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Math Help - Normal Distribution problems

  1. #1
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    Post Normal Distribution problems

    Mean = 40 and Sd = 10
    Need to find the value of m for the area given:

    From left to right the normal dist has an area of 0.97

    How do I find this?

    This is what I did, normalcdf (-100,-3) = 0.00135
    (Put -100 as some big number [Should be infinity])
    then add to 0.97 = 0.97135, then I used invNorm (0.97135)

    Don't know where I went wrong! What is the mean and sd for in this question O.o?

    The answer is 58.81 but I cant get it

    Please help!
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  2. #2
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    Quote Originally Posted by Modernized View Post
    Mean = 40 and Sd = 10
    Need to find the value of m for the area given:

    From left to right the normal dist has an area of 0.97

    How do I find this?

    This is what I did, normalcdf (-100,-3) = 0.00135
    (Put -100 as some big number [Should be infinity])
    then add to 0.97 = 0.97135, then I used invNorm (0.97135)

    Don't know where I went wrong! What is the mean and sd for in this question O.o?

    The answer is 58.81 but I cant get it

    Please help!
    It's an inverse normal problem. You're told mean and standard deviation so you know that z_m = \frac{m - 40}{10} where \Pr(Z > z_m) = 0.97 \Rightarrow \Pr(Z < z_m) = 0.03. Get z_m from four figure math tables. (If you're using technology, surely you just enter the given numbers into the program).
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  3. #3
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    Smile

    No. I still can't get the answer

    invNorm (0.03) nor invNorm (0.97) give me the correct answer.
    What's wrong here?

    and I don't understand when do you put the smaller than or bigger than signs, can you please teach me?
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  4. #4
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    Quote Originally Posted by Modernized View Post
    No. I still can't get the answer

    invNorm (0.03) nor invNorm (0.97) give me the correct answer.
    What's wrong here?

    and I don't understand when do you put the smaller than or bigger than signs, can you please teach me?
    You have to understand what the output from technology means. Otherwise the technology is useless.

    In this case the technology is giving you a z-value which you have to convert into an x-value in the way I have already explained. Furthermore, you need to know which of invNorm(0.97) or invNorm(0.03) will give you the correct z-value.
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