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    (Original post by shuu00)
    What were the values for p and q
    q = 0.15
    p = 0.175
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    did anyone get 0.81 and 0.405 for the P(DIBnD) and P(BnD')??
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    (Original post by veldt127)
    q = 0.15
    p = 0.175
    I got 0.22 and 0.03
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    (Original post by NickDacre)
    I got 0.22 and 0.03
    This is for discrete random variables. T and U were 0.03 and 0.22 for the probability question
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    (Original post by googleit)
    The mean DECREASED guys, not stayed the same, we're dealing with classes here. The value of 3.43 will add one to the frequency of the 3-3.5 class. The way we work out the estimate is to multiply the midpoint of the class by the frequency. Hence, since the midpoint of this class is 3.25, which is lower than the mean, the mean decreases.
    You could have also done the actual calculation to check, the mean decreased

    EDIT: I'll explain this better. The mean before the added values was 3.43. Now there's an added value of 3.43. This adds one to the frequency of 3-3.5 class. In working out the value of the mean, you have to multiply mid point by frequency. Hence, you don't actually deal with 3.43 but rather 3.25, the mid point of the class. Since this value is less than 3.43 the mean decreases.
    Try the calculation out for yourself and please amend the unofficial mark schemes.
    But the question said 'no further calculations' so I imagine it wants you to treat the 3.43 how it is when adding to the data and not include the tables, which would keep the mean the same. I think your method, then doing the same for Standard Deviation and explaining them both would be too much for 3 marks.
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    (Original post by veldt127)
    q = 0.15
    p = 0.175
    thanks
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    (Original post by NickDacre)
    did anyone get 0.81 and 0.405 for the P(DIBnD) and P(BnD'??
    I got P(D|BnD) = 0.45
    P(D|BnD') = 15/37
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    The DRV question (the final bit with the E(R), etc) I got these answers (in this order)

    E(r) = 0.45
    Prob (whoever was the first one) = 0.475
    Prob (whoever was the second one) = 0.375

    I am 100% sure on these answers
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    (Original post by hogree)
    Remember you did that one yesterday didn't you lol. Time is a naughty one to throw at you.

    Posted from TSR Mobile
    Yep!! I saw it and thought yess!! I remembered the one yesterday
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    anyone got the paper??
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    How to get 33 customers model answer!

    P(D|RnB) = 0.45 and P(D|RnB') = 15/37

    It says that 40 customers have rented a room and eaten breakfast
    It also says that 37 have rented a room and not had breakfast

    Well, we worked out the chance of them having dinner given those conditions in the two previous parts, so all we needed to do was 40*0.45 + 37 * 15/37 to come to our answer of 18+15 = 33 customers.

    Hope this clears it up
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    I got P=0.1626 and Q=0.175 no idea how !! 😰😰 how many marks would I get and also would I lose marks on the rest of the question or get follow through ????
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    I have a question about the Sarah and Rebecca winnings one.

    Basically how do you do it

    Do they each spin individually or is it 1 spin only tat determine both of their scores if that makes sense?

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    (Original post by thelegend99)
    Wahoo! Thanks for your help from yesterday as well!

    How did it go for you?
    That's alright. Think it was a pretty rough paper. No coding after all that revision lol. I think I've probably dropped 5-10, if I misunderstood the Sarah question. So pretty well thanks

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    (Original post by stressed16)
    I got P=0.1626 and Q=0.175 no idea how !! 😰😰 how many marks would I get and also would I lose marks on the rest of the question or get follow through ????
    I think you may lose a lot because maybe you'd of gained follow through had the probabilities all add up to one (which was a simple check to see you're along the right lines) but I don't think your values do
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    (Original post by iMacJack)
    How to get 33 customers model answer!

    P(D|RnB) = 0.45 and P(D|RnB' = 15/37

    It says that 40 customers have rented a room and eaten breakfast
    It also says that 37 have rented a room and not had breakfast

    Well, we worked out the chance of them having dinner given those conditions in the two previous parts, so all we needed to do was 40*0.45 + 37 * 15/37 to come to our answer of 18+15 = 33 customers.

    Hope this clears it up
    do you remember the numbers in the venn diagram?
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    (Original post by hogree)
    That's alright. Think it was a pretty rough paper. No coding after all that revision lol. I think I've probably dropped 5-10, if I misunderstood the Sarah question. So pretty well thanks

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    Hiw did you do Sarah question?

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    (Original post by Themathgeek)
    I have a question about the Sarah and Rebecca winnings one.

    Basically how do you do it

    Do they each spin individually or is it 1 spin only tat determine both of their scores if that makes sense?

    Posted from TSR Mobile
    I dont remember my answer. I understood the question last minute. Can someone tell me if i was correct. It said a computer randomly generates their score so that means you look at the 2 charts and compare when Rachel had a higher score than the other one. Then you just add the probabilities together.
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    (Original post by Raees_Sharif)
    do you remember the numbers in the venn diagram?
    0.22- 0.03- 0.15- 0(on outside)
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    (Original post by Raees_Sharif)
    do you remember the numbers in the venn diagram?
    Unfortunately not, sorry
 
 
 
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