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    Not totally sure if some of my answers are correct but i thought i'd share them before i forget
    1.) Swt = -49 Stt = 74.8333
    S ss = 50,000,000 Sst = -49000
    r = -0.801
    Regression line is suitable as strong negative linear correlation
    b = -0.655 a= 20.0
    s = 20000 - 655t
    For an increase in 1 degrees sales decrease by £655

    Venn diagram Q : p= 0.15 q= 0.1 r=0.22 s= 0.28 t= 0.17
    Events were not independent
    For the given probability 0.43 / 0.75

    Measures of location Q :
    Width = 4 cm height = 1.5 cm
    Median : 49 mean : 49.1 SD : 10.32
    Almost no skew as mean and median take similar values , or slight positive skew as mean > median
    Normal distribution is suitable
    80th percentile = 58.416

    First discrete random variables Q :
    a = 1/3 b= 1/4 c = 1/6 d= 7/12 e = 1
    P( x^2 = 1 ) = 1/3 + 1/4 = 7/12

    Normal distribution Q :
    0.3446 for part a.)
    Can only remember my answer for part c.) Was 19.75

    Last discrete random variables Q :
    P= 1 /3 so P(2) = 2/3
    Only way of getting p(s=30 ) = 6x5 , multiply these probabilities from first and second table together
    For distribution of S there were 6 outcomes including 30, 25 , 15 , 12 , 6 ,4 ( i think???)
    Charlotte
    Feel free to comment your answers!
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    Ahhhh i literally recognise about three of yours answer. Looking forward to my non existent A grade
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    For the last question I got for the probability outcomes 0,6,12,15,30
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    (Original post by ErinMei)
    Not totally sure if some of my answers are correct but i thought i'd share them before i forget
    1.) Swt = -49 Stt = 78.3
    S ss = 50,000,000 Sst = -49000
    r = -0.801
    Regression line is suitable as strong negative linear correlation
    b = -0.655 a= 20.0
    s = 20000 - 655b
    For an increase in 1 degrees sales decrease by £655

    Venn diagram Q : p= 0.15 q= 0.1 r=0.22 s= 0.28
    Events were not independent
    For the given probability 0.43 / 0.75

    Measures of location Q :
    Width = 4 cm height = 1.5 cm
    Median : 49 mean : 49.1 SD : 10.32
    Almost no skew as mean and median take similar values , or slight positive skew as mean > median
    Normal distribution is suitable

    First discrete random variables Q :
    a = 1/3 b= 1/4 c = 1/6 d= 7/12 e = 1
    P( x^2 = 1 ) = 1/3 + 1/4 = 7/12

    Normal distribution Q :
    Can only remember my answer for part c.) Was 19.75

    Last discrete random variables Q :
    P= 1 /3 so P(2) = 2/3
    Only way of getting p(s=30 ) = 6x5 , multiply these probabilities from first and second table together
    For distribution of S there were 6 outcomes including 30, 25 , 15 , 12 , 6 ,4 ( i think???)
    Feel free to comment your answers!
    i do not agree with your regression line , it should be s=20000-655t, but I agree with everything else, also last question, it was charlotte because she had a higher expected score ( E(X^2) is greater than E( S)
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    (Original post by ihatehannah)
    i do not agree with your regression line , it should be s=20000-655t,
    Woops i meant to put that 😜

    Posted from TSR Mobile
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    I got that 655 for b but I hear some people are saying 900 something. Can you remember the working out because I can't
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    Thanks! Got nearly all of the same answers - did you get Charlotte winning at the end because E(x^2) > E(S)
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    how can your regression line be -655 when b is Sst/ Sss and that gives like -980????
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    (Original post by Chris_isaacs)
    For the last question I got for the probability outcomes 0,6,12,15,30
    I may be wrong but pretty sure 0 wasn't one of them
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    Did anyone else get charlotte for last part??
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    (Original post by ErinMei)
    Not totally sure if some of my answers are correct but i thought i'd share them before i forget
    1.) Swt = -49 Stt = 78.3
    S ss = 50,000,000 Sst = -49000
    r = -0.801
    Regression line is suitable as strong negative linear correlation
    b = -0.655 a= 20.0
    s = 20000 - 655t
    For an increase in 1 degrees sales decrease by £655

    Venn diagram Q : p= 0.15 q= 0.1 r=0.22 s= 0.28
    Events were not independent
    For the given probability 0.43 / 0.75

    Measures of location Q :
    Width = 4 cm height = 1.5 cm
    Median : 49 mean : 49.1 SD : 10.32
    Almost no skew as mean and median take similar values , or slight positive skew as mean > median
    Normal distribution is suitable

    First discrete random variables Q :
    a = 1/3 b= 1/4 c = 1/6 d= 7/12 e = 1
    P( x^2 = 1 ) = 1/3 + 1/4 = 7/12

    Normal distribution Q :
    Can only remember my answer for part c.) Was 19.75

    Last discrete random variables Q :
    P= 1 /3 so P(2) = 2/3
    Only way of getting p(s=30 ) = 6x5 , multiply these probabilities from first and second table together
    For distribution of S there were 6 outcomes including 30, 25 , 15 , 12 , 6 ,4 ( i think???)
    Feel free to comment your answers!
    Think i dropped a total of 5-6 marks unless i get method marks
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    For the yuto did you do p (median >18 | TIME > 15) = 0.5?
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    (Original post by stephaniecheung)
    how can your regression line be -655 when b is Sst/ Sss and that gives like -980????
    Because they wanted it for w on t, not s on t
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    (Original post by ErinMei)
    Not totally sure if some of my answers are correct but i thought i'd share them before i forget
    1.) Swt = -49 Stt = 78.3
    S ss = 50,000,000 Sst = -49000
    r = -0.801
    Regression line is suitable as strong negative linear correlation
    b = -0.655 a= 20.0
    s = 20000 - 655t
    For an increase in 1 degrees sales decrease by £655

    Venn diagram Q : p= 0.15 q= 0.1 r=0.22 s= 0.28
    Events were not independent
    For the given probability 0.43 / 0.75

    Measures of location Q :
    Width = 4 cm height = 1.5 cm
    Median : 49 mean : 49.1 SD : 10.32
    Almost no skew as mean and median take similar values , or slight positive skew as mean > median
    Normal distribution is suitable

    First discrete random variables Q :
    a = 1/3 b= 1/4 c = 1/6 d= 7/12 e = 1
    P( x^2 = 1 ) = 1/3 + 1/4 = 7/12

    Normal distribution Q :
    Can only remember my answer for part c.) Was 19.75

    Last discrete random variables Q :
    P= 1 /3 so P(2) = 2/3
    Only way of getting p(s=30 ) = 6x5 , multiply these probabilities from first and second table together
    For distribution of S there were 6 outcomes including 30, 25 , 15 , 12 , 6 ,4 ( i think???)
    Feel free to comment your answers!
    I agree with everything
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    I agree with those answers
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    (Original post by stephaniecheung)
    how can your regression line be -655 when b is Sst/ Sss and that gives like -980????
    It is
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    I had Sam got a higher score from E(S)> E(X^2)? maybe I just calculated it wrong
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    I got 0.3446 for 5.a) and 0.4749 5.b)
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    I got almost all your answers
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    So I definitely know I got q1 b and q5 c wrong but I got everything else right.

    That would be pretty much 0.906

    Now I'm just hoping for low grade boundaries. Thanks for the answers
 
 
 
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