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    Need some help with a qs.

    The heights of a population of male students are distributed normally with mean 178cm and standard deviation 5cm. The heights of a population of female students are distributed normally with mean 168cm and standard deviation 4cm.
    Find the probability that a randomly chosen female is taller than a randomly chosen male.


    So I tried subtracting the means and adding the s.d's.

    I got F-M - (-10, 9)
    Then idk. I tried F-m>0
    But that didn't work.

    Thanks
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    (Original post by Super199)
    Need some help with a qs.

    The heights of a population of male students are distributed normally with mean 178cm and standard deviation 5cm. The heights of a population of female students are distributed normally with mean 168cm and standard deviation 4cm.
    Find the probability that a randomly chosen female is taller than a randomly chosen male.


    So I tried subtracting the means and adding the s.d's.

    I got F-M - (-10, 9)
    Then idk. I tried F-m>0
    But that didn't work.

    Thanks
    Check the variance of F-M again.
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    (Original post by SeanFM)
    Check the variance of F-M again.
    4-5?

    Surely that would give a really big z value?

    0-10/-1
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    (Original post by Super199)
    4-5?

    Surely that would give a really big z value?

    0-10/-1
    Var(aX+bY) = ?

    (where X and Y are indep.)
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    (Original post by SeanFM)
    Var(aX+bY) = ?

    (where X and Y are indep.)
    a^2var(x)+b^2var(y)

    What are the values of a and b?

    Is it 16 and 25. idek sorry
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    (Original post by Super199)
    a^2var(x)+b^2var(y)

    What are the values of a and b?

    Is it 16 and 25. idek sorry
    That is correct. In this case (as you have worked out already, to give Var(F-M) = 16 + 25) X = F, a = 1, Y = M, b = -1.
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    (Original post by SeanFM)
    That is correct. In this case (as you have worked out already, to give Var(F-M) = 16 + 25) X = F, a = 1, Y = M, b = -1.
    Im confused.

    a^2var(x) +b^2var(y)

    In the qs is a = 4 and b= 5. But whats var(x) and var(y)?

    I don't get what you have wrote in bold, because I thought you add the variances anyway.
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    (Original post by Super199)
    Im confused.

    a^2var(x) +b^2var(y)

    In the qs is a = 4 and b= 5. But whats var(x) and var(y)?

    I don't get what you have wrote in bold, because I thought you add the variances anyway.
    Forget about the X's and Y's, I think it is too confusing - let's stick with M and F.

    Var(M-F) = 1^2 * Var(M) + (-1)^2 * Var(F) = Var(M) + Var(F), and remember that you are given standard deviation rather than variance.
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    (Original post by SeanFM)
    Forget about the X's and Y's, I think it is too confusing - let's stick with M and F.

    Var(M-F) = 1^2 * Var(M) + (-1)^2 * Var(F) = Var(M) + Var(F), and remember that you are given standard deviation rather than variance.
    P(z> (0-10)/ root 41)?
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    (Original post by Super199)
    P(z> (0-10)/ root 41)?
    Correct :borat:
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    (Original post by SeanFM)
    Correct :borat:
    That doesn't give me the right answer lol. It gives me 0.94083.
    The answer in the book is 0.0592
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    (Original post by Super199)
    That doesn't give me the right answer lol. It gives me 0.94083.
    The answer in the book is 0.0592
    Notice the relationship between those two numbers, and that your probability is far too big (probability of getting 0 or greater when the mean is -10 is the warning sign)

    You're looking for P(Z>z) and you have read off the P(Z=<z) value from tables...
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    (Original post by SeanFM)
    Notice the relationship between those two numbers, and that your probability is far too big (probability of getting 0 or greater when the mean is -10 is the warning sign)

    You're looking for P(Z>z) and you have read off the P(Z=<z) value from tables...
    I just chucked it into the calc. Do you have a casio fx-991e plus by any chance?
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    (Original post by Super199)
    I just chucked it into the calc. Do you have a casio fx-991e plus by any chance?
    Afraid not :lol: but your calculator most likely has found  P(Z \leq z).

    Another hint is that 0.9408 + 0.0592 = 1.
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    (Original post by SeanFM)
    Afraid not :lol: but your calculator most likely has found  P(Z \leq z).

    Another hint is that 0.9408 + 0.0592 = 1.
    Lol no worries. Guess I will stick to the tables.
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    (Original post by Super199)
    Lol no worries. Guess I will stick to the tables.
    You can use your calculator if you want, you just need to know what it's actually calculating.

    Your calculation to find 0.9408 is correct - you just need 1 final calculation/step to get the correct answer.
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    (Original post by SeanFM)
    You can use your calculator if you want, you just need to know what it's actually calculating.

    Your calculation to find 0.9408 is correct - you just need 1 final calculation/step to get the correct answer.
    Do you mind helping me with a qs.

    https://f1f559d498e385d34687bce0088e...20S3%20OCR.pdf

    q3i.

    Is the confidence interval : x( + or -) z*(s/square root of n)?
 
 
 
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