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Math Help - The Normal distribution

  1. #1
    Senior Member chella182's Avatar
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    The Normal distribution

    The question goes...

    An athlete finds that in the long jump his distances form a Normal distribution with mean 8.2 metres and standard deviation 0.24 metres. In a competition he is currently in second place with one more jump left to take. The current leader has jumped a distance of 8.5 metres. Find the probability that, on his next jump, he:

    (a) takes the gold medal; (I've done this and got an answer of 0.1056)

    (b) takes the gold medal, but doesn't quite manage to break the long jump world record of 8.9 metres.

    (c) What distance can this athlete expect to exceed once in 500 jumps?

    It's part (b) that I'm stuck on really. I know it's a P(X\leq 8.9|X\geq 8.5) kind of thing, and I know (or I think I do!) that is just...

    \frac{P(X\leq 8.9\cap X\geq 8.5)}{P(X\geq 8.5)}

    ...but then I'm not sure how to go about calculating the numerator (since I already know the denominator from part (a)) of that from there.
    I did think for a while that the numerator was the same as P(8.5\leq X\leq8.9), but that got a silly answer i.e. an answer bigger than 1.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by chella182 View Post
    The question goes...

    An athlete finds that in the long jump his distances form a Normal distribution with mean 8.2 metres and standard deviation 0.24 metres. In a competition he is currently in second place with one more jump left to take. The current leader has jumped a distance of 8.5 metres. Find the probability that, on his next jump, he:

    (a) takes the gold medal; (I've done this and got an answer of 0.1056)

    (b) takes the gold medal, but doesn't quite manage to break the long jump world record of 8.9 metres.

    (c) What distance can this athlete expect to exceed once in 500 jumps?

    It's part (b) that I'm stuck on really. I know it's a P(X\leq 8.9|X\geq 8.5) kind of thing, and I know (or I think I do!) that is just...

    \frac{P(X\leq 8.9\cap X\geq 8.5)}{P(X\geq 8.5)}

    ...but then I'm not sure how to go about calculating the numerator (since I already know the denominator from part (a)) of that from there.
    I did think for a while that the numerator was the same as P(8.5\leq X\leq8.9), but that got a silly answer i.e. an answer bigger than 1.
    Then you did it wrong, since:

    P(8.5\leq X\leq8.9)=P(X\ge 8.5)-P(X\ge 8.9)<P(X\ge 8.5)

    CB
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chella182 View Post
    The question goes...

    An athlete finds that in the long jump his distances form a Normal distribution with mean 8.2 metres and standard deviation 0.24 metres. In a competition he is currently in second place with one more jump left to take. The current leader has jumped a distance of 8.5 metres. Find the probability that, on his next jump, he:

    (a) takes the gold medal; (I've done this and got an answer of 0.1056)

    (b) takes the gold medal, but doesn't quite manage to break the long jump world record of 8.9 metres.

    (c) What distance can this athlete expect to exceed once in 500 jumps?

    It's part (b) that I'm stuck on really. I know it's a P(X\leq 8.9|X\geq 8.5) kind of thing, and I know (or I think I do!) that is just...

    \frac{P(X\leq 8.9\cap X\geq 8.5)}{P(X\geq 8.5)}

    ...but then I'm not sure how to go about calculating the numerator (since I already know the denominator from part (a)) of that from there.
    I did think for a while that the numerator was the same as P(8.5\leq X\leq8.9), but that got a silly answer i.e. an answer bigger than 1.
    For part (b), you did think correctly! It is \frac{P\!\left(8.5\leqslant X\leqslant8.9\right)}{P\!\left(X\geqslant 8.5\right)}. This turns out to be about .9833.
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  4. #4
    Senior Member chella182's Avatar
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    I think I've done P(X\leq8.9)-P(X\leq8.5) maybe, instead of the other way around.
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  5. #5
    Senior Member chella182's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    For part (b), you did think correctly! It is \frac{P\!\left(8.5\leqslant X\leqslant8.9\right)}{P\!\left(X\geqslant 8.5\right)}. This turns out to be about .9833.
    Care to tell me how so?
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