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    I think I didn't get the show that P(x=6)=91/216 one because I excluded the cases where more than one of the dices were 6, as I thought that wouldn't count because X=The highest score on any of the three dice so i assumed if more than one of them rolled a 6 then 6 wouldn't be the 'highest'
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    For 8v, I think I probably misinterpreted the question - I get the same as iamJOM from 1 - (14/20) - (6/20) * (14/19), I just added an extra turn thanks to the wording :P

    That makes 3/38 = 0.0789 the most likely correct answer IMO.
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    (Original post by AlexParmenter)
    I think I didn't get the show that P(x=6)=91/216 one because I excluded the cases where more than one of the dices were 6, as I thought that wouldn't count because X=The highest score on any of the three dice so i assumed if more than one of them rolled a 6 then 6 wouldn't be the 'highest'
    That's what I thought at first, just that it was a 'show' question, so you could fiddle with the method until you found what they wanted you to do :P
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    (Original post by Bellbird)
    your answer would be getting other chocolate exactly after 2 cherries?

    because it asked at least 2 cherries before other chocolates that he likes..
    Let cherries be C and others be B
    the possibilities of not getting at least 2 cherries be4 other chocolate would be
    CB
    BC
    BB
    i did 1- P(CB+BC+BB)
    I am probably wrong ..
    I did something like this too. But didn't the question say more than 2 cherries before he finds the one he likes? so I included the p(CC) aswell in the sum and took it away from one.


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    Okay cheers guys aha, realised I got 8ii B) right, just misread the guys answer^


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    (Original post by FM2196)
    I did something like this too. But didn't the question say more than 2 cherries before he finds the one he likes? so I included the p(CC) aswell in the sum and took it away from one.


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    what did you get as a final answer? i did the same thing
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    What percentage of as maths is statistics?
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    1. i) Mean x = 1499kWH; SD x = 419kWh [would have done 3sf for mean but felt uncomfortable rounding to 1500!]

    ii) Mean y = £258.90; SD y = £68.32

    2. i) A) 15/250 = 0.06
    B) 148/250 = 0.592

    ii) 15/67 = 0.2239

    3. i) 144/169 = 0.8521
    ii) 8.52

    4. i) 3268760
    ii) 1019304
    iii) (ii)/(i) from above = 0.3118

    5. i) 0 | 6
    1 | 5 8
    2 | 1 5 8
    3 | 1 1 3 5 8 9

    Key: 1 | 5 = 15 people

    ii) Negative skewness

    iii) Median = 29.5; mean = 26.7; mode = 31
    I don't have any reasonable idea of what is wanted here. Others have suggested a reference to effect of negative skew on mode. I mentioned something about the mode not being particularly significant since it only occurs one time more frequently than all other values.

    6. i) 1 – (5/6)3 = 91/216

    ii) E(X) = 4.96
    Var(X) = 1.31

    7. i) A) 0.0320
    B) 0.9539
    C) 15.6

    ii) Let p = probability that randomly selected patient is cured by drug
    H0: p = 0.78
    H1: p > 0.78

    X ~ B(20, 0.78)

    P(X>=19) = 1 – P(X<=18) = 1 – 0.9539 = 0.0461

    0.0461 > 0.01
    therefore not significant (reject H1, accept H0)
    insufficient evidence to suggest that the new drug is more effective than the old

    iii) 0.0461 < 0.05
    therefore significant (accept H1, reject H0)
    at the 5% significance level there is sufficient evidence to suggest that new
    drug is more effective than the old

