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    I just don't know how to go about answering this question. I've played around with it a bit and seem to be getting no where. Would appreciate it if someone could help me out.

    The question:
    The mean of the IQ of a class of 9 students is 121 and the variance is 226. Another student joins the class and the variance changes to 239.4 .
    What are the possible values of the IQ of the new student?
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    Stuck on the same question! Did you figure it out?
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    (Original post by Ray09)
    Stuck on the same question! Did you figure it out?
    Have you tried writing out the variance formula? Please post what you've tried.
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    (Original post by Notnek)
    Have you tried writing out the variance formula? Please post what you've tried.
    So I wrote that the variance is sigma x^2 -121^2 = 226 so sigma x^2 = (226+121^2) x 9 so sigma x^2 = 122803. I also just wrote down that the total IQ is 121 x 9 so sigma x is 1089. These are all the calculations I managed to do with the first set of data.

    I then wrote down that new variance is sigma x^2 (different from the previous one) divided by 10 (as there are now 10 valuues) - (sigma x divided by 10)^2 is equal to 239.4.

    I don't know where to go from there
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    (Original post by Ray09)
    So I wrote that the variance is sigma x^2 -121^2 = 226 so sigma x^2 = (226+121^2) x 9 so sigma x^2 = 122803. I also just wrote down that the total IQ is 121 x 9 so sigma x is 1089. These are all the calculations I managed to do with the first set of data.

    I then wrote down that new variance is sigma x^2 (different from the previous one) divided by 10 (as there are now 10 valuues) - (sigma x divided by 10)^2 is equal to 239.4.

    I don't know where to go from there
    Call the new student's IQ y.

    Then the new sum of the squares is going to be \Sigma x^2 + y^2. Does this make sense?

    See if you can continue from here.
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    (Original post by Notnek)
    Call the new student's IQ y.

    Then the new sum of the squares is going to be \Sigma x^2 + y^2. Does this make sense?

    See if you can continue from here.
    That does make sense! I have tried but I don't know whether I am approaching the question incorrectly?

    So I rewrote out the variance as (sigma x^2 + y^2)/10 - (1089 + y)/10 is equal to 239.4 so (sigma x^2 + y^2) - (1089 + y) is equal to 2394. As you can see above, from my previous calculations I got the initial sum of squares as 133803 so I substituted this in. So 133803 + y^2 - 1089 - y = 2394 so y^2 - y = -130320 but as soon as I got a negative I was pretty sure I am doing it incorrectly?
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    (Original post by Ray09)
    That does make sense! I have tried but I don't know whether I am approaching the question incorrectly?

    So I rewrote out the variance as (sigma x^2 + y^2)/10 - (1089 + y)/10 is equal to 239.4 so (sigma x^2 + y^2) - (1089 + y) is equal to 2394. As you can see above, from my previous calculations I got the initial sum of squares as 133803 so I substituted this in. So 133803 + y^2 - 1089 - y = 2394 so y^2 - y = -130320 but as soon as I got a negative I was pretty sure I am doing it incorrectly?
    At first glance I can't see anything wrong but that quadratic doesn't have solutions. I recommend trying it again and see if you can correct your mistake.

    I can try the question myself a little bit later but I'm trying to do a few things at once right now - sorry for the delayed replies
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    (Original post by 1Lucia)
    I just don't know how to go about answering this question. I've played around with it a bit and seem to be getting no where. Would appreciate it if someone could help me out.

    The question:
    The mean of the IQ of a class of 9 students is 121 and the variance is 226. Another student joins the class and the variance changes to 239.4 .
    What are the possible values of the IQ of the new student?
    See if you can calculate various statistics. For example, the sum of the original class's IQs is 9x126. See if you can find the sum of the squares of the IQs both before and after the 10th student joins.
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    (Original post by Notnek)
    At first glance I can't see anything wrong but that quadratic doesn't have solutions. I recommend trying it again and see if you can correct your mistake.

    I can try the question myself a little bit later but I'm trying to do a few things at once right now - sorry for the delayed replies
    I shall try again later as well! Don't worry about the delayed replies; I very much appreciate the help you've given already
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    (Original post by Ray09)
    That does make sense! I have tried but I don't know whether I am approaching the question incorrectly?

    So I rewrote out the variance as (sigma x^2 + y^2)/10 - (1089 + y)/10 is equal to 239.4 so (sigma x^2 + y^2) - (1089 + y) is equal to 2394. As you can see above, from my previous calculations I got the initial sum of squares as 133803 so I substituted this in. So 133803 + y^2 - 1089 - y = 2394 so y^2 - y = -130320 but as soon as I got a negative I was pretty sure I am doing it incorrectly?
    Approach is good, but:

    Var(X)=\mathbb{E}(X^2)-(\mathbb{E}(X))^2

    You seem to have done:

    Var(X)=\mathbb{E}(X^2)-\mathbb{E}(X)
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    (Original post by ghostwalker)
    Approach is good, but:

    Var(X)=\mathbb{E}(X^2)-(\mathbb{E}(X))^2

    You seem to have done:

    Var(X)=\mathbb{E}(X^2)-\mathbb{E}(X)
    OMG THANK YOU SO MUCH IT WORKED THIS TIME!! ^^ Thank you to everyone in this thread <3
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    Thanks to everyone who had responded! I understand it now. In fact, I had made the same mistake as Ray09. Thanks again
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