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Math Help - [SOLVED] Normal Distrubtion Question...

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

    Question
    X has a normal distribution with mean 32 and variance \sigma^2. Given that the probability X is less than 33.14 is 0.6406, find \sigma^2. Give your answer correct to 2 decimal places.

    Attempt:
     X \sim N (32,\sigma^2)

     P\left( \frac{X-\mu}{\sigma} \leq \frac{33.14-32}{\sigma}  \right) = 0.6046

     P\left( Z \leq \frac{33.14 - 32}{ \sigma } \right) = 0.6046

    \frac{33.14 - 32}{ \sigma } = 0.27

    \frac{33.14 - 32}{0.27} = \sigma

    \sigma = 4.22

    \sigma^2 = 17.83 (2 DP)

    Where did I go wrong?
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  2. #2
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    Fresh eyes often win the day.

    0.6406 \neq 0.6046
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