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Math Help - Another question on normal distribution

  1. #1
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    Another question on normal distribution



    I have no idea how to do these questions,I only know parts a(i) and b(i).
    Also,are a(iii) and b(ii) conditional probability questions?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Find the probability (percentage) of staff rated as poor.

    Take 1 - P(poor)

    This gives you the sum percentages of staff with fair and good ratings.

    Then, divide this by 2 to get the percentage of fair and good ratings.

    So, now depending on what type of z table you have, you can proceed with two different ways.
    Find the z value for P(X > z) = P(Good)

    or Find the z value for P(X < z) = P(Poor) + P(Fair)

    When you got the z value, convert it into X.
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  3. #3
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    a(ii) Solving a(i) you know that the probability for a salesman to be rated poor equals 0.1288. Hence, 1-0.1288 of chance is left to be split among rates "fair" and "good". Meaning that there is an equal chance for a salesman to be rated "fair" and "good", each rate belonging the same probability (1-0.1288)/2=0.4356. So the surface under the probability density function is split into three parts: the first (leftmost) one measures 0.1288, and middle and the rightmost part each measuring 0.4356. All you have to do is find the value a having the property that 1-P(X<a)=0.4356. Can you do that?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    For the second one, you know that:

    X~B(500, 0.98)

    You get Sd(x) = \sqrt{500(0.98)(0.02)}

    Hence,

    X~N(np, npq)

    X~N(490, 9.8)

    This means, on average, 490 persons get the SMS within 3.5 seconds with a standard deviation of 9.8 people. If there are a total of 500 persons, more than 12 get the SMS after 3.5 seconds, that means there less than 488 people get the SMS within 3.5 s.

    P(X < 488) = P(Z < \frac{488 - 490}{\sqrt{9.5}})

    I think you can complete this now
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  5. #5
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    The value of Z is (1-P(poor))/2 right?
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  6. #6
    MHF Contributor Unknown008's Avatar
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    hmm...

    P(Poor) = 0.1288

    P(Fair) + P(Good) = 1- 0.1288 = 0.8712

    P(Good) = 0.8712/2 = 0.4356

    P(X > z) = 0.4356

    P(X < z) = 1-0.4356 = 0.5644

    The z value corresponding to this probability is 0.162 from my z table.

    Then, I use:

    z = \dfrac{x - \mu}{\sigma}

    0.162 = \dfrac{x-50}{5.3}

    x = 50.85

    Hence, a staff needs at least a score of 50.85 to be classified as 'Good'.
    Last edited by Unknown008; September 15th 2010 at 02:16 AM. Reason: Took the wrong value of mu
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  7. #7
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    ooh.I get it now.Didn't know it could be done that way.
    Can I use X~N(500,0.02) and calculate by using P(X>12)?
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  8. #8
    MHF Contributor Unknown008's Avatar
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    For the second one, yes, but be careful!

    X ~ B(500, 0.02)

    Thus,

    X ~ N(10, 9.8)
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  9. #9
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    Quote Originally Posted by Unknown008 View Post
    hmm...

    P(Poor) = 0.1288

    P(Fair) + P(Good) = 1- 0.1288 = 0.8712

    P(Good) = 0.8712/2 = 0.4356

    P(X > z) = 0.4356

    P(X < z) = 1-0.4356 = 0.5644

    The z value corresponding to this probability is 0.162 from my z table.

    Then, I use:

    z = \dfrac{x - \mu}{\sigma}

    0.162 = \dfrac{x-44}{5.3}

    x = 44.9

    Hence, a staff needs at least a score of 44.9 to be classified as 'Good'.

    I got that answer too,but its wrong.The answer key says its 50.8
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  10. #10
    MHF Contributor Unknown008's Avatar
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    Oops, I typed in the wrong mean in my previous post.

    0.162 = \frac{x - 50}{5.3}

    x = 50.85
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  11. #11
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    Oh ok.One more thing I'm not sure about,if P(Good)=0.4356,why can't I find the x value of that?Why is P(X < z) = 1-0.4356 = 0.5644 required?
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  12. #12
    MHF Contributor Unknown008's Avatar
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    My z table gives the probabilities of z values to the left of the z value. I cannot read the probability of the z value given the probability after that value.

    That is, I cannot find the z value for P(X > z), but I can find for P(X < z).

    However, I could take the value directly, but this will require that I also invert the sign of the z value later. To avoid confusion, I already take the precautions beforehand.

    The third part is some sort of conditional probability.

    P(Good| >44) = \dfrac{P(Good)}{P(X > 44)}
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  13. #13
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    Oh,no wonder it was like that.Thanks a bunch!
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