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Thread: [SOLVED] Normal Distrubtion Question...

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

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

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

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

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

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

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

    $\displaystyle \sigma = 4.22$

    $\displaystyle \sigma^2 = 17.83 (2 DP)$

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

    $\displaystyle 0.6406 \neq 0.6046$
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