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    did you sneak it out of exam hall or are you a teacher :O
    itd be better if u post the whole thing
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    (Original post by Attz_09)
    hoa post picture or email them me i need to find out how i dun can get my bro to do the maths and then feel bad about how **** i dun
    they're up mate
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    I need the tree diagram if you dont mind
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    (Original post by Attz_09)
    did you sneak it out of exam hall or are you a teacher :O
    itd be better if u post the whole thing
    posted the whole thing just in 3 different posts by accident haha. Nope our school lets us take them out
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    do it landscape plz i have to tilt my head sideways and the shade makes it difficult to read black ink pleasee
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    (Original post by Attz_09)
    UNOFFICIAL MARK SCHEME

    1i) mean = 249.4, standard deviation = 14.5 [3]
    ii) new mean = 209.46, new standard deviation = 13.0 [3]

    2i) P(2 women, 1 man) = (5C2x5C1)/10C3 = 5/12 [4]
    ii) P(four evenings with 2 women, 1 man) = (5/12)^4 + 4C3(5/12)^3(7/12) = 0.1989 [4]

    3i) P(5underweight bags) = 50C5(0.1)^5(0.9)^45 = 0.1849 [3]
    ii) P(at least one underweight bag in 20) = 1-(0.9)^20 = 0.8784 [2]
    iii) E(X) = np = 48x0.8784=42.2 [2]

    4i) 3! x 1/6 x 1/5 x 1/4 = 1/20 [2]
    ii) E(X)= sum xP(X=x) = £1507.50, Var(X)=E(X^2)-E(X)^2 = £445511.25 [5]

    5i) Because the researcher has assumed that the probability that a randomly selected person makes a correct identification is 0.5, as she assumes that there is a 50:50 chance of guessing the difference between tap and bottled water [2]
    ii) Because the point of the test is to determine if the probability that a randomly selected person makes a correct identification is in fact greater than 0.5 [1]
    iii) P(X≥13)=1-P(X≤12)=1-0.8684=0.1316>0.05 Therefore, 13 is not a member of the critical region and it is not significant. We accept H0. There is insufficient evidence from the data to suggest the probability that a randomly selected person makes a correct identification is greater than 0.5 [5]

    6i) Median=3.32, LQ=(2.82+2.84)/2=2.83, UQ=(3.70+3.72)/2=3.71. IQR=0.88 [3]
    ii) Draw box and whisker plot [3]
    iii) 1step = 1.5xIQR=1.32. UQ+1.32=5.03, therefore no outliers above. LQ-1.32=1.51. Therefore 1.39 is an outlier below. It shouldn't be removed as there is no evidence to suggest that it forms a separate pool of data or is not a genuine data item. [4]
    iv) Median=3.5. UQ=3.84, LQ=3.13. IQR=0.71 [3]
    v) Males have a higher average weight (higher median). Males have less variation in birth weight (lower interquartile range). [2]
    vi) (8/200)x(7/199). You can probably use any values from 7-10/200 (it depends how you read the graph). [3]

    7i) Draw probability tree diagram [4]
    iiA) 1-0.9x0.95x0.95=0.1878 (4 d.p.) [3]
    iiB) 0.1x0.8x0.95+0.9x0.05x0.8+0.9x0. 95x0.05=0.1548 (4 d.p.) [4]
    iii) 0.1548/0.1878=0.8242 [3]
    iv) P(Hit,Hit,Hit) + P(Miss,Miss,Miss,Hit,Hit,Hit) = 0.0056 (4 d.p.) [4]


    bold is the things am unsure/ lost marks on

    im not sure if you did 3ii right because if i remember correctly it said number of bags in box 0.1 faulty. and 48 boxes in crate. so i did 20 times 0.1 which =2
    then done 2 time 48
    Nope i'm sorry, you needed to find the probability that it's greater than 1, which is 1-P(0 underweight bags)
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    (Original post by hoa123)
    posted the whole thing just in 3 different posts by accident haha. Nope our school lets us take them out
    thats wicked i wouldnt be waiting for an unnofficial mark scheme if i had the paper. do u live in liverpool by any chance lol
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    (Original post by Attz_09)
    do it landscape plz i have to tilt my head sideways and the shade makes it difficult to read black ink pleasee
    cbaaaaaaaa
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    (Original post by Attz_09)
    thats wicked i wouldnt be waiting for an unnofficial mark scheme if i had the paper. do u live in liverpool by any chance lol
    nope, newcastle haha
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    No 7 tree diagram page? Thank you.
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    (Original post by hoa123)
    Nope i'm sorry, you needed to find the probability that it's greater than 1, which is 1-P(0 underweight bags)
    i meant iii sorry it said 20 bags in a box 48 box in crate
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    (Original post by Attz_09)
    i meant iii sorry it said 20 bags in a box 48 box in crate
    yeah so you have P(≥1 underweight bag in a box)=0.8784 from ii

