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    help me pliz

    how to derive mean ang variance for normal distribution???????????
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    Quote Originally Posted by students View Post
    how to derive mean ang variance for normal distribution???????????
    E(X) = \frac{1}{\sigma \sqrt{\pi}} \int_{-\infty}^{+\infty} x \, e^{\frac{(x - \mu)^2}{2 \sigma^2}} \, dx = \mu.


    \text{Var}(X) = E\left(X^2\right) - \mu^2 = \sigma^2.


    Note that E\left(X^2\right) = \frac{1}{\sigma \sqrt{\pi}} \int_{-\infty}^{+\infty} x^2 \, e^{\frac{(x - \mu)^2}{2 \sigma^2}} \, dx = \sigma^2 + \mu^2.


    You'll find the details of the actual calculations in most mathematical statistics textbooks.
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