    8. i) IQR = 18.1 – 17.8 = 0.3
    Upper limit = 18.55
    Lower limit = 17.35

    Therefore no outliers at lower end since 17.4 is lowest value and is higher than
    17.35, but there is at least one outlier at upper end since 18.6 is highest value and
    is greater than upper limit of 18.55

    ii) A: 7/228 = 0.03070
    B: 1/19 = 0.05263

    iii) P(A|B) = 7/12 = 0.5833
    P(B|A = 1

    iv) 49/51984 = 0.0009426

    v) 3/38 = 0.07895

    I'm not confident of the answers here, there must be some errors. Please correct me!
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    (Original post by ZoeWoah)
    What percentage of as maths is statistics?
    33.3333...% I think if you're just doing C1, C2 and S1.
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    (Original post by otherdan)
    1. i) Mean x = 1499kWH; SD x = 419kWh [would have done 3sf for mean but felt uncomfortable rounding to 1500!]

    ii) Mean y = £258.90; SD y = £68.32

    2. i) A) 15/250 = 0.06
    B) 148/250 = 0.592

    ii) 15/67 = 0.2239

    3. i) 144/169 = 0.8521
    ii) 8.52

    4. i) 3268760
    ii) 1019304
    iii) (ii)/(i) from above = 0.3118

    5. i) 0 | 6
    1 | 5 8
    2 | 1 5 8
    3 | 1 1 3 5 8 9

    ii) Negative skewness

    iii) Median = 29.5; mean = 26.7; mode = 31
    The mode isn’t particularly useful because it only occurs one time more frequently than all other values, therefore it isn’t particularly significant. [I have no idea if this reasoning is valid!]

    6. i) 1 – (5/6)3 = 91/216

    ii) E(X) = 4.96
    Var(X) = 1.31

    7. i) A) 0.0.0320
    B) 0.9539
    C) 15.6

    ii) Let p = probability that randomly selected patient is cured by drug
    H0: p = 0.78
    H1: p > 0.78

    X ~ B(20, 0.78)

    P(X>=19) = 1 – P(X<=18) = 1 – 0.9539 = 0.0461

    0.0461 > 0.01
    therefore not significant, reject H1, accept H0
    insufficient evidence to suggest that the new drug is more effective than the old

    iii) 0.0461 < 0.05
    therefore significant, accept H1, reject H0
    at the 5% significance level there is sufficient evidence to suggest that new
    drug is more effective than the old

    8. i) IQR = 18.1 – 17.8 = 0.3
    Upper limit = 18.55
    Lower limit = 17.35

    Therefore no outliers at lower end since 17.4 is lowest value and is higher than
    17.35, but there is at least one outlier at upper end since 18.6 is highest value and
    is greater than upper limit of 18.55

    ii) A: 7/228 = 0.03070
    B: 1/19 = 0.05263

    iii) P(A|B) = 7/12 = 0.5833
    P(B|A = 1

    iv) 49/51984 = 0.0009426

    v) 3/38 = 0.07895

    I'm not confident of the answers here, there must be some errors. Please correct me!
    Exactly what I got! Bit of a typing error on 7i A) though?

    For the mean I said it would be useful because the data is unimodal, and the mode is also not skewed by the extreme low value of 6 as the mean is.

    The only other thing I would say is that in hypothesis tests you are not allowed to say 'accept ...', since all you are testing is which one it is most likely to be - you can't say for definite which one it is.
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    For 1i) I put 1499.3 and 419.1
    For 1ii) I put £289.9 and £68.32

    Will I lose any marks for over-specification???
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    (Original post by otherdan)
    1. i) Mean x = 1499kWH; SD x = 419kWh [would have done 3sf for mean but felt uncomfortable rounding to 1500!]

    ii) Mean y = £258.90; SD y = £68.32

    2. i) A) 15/250 = 0.06
    B) 148/250 = 0.592

    ii) 15/67 = 0.2239

    3. i) 144/169 = 0.8521
    ii) 8.52

    4. i) 3268760
    ii) 1019304
    iii) (ii)/(i) from above = 0.3118

    5. i) 0 | 6
    1 | 5 8
    2 | 1 5 8
    3 | 1 1 3 5 8 9

    Key: 1 | 5 = 15 people

    ii) Negative skewness

    iii) Median = 29.5; mean = 26.7; mode = 31
    The mode isn’t particularly useful because it only occurs one time more frequently than all other values, therefore it isn’t particularly significant. [I have no idea if this reasoning is valid!]

