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Given a normal random variable with mean 90 and a standard deviation 3, find the probability that x is greater than or equal to 86.67 and less than or equal to 96.33

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    $\begingroup$ Do you know how to normalise to standard normal distribution? $\endgroup$ – peterwhy Nov 18 '13 at 3:56
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Hint: If $Y\text{~}N(\mu,\sigma^2)$, then $X=\frac{Y-\mu}{\sigma}\text{~}N(0,1)$.

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