    Then iii asks for the number of boxes with ≥1underweight bags. So you do E(X)=np=48x0.8784
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    (Original post by hoa123)
    cbaaaaaaaa
    Hey any chance you could post the info for me to draw the tree diagram again?
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    (Original post by Blobar)
    No 7 tree diagram page? Thank you.
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    (Original post by hoa123)
    Name:  IMG_2159.jpg
Views: 245
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    Sorry it's upside down haha
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    its great you posting all these pictures and everthing but am turning my laptop screen sideways and upside down to read it haha
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    (Original post by Blobar)
    Hey any chance you could post the info for me to draw the tree diagram again?
    probability of bull's-eye with first throw is 0.1, for each subsequent throw it's 0.2 if she hit with the previous dart and 0.05 if she missed with the previous dart
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    (Original post by Attz_09)
    its great you posting all these pictures and everthing but am turning my laptop screen sideways and upside down to read it haha
    sorry haha
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    (Original post by hoa123)
    UNOFFICIAL MARK SCHEME

    1i) mean = 249.4, standard deviation = 14.5 [3]
    ii) new mean = 209.46, new standard deviation = 13.0 [3]

    2i) P(2 women, 1 man) = (5C2x5C1)/10C3 = 5/12 [4]
    ii) P(four evenings with 2 women, 1 man) = (5/12)^4 + 4C3(5/12)^3(7/12) = 0.1989 [4]

    3i) P(5underweight bags) = 50C5(0.1)^5(0.9)^45 = 0.1849 [3]
    ii) P(at least one underweight bag in 20) = 1-(0.9)^20 = 0.8784 [2]
    iii) E(X) = np = 48x0.8784=42.2 [2]

    4i) 3! x 1/6 x 1/5 x 1/4 = 1/20 [2]
    ii) E(X)= sum xP(X=x) = £1507.50, Var(X)=E(X^2)-E(X)^2 = £445511.25 [5]

    5i) Because the researcher has assumed that the probability that a randomly selected person makes a correct identification is 0.5, as she assumes that there is a 50:50 chance of guessing the difference between tap and bottled water [2]
    ii) Because the point of the test is to determine if the probability that a randomly selected person makes a correct identification is in fact greater than 0.5 [1]
    iii) P(X≥13)=1-P(X≤12)=1-0.8684=0.1316>0.05 Therefore, 13 is not a member of the critical region and it is not significant. We accept H0. There is insufficient evidence from the data to suggest the probability that a randomly selected person makes a correct identification is greater than 0.5 [5]

    6i) Median=3.32, LQ=(2.82+2.84)/2=2.83, UQ=(3.70+3.72)/2=3.71. IQR=0.88 [3]
    ii) Draw box and whisker plot [3]
    iii) 1step = 1.5xIQR=1.32. UQ+1.32=5.03, therefore no outliers above. LQ-1.32=1.51. Therefore 1.39 is an outlier below. It shouldn't be removed as there is no evidence to suggest that it forms a separate pool of data or is not a genuine data item. [4]
    iv) Median=3.5. UQ=3.84, LQ=3.13. IQR=0.71 [3]
    v) Males have a higher average weight (higher median). Males have less variation in birth weight (lower interquartile range). [2]
    vi) (8/200)x(7/199). You can probably use any values from 7-10/200 (it depends how you read the graph). [3]

    7i) Draw probability tree diagram [4]
    iiA) 1-0.9x0.95x0.95=0.1878 (4 d.p.) [3]
    iiB) 0.1x0.8x0.95+0.9x0.05x0.8+0.9x0. 95x0.05=0.1548 (4 d.p.) [4]
    iii) 0.1548/0.1878=0.8242 [3]
    iv) P(Hit,Hit,Hit) + P(Miss,Miss,Miss,Hit,Hit,Hit) = 0.0056 (4 d.p.) [4]
    With the last part, i got 0.0085.

    P( Hit, Hit, Hit ) = 0.1 x 0.2 x 0.2 = 0.004

    P ( Miss, Miss, Miss, Hit, Hit, Hit ) = 0.9 x 0.5 x 0.5 x 0.5 x 0.2 x 0.2 = 0.0045

    0.004 + 0.0045 = 0.0085

    ?
    How is it 0.0056 ?
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    (Original post by RomanMarkOliver)
    With the last part, i got 0.0085.

    P( Hit, Hit, Hit ) = 0.1 x 0.2 x 0.2 = 0.004

    P ( Miss, Miss, Miss, Hit, Hit, Hit ) = 0.9 x 0.5 x 0.5 x 0.5 x 0.2 x 0.2 = 0.0045

    0.004 + 0.0045 = 0.0085

    ?
    How is it 0.0056 ?
    Hit, hit, hit = 0.1x0.2x0.2 yes
    Miss, miss, miss, hit, hit, hit = 0.9 x 0.95 x 0.95 x 0.05 x 0.2 x 0.2
 
 
 
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