    6. i) 1 – (5/6)3 = 91/216

    ii) E(X) = 4.96
    Var(X) = 1.31

    7. i) A) 0.0.0320
    B) 0.9539
    C) 15.6

    ii) Let p = probability that randomly selected patient is cured by drug
    H0: p = 0.78
    H1: p > 0.78

    X ~ B(20, 0.78)

    P(X>=19) = 1 – P(X<=18) = 1 – 0.9539 = 0.0461

    0.0461 > 0.01
    therefore not significant, reject H1, accept H0
    insufficient evidence to suggest that the new drug is more effective than the old

    iii) 0.0461 < 0.05
    therefore significant, accept H1, reject H0
    at the 5% significance level there is sufficient evidence to suggest that new
    drug is more effective than the old

    8. i) IQR = 18.1 – 17.8 = 0.3
    Upper limit = 18.55
    Lower limit = 17.35

    Therefore no outliers at lower end since 17.4 is lowest value and is higher than
    17.35, but there is at least one outlier at upper end since 18.6 is highest value and
    is greater than upper limit of 18.55

    ii) A: 7/228 = 0.03070
    B: 1/19 = 0.05263

    iii) P(A|B) = 7/12 = 0.5833
    P(B|A = 1

    iv) 49/51984 = 0.0009426

    v) 3/38 = 0.07895

    I'm not confident of the answers here, there must be some errors. Please correct me!
    Your answers are the same as mine except for the mode (which I just argued the opposite way) and the last part of the last question, but I've decided I'm wrong and that answer is right :P
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    (Original post by Connorbwfc)
    For 1i) I put 1499.3 and 419.1
    For 1ii) I put £289.9 and £68.32

    Will I lose any marks for over-specification???
    Not typically, unless you go ott. For £/$ values I would go to 2 dp as a rule of thumb.
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    (Original post by iamJOM)
    Exactly what I got! Bit of a typing error on 7i A) though?

    For the mean I said it would be useful because the data is unimodal, and the mode is also not skewed by the extreme low value of 6 as the mean is.

    The only other thing I would say is that in hypothesis tests you are not allowed to say 'accept ...', since all you are testing is which one it is most likely to be - you can't say for definite which one it is.
    Yes, definitely a typing error! Will correct.

    I only commented on the mode and not the other averages. Hopefully that is okay.

    We were told to say accept or reject in our lessons, but I noticed the mark schemes seemed to vary, so I always put both. Hopefully that doesn't negate too many marks.
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    (Original post by Connorbwfc)
    For 1i) I put 1499.3 and 419.1
    For 1ii) I put £289.9 and £68.32

    Will I lose any marks for over-specification???
    You might for putting 1499.3, though it's not definite - often they say to penalise anything that is to 4sf and not a probability. The others should be alright though.
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    (Original post by Connorbwfc)
    For 1i) I put 1499.3 and 419.1
    For 1ii) I put £289.9 and £68.32

    Will I lose any marks for over-specification???
    The only potential problem there would be 1499.3 is 5sf and non-probabilities are usually given to 4sf. You might be fine though, as not all over-specification is penalised.
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    did anyone else say that the mode is useful because it's not affected by the negative skewness of the data?
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    (Original post by corey7695)
    did anyone else say that the mode is useful because it's not affected by the negative skewness of the data?
    Ha, uum... I said the mode was useful in this case in part because it is representative of the skew :P
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    omg thanks you so much I love u
    I genuinely thought I was going to get a D
    turns out I only lost 5 marks , that's if youre answers are correct
    I missed out the 91/216 question, so I lost 3 there
    for the variance I did the correct method, but I substracted the rounded mean, so my answer is less accurate so i got 1.29. how many marks would i lose? I should have known to not use the rounded mean to calculate variance.
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    Of course I would forget to put the £ sign in for part 2 with the new SD/mean.. Let the panic set in now.
 
 
 
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