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Math Help - Probability Problem

  1. #1
    Newbie ayato's Avatar
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    Probability Problem

    Ok, here's the problem i've been having some trouble with:

    Ann and Bob have agreed to meet for dinner "around four". The number of minutes after 4:00 that Ann will arrive is approximately normally distributed with mean 10 and standard deviation 3.1, and the number of minutes after 4:00 that Bob will arrive is approximately normally distributed with mean 20 and standard deviation 4.2.

    If the longest each will wait is 10 minutes, find the probability that they'll meet?

    I know the answer will be around the 1 range but i don't really know how to show the work for it, at least for the work that i showed for it, i'm not confident about it. I'll be very appreciated if someone can help me out. Thank You in advance.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ayato View Post
    Ok, here's the problem i've been having some trouble with:

    Ann and Bob have agreed to meet for dinner "around four". The number of minutes after 4:00 that Ann will arrive is approximately normally distributed with mean 10 and standard deviation 3.1, and the number of minutes after 4:00 that Bob will arrive is approximately normally distributed with mean 20 and standard deviation 4.2.

    If the longest each will wait is 10 minutes, find the probability that they'll meet?

    I know the answer will be around the 1 range but i don't really know how to show the work for it, at least for the work that i showed for it, i'm not confident about it. I'll be very appreciated if someone can help me out. Thank You in advance.
    Consider the difference D in arrival times, this will be normally distributed with
    mean 10 minutes, and sd sqrt(3.1^2+4.2^2). Then you want: p(-10<D<10).

    RonL
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  3. #3
    Newbie ayato's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Consider the difference D in arrival times, this will be normally distributed with
    mean 10 minutes, and sd sqrt(3.1^2+4.2^2). Then you want: p(-10<D<10).

    RonL
    Thanks alot, so let me get this straight:

    sqrt(3.1^2+4.2^2)
    p(-10<D<10)

    z> (-10-10)/sqrt(3.1^2+4.2^2)
    = -3.83
    P= practically 1

    Z<(10-10)/sqrt(3.1^2+4.2^2)
    =0
    P= .5000

    1-.5000
    =.5000
    Is this right?

    Thanks again for your help
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ayato View Post
    Thanks alot, so let me get this straight:

    sqrt(3.1^2+4.2^2)
    p(-10<D<10)

    z> (-10-10)/sqrt(3.1^2+4.2^2)
    = -3.83
    P= practically 1

    Z<(10-10)/sqrt(3.1^2+4.2^2)
    =0
    P= .5000

    1-.5000
    =.5000
    Is this right?

    Thanks again for your help
    That looks OK to me

    RonL
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