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    1) peice of data from 9 people. data mean was 35.5 but one of the 9 people gave a value of 42 instead of 32

    a) write down effect - obv the mean will increase since 42 is > mean

    b) calculate new mean


    so i thought b) was 32x8 + 42 divide by 9 but its not. confused!
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    (Original post by seconda1)
    1) peice of data from 9 people. data mean was 35.5 but one of the 9 people gave a value of 42 instead of 32

    a) write down effect - obv the mean will increase since 42 is > mean

    b) calculate new mean


    so i thought b) was 32x8 + 42 divide by 9 but its not. confused!
    We have 35.5 = \dfrac{T_8 + 42}{9} where T_8 is the total data from 8 people and the 42 is the leftover data from the ninth person who got it wrong.

    You want to use this equation to determine the value of \dfrac{T_8+32}{9}
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    (Original post by RDKGames)
    We have 35.5 = \dfrac{T_8 + 42}{9} where T_8 is the total data from 8 people and the 42 is the leftover data from the ninth person who got it wrong.

    You want to use this equation to determine the value of \dfrac{T_8+32}{9}
    im confused was is t?
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    (Original post by seconda1)
    im confused was is t?
    I've just explained it. It's a variable to denote the total of the other 8 people.

    Like y'know, when calculating the mean of like 4 people you'd do something like \dfrac{1+2+3+4}{4} but you can freely take the sum of the first three numbers 1+2+3 = 6 and just rewrite the mean as  \dfrac{6+4}{4}. Same idea here, except we do not know the mean of the 8 people so we just denote it by T_8
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    (Original post by RDKGames)
    I've just explained it. It's a variable to denote the total of the other 8 people.

    Like y'know, when calculating the mean of like 4 people you'd do something like \dfrac{1+2+3+4}{4} but you can freely take the sum of the first three numbers 1+2+3 = 6 and just rewrite the mean as  \dfrac{6+4}{4}. Same idea here, except we do not know the mean of the 8 people so we just denote it by T_8
    still confused

    the mark scheme says 9x35.5 -10 / 9
    why is it -10?
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    (Original post by seconda1)
    still confused

    the mark scheme says 9x35.5 -10 / 9
    why is it -10?
    Because if you take the eq. I said and proceed to say

    \begin{aligned} 35.5 & = \frac{T_8 + 42}{9} \\ & = \frac{T_8 + 32 + 10}{9} \\ & = \frac{T_8+32}{9} + \dfrac{10}{9}

    then just subtract 10/9 from both sides to get the answer they're looking for.
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    (Original post by RDKGames)
    Because if you take the eq. I said and proceed to say

    \begin{aligned} 35.5 & = \frac{T_8 + 42}{9} \\ & = \frac{T_8 + 32 + 10}{9} \\ & = \frac{T_8+32}{9} + \dfrac{10}{9}

    then just subtract 10/9 from both sides to get the answer they're looking for.
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    so the data was said to have found a mean of 35.5 and then it says one person out of the 9 had the 42 instead of 32 so cant you work out the mean of 8 people because its 32x8/8?

    cant seem to get my head around this
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    (Original post by seconda1)

    so the data was said to have found a mean of 35.5 and then it says one person out of the 9 had the 42 instead of 32 so cant you work out the mean of 8 people because its 32x8/8?

    cant seem to get my head around this
    Why would you want the mean of 8 people? Also no, the mean of the 8 people is given by \dfrac{T_8}{8} = \dfrac{9\cdot 35.5-42}{8} instead.

    Anyway, I don't know how else to explain it so I'll leave it to someone else. All I can do is repeat myself by saying that each person has data values of a_1, a_2, a_3, ..., a_8, 42 initially. We don't know the values of the first 8 people so we just denote them by a's. Then the mean is given by \dfrac{a_1 + a_2 + a_3 + ... + a_8 + 42}{9} = 35.5 however we know that the 42 was incorrectly entered instead of 32 therefore we must find the value of \dfrac{a_1+a_2+a_3 + ... a_8 + 32}{9} instead. I simply denoted the sum a_1+ a_2 + ... + a_8 = T_8 since we don't need to do anything with it and it takes up too much space.
 
 
 
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