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Thread: [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:

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

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

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

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

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

    $\displaystyle 63 - \mu = 1.036\sigma$

    $\displaystyle 1.036\sigma + \mu = 63$



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

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

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

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

    $\displaystyle 32 - \mu = -0.282\sigma$



    $\displaystyle -0.282\sigma + \mu = 32$

    $\displaystyle 1.036\sigma + \mu = 63$

    $\displaystyle -0.282\sigma + \mu = 32$

    $\displaystyle 1.036\sigma = 63$

    $\displaystyle -0.282\sigma = 32$

    $\displaystyle 1.318\sigma = 31$

    $\displaystyle \sigma = \frac{31}{1.318}$

    $\displaystyle \sigma = 23.52$

    $\displaystyle 1.036\sigma + \mu = 63$

    $\displaystyle 1.036 \times 23.52 + \mu = 63$

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

    $\displaystyle \mu = 38.44$



    $\displaystyle \mu = 38.44$

    $\displaystyle \sigma = 23.52$

    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:

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

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

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

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

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

    $\displaystyle 63 - \mu = 1.036\sigma$

    $\displaystyle 1.036\sigma + \mu = 63$



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

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

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

    Then $\displaystyle {\color{red}\left(\frac{32 - \mu}{\sigma}\right) = -1.282}$ 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 $\displaystyle z_{0.10}$ corresponding to $\displaystyle \Pr(Z < z_{0.10}) = 0.10$

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

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

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

    from the tables

    $\displaystyle \Rightarrow z_{0.10} = -1.28$.

    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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