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Math Help - [SOLVED] Normal Distribution

  1. #1
    Member looi76's Avatar
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    [SOLVED] Normal Distribution

    Question:
    In a statistics examinations, 15% of the candidates scored more than 63 marks and 10% of the candidates scored less than 32 marks. Assuming that the marks were distributed normally find the mean marks and the standard deviation.

    Attempt:

    P(X \geq 63) = 15\%
    P(X \geq 63) = -85\%████ 15\% - 100\% = -85\%

    P\left(Z \geq \frac{63 - \mu}{\sigma} \right) = 1 - \phi^{-1}(0.85)

    P\left(Z \geq \frac{63 - \mu}{\sigma} \right) = 0.036

    1 - \left( \frac{63 - \mu}{\sigma} \right) = 0.036

    \left( \frac{63 - \mu}{\sigma} \right) = 0.036 + 1

    63 - \mu = 1.036\sigma

    1.036\sigma + \mu = 63



    P(X \leq 32) = 10\%
    P(X \leq 32) = -90\%████ 10\% - 100\% = 90\%

    P\left(Z \leq \frac{32 - \mu}{\sigma}\right) = 1 - \phi^{-1}(90)

    P\left(Z \leq \frac{32 - \mu}{\sigma}\right) = 1 - 1.282

    \left(\frac{32 - \mu}{\sigma}\right) = -0.282

    32 - \mu = -0.282\sigma



    -0.282\sigma + \mu = 32

    1.036\sigma + \mu = 63

    -0.282\sigma + \mu = 32

    1.036\sigma = 63

    -0.282\sigma = 32

    1.318\sigma = 31

    \sigma = \frac{31}{1.318}

    \sigma = 23.52

    1.036\sigma + \mu = 63

    1.036 \times 23.52 + \mu = 63

    \mu = 63 - (1.036\times23.52)

    \mu = 38.44



    \mu = 38.44

    \sigma = 23.52

    Answer in textbook is 49.1 & 13.4... Where did I go wrong?
    Attached Thumbnails Attached Thumbnails [SOLVED] Normal Distribution-normal_distrubtion_table.gif  
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  2. #2
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    Quote Originally Posted by looi76 View Post
    Question:
    In a statistics examinations, 15% of the candidates scored more than 63 marks and 10% of the candidates scored less than 32 marks. Assuming that the marks were distributed normally find the mean marks and the standard deviation.

    Attempt:

    P(X \geq 63) = 15\%
    P(X \geq 63) = -85\%████ 15\% - 100\% = -85\%

    P\left(Z \geq \frac{63 - \mu}{\sigma} \right) = 1 - \phi^{-1}(0.85)

    P\left(Z \geq \frac{63 - \mu}{\sigma} \right) = 0.036

    1 - \left( \frac{63 - \mu}{\sigma} \right) = 0.036

    \left( \frac{63 - \mu}{\sigma} \right) = 0.036 + 1

    63 - \mu = 1.036\sigma

    1.036\sigma + \mu = 63



    P(X \leq 32) = 10\%
    P(X \leq 32) = -90\%████ 10\% - 100\% = 90\%

    P\left(Z \leq \frac{32 - \mu}{\sigma}\right) = 1 - \phi^{-1}(90)

    P\left(Z \leq \frac{32 - \mu}{\sigma}\right) = {\color{red}1 - 1.282} Mr F says: Your mistake is in red. It should be -1.282. See main reply below.

    Then {\color{red}\left(\frac{32 - \mu}{\sigma}\right) = -1.282} etc.

    [snip]

    Answer in textbook is 49.1 & 13.4... Where did I go wrong?
    Again, well done on taking the time and trouble to show all your working.

    You want the value of z_{0.10} corresponding to \Pr(Z < z_{0.10}) = 0.10

     \Rightarrow \Pr(Z > - z_{0.10}) = 0.10

    (note that z_{0.10} < 0 \Rightarrow -z_{0.10} > 0 so this is not as wrong as it first looks)

    \Rightarrow \Pr(Z < - z_{0.10}) = 0.90 \Rightarrow - z_{0.10} = 1.28

    from the tables

    \Rightarrow z_{0.10} = -1.28.

    This would be a good time to re-inforce the importance of drawing simple normal distribution pictures with the required area shaded ......
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