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    Feel free to post what you got for May 25th 2016 mei ocr s1
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    Someone relieve us of the answers for the last question
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    this is what i got:
    1i). For the stem and leaf diagram. Median = 29g
    Q1= 27.8 g
    Q2= 31.8g
    IQR = 4g
    ii) 1.5 x 4= 6
    31.8 + 6= 37.8g
    27.8-6= 21.8g
    So, no outliers.

    2i) The probability that team A does not lose is 0.5 * 0.5 * 0.5 = 0.125
    ii) 0.125 + (0.2)^3 +(0.3)^3=
    iii) 3(0.3*0.3*0.2) + 3(0.3*0.3*0.5) + 3(0.5*0.5*0.3) +3(0.5*0.5*0.2) etc you get the gist...

    3i) 5!= 120
    ii) 5P3= 60
    iii) 5C3= 10

    anyone wants to add to this so far?
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    sure! the last question was

    X~B( 20??? ,0.1)
    i) P(X=3) = 20C3 * (0.1)^3 * (0.9)^17
    ii) P(X less that or equal to 2)=
    use the tables
    iii) 0.1 * 20 =
    iv)Ho : p= 0.1 H1 : p<0.1v) there would be no critical region for that value of 20. 20 is too small. 12.16 % would have to be the significance levelvi) X~B (65,0.1)2 people use the pin 1234P(X less than or equal to 2)= 0.0531 > 5 percent Accept the Ho. There's insuffient evidence to suggest number of people using 1234 as PIN lowered.anyone want to add to this?
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    Yeah i got similar to that...Dont remember much...hopefully all good
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    (Original post by chocolatecookieM)
    this is what i got:
    1i). For the stem and leaf diagram. Median = 29g
    Q1= 27.8 g
    Q2= 31.8g
    IQR = 4g
    ii) 1.5 x 4= 6
    31.8 + 6= 37.8g
    27.8-6= 21.8g
    So, no outliers.

    2i) The probability that team A does not lose is 0.5 * 0.5 * 0.5 = 0.125
    ii) 0.125 + (0.2)^3 +(0.3)^3=
    iii) 3(0.3*0.3*0.2) + 3(0.3*0.3*0.5) + 3(0.5*0.5*0.3) +3(0.5*0.5*0.2) etc you get the gist...

    3i) 5!= 120
    ii) 5P3= 60
    iii) 5C3= 10

    anyone wants to add to this so far?
    Alright so pretty sure question 1 for me is wrong lol
    But for your part 2, I don't think you've taken into account them drawing...I did 1-(0.2*0.2*0.2) = 0.992
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    For 2 i) did it not ask for the probability that the team doesn't lose at all in any of the 3 matches? So didn't it require more working out than that? As its the probabilities that it's win win draw or win draw draw etc.
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    And I think for question 3 i) and ii) they wanted probability's so the answer can't be bigger than 1
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    (Original post by chocolatecookieM)
    2i) The probability that team A does not lose is 0.5 * 0.5 * 0.5 = 0.125
    no you have to include the probability to draw also
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    Question 2 part i, the probability of not losing is 1-0.2=0.8, so not losing would be 0.8 cubed....
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    (Original post by chocolatecookieM)
    sure! the last question was

    X~B( 20??? ,0.1)
    i) P(X=3) = 20C3 * (0.1)^3 * (0.9)^17
    ii) P(X less that or equal to 2)=
    use the tables
    iii) 0.1 * 20 =
    iv)Ho : p = 0.1 H1 : p<0.1v) there would be no critical region for that value of 20. 20 is too small. 12.16 % would have to be the significance levelvi) X~B (65,0.1)2 people use the pin 1234P(X less than or equal to 2)= 0.0531 > 5 percent Accept the Ho. There's insuffient evidence to suggest number of people using 1234 as PIN lowered.anyone want to add to this?
    i) A) X~B (16,0.1), 16C3 * 0.1^3 * 0.9^13 = 0.1423

    i) B) ATLEAST 3 so 1 - P(x<=2) = 1- 0.7892 = 0.2108

    i) C) 0.1*16 = 1.6, 2


    ii) H0=0.1, H1<0.1, p is the probability of a person having 1234 as their PIN. H1 has this form as we are testing to see if the number of people using the PIN 1234 has reduced making it less likely to find someone with the PIN 1234


    iii) A) X~B (20,0.1) using a significance level of 10% for a lower tail test will not work since P(X<=0) is greater than 10%, so even if no one uses 1234 we'd still be accepting H1


    iv) P(X<=2)=0.0530 > 5% so accept Ho as 2 is not in the critical region (0 & 1) for the lower tail AND/OR accept H0 as there is no significant evidence to show there is a reduced amount of people using 1234 as their PIN so reject H1/accept H0
    (for full marks I believe you'd need to APPLY it to the scenario as you've done since there is no other place for the 4 marks to be allocated)
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    (Original post by malderson999)
    Alright so pretty sure question 1 for me is wrong lol
    But for your part 2, I don't think you've taken into account them drawing...I did 1-(0.2*0.2*0.2) = 0.992
    yeah...for part 2 instead of doing 3 multiplied by all outcomes, I did 1 - the probability of all 3 and 3 times the probability of getting each one once
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    (Original post by Haneenamer_)
    And I think for question 3 i) and ii) they wanted probability's so the answer can't be bigger than 1
    I think the numbers were right but it was 1 over

    so, it would be 1/60 and 1/10
    I remember doubting myself but there is another method that gets you these numbers too
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    Name:  Screen Shot 2016-05-25 at 16.14.46.png
Views: 972
Size:  34.6 KBAttachment 538369538371Attachment 538369538371538370
    Attached Images
      
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    For the last question, I looked up p(x=0) in the table when x~B (20, 0.1) and the number was bigger than 0.1, something like 0.16??, the number is in the formula book.
    Because the probability of getting 0, so no people having a pin of 1234 was bigger than the 0.1 significant level.
    I think that means that we couldn't necessarily reject the h0 at a 10% sig fig level????
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    (Original post by chocolatecookieM)
    sure! the last question was

    X~B( 20??? ,0.1)
    i) P(X=3) = 20C3 * (0.1)^3 * (0.9)^17
    ii) P(X less that or equal to 2)=
    use the tables
    iii) 0.1 * 20 =
    iv)Ho : p= 0.1 H1 : p<0.1v) there would be no critical region for that value of 20. 20 is too small. 12.16 % would have to be the significance levelvi) X~B (65,0.1)2 people use the pin 1234P(X less than or equal to 2)= 0.0531 > 5 percent Accept the Ho. There's insuffient evidence to suggest number of people using 1234 as PIN lowered.anyone want to add to this?
    13%, it asked for an integer
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    (Original post by maarz)
    For the last question, I looked up p(x=0) in the table when x~B (20, 0.1) and the number was bigger than 0.1, something like 0.16??, the number is in the formula book.
    Because the probability of getting 0, so no people having a pin of 1234 was bigger than the 0.1 significant level.
    I think that means that we couldn't necessarily reject the h0 at a 10% sig fig level????
    It means H0 would never have enough evidence to be rejected
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    His graph shows a negative skew.… hmm?
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    (Original post by Daniel9998)
    no you have to include the probability to draw also
    Thats what I thought when i first did the question i put 0.125 but then read it again and added the probability to draw so hopefully that is right
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    For the hypothesis testing question, i still don't understand how to worked out that 'k' was 13????
    also what p(x<=???) would you write down to compare it to the significance level % to reject the null hypothesis?? as you would need to compare it at the end, right????
    I'm so confused!!!!
    Please can somebody explain it to me.
    Thanks!
 
 
 
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