told they were independent so P(AnB)= P(A) x P(B) then rearrange. I almost forgot to add the value at the bottom of the circle, maybe you did that?
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    (Original post by Einstein101)
    How did you get 0.03 for t in the Venn diagrams one? I got 0.17 and for u I got 0.08
    P (BnD)= P(B) x P(D) because they are independent.

    P(D) was 0.45 I think and then you subtract everything inside D from 0.45 giving you T to be 0.03.

    And then to find U, we subtract everything else from 1.
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    (Original post by ShatnersBassoon)
    But the new value was exactly the same as the mean, so the SD (average distance from mean) can't increase. Others said it stayed the same, but I wrote that it decreases, because now the average distance from the mean is slightly less (as n has increased by 1 but deviance hasn't changed).
    Im pretty sure i got that wrong but ill add your answer to it.
    http://www.thestudentroom.co.uk/show....php?t=4166725
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    (Original post by SANTR)
    Height for the bar is 17cm and the width is 0.5cm.
    E(R) should be 2.5. Because 1/.4 is 2.5
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    (Original post by SANTR)
    Height for the bar is 17cm and the width is 0.5cm.

    I got the same
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    (Original post by neyjunior)
    for part e can you mention that the data was very slightly negatlively skewed and median and mean were close together therefore it can be modelled by a normal?
    The question was 3 marks. In the worst case, no, you'd get 0/3 (1 mark for saying it's a bad model; another for saying mean =/= median; a third for comparing d to the actual answer).

    You didn't compare your answer to part (d) to anything, but the question explicitly told you to, so you've definitely lost a mark there. I'd give you 1/3 but maybe if they're generous you'd get 2/3.
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    (Original post by Florent venhari)
    I got these wrong:
    2 f ii)
    3 e) maybe I got one mark
    4 d i)
    4 e)
    5 e) lost 1-2 marks
    5 f ii)
    6 c) maybe I got one mark for getting 0.5 as one of the probabilities.

    Oh wells

    Posted from TSR Mobile

    Im super super confused with question 6c, i had NO idea it completely threw me off, could anyone explain?
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    (Original post by ShatnersBassoon)
    But the new value was exactly the same as the mean, so the SD (average distance from mean) can't increase. Others said it stayed the same, but I wrote that it decreases, because now the average distance from the mean is slightly less (as n has increased by 1 but deviance hasn't changed).
    I'm pretty sure I got it wrong haha but ill add your answer and others' so people can discuss it.
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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:
    Show
    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:
    Show
    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:
    Show
    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:
    Show
    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:
    Show
    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
    ii. The standard deviation decreases as the variance decreases with a new value close to the mean
    Question 6
    Spoiler:
    Show
    a. 0.0668
    b. 206 - The lowest 20% of times were the fastest 20% of runners.
    c. 0.36
    I said all of these except 5e! I put yes because the mean and median are similar but I didn't even understand the question lol
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    (Original post by pranto96)
    E(R) should be 2.5. Because 1/.4 is 2.5
    I thought that initially but worked through:

    Name:  13441821_10207178478156340_594191193_o.jpg
Views: 645
Size:  127.9 KB
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    (Original post by veldt127)
    I thought that initially but worked through:

    Name:  13441821_10207178478156340_594191193_o.jpg
Views: 645
Size:  127.9 KB
    I totally get your point. But why don't we get the right answer if we do 1/.4?
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    (Original post by pranto96)
    I totally get your point. But why don't we get the right answer if we do 1/.4?
    Well, E(x) is the sum of x over n, right?

    E(1/x) is the sum of 1/x over n.

    (p(X = x)( (Σ1/x) )/n is not the same as (p(X = x)(1/( (Σx)/n ))
 
 
 
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