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Suppose that $X$ is a random variable that has a normal distribution with mean = 5 and standard deviation = 10. Evaluate the following probabilities:

$\mathrm{Prob}((X-10)^2 < 12)$

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What are your thoughts about this question? – Did Sep 26 '12 at 16:16
My first algebraic instinct is to take the square root of both sides, add ten, etc. and then evaluate the probability using a normal distribution function. However, I am unsure if these transformations change the mean or standard deviation, and if so, how. – Emily Sep 26 '12 at 16:18
up vote 4 down vote accepted

You are free to make changes like the one you discuss in your comment as long as they replace $(X-10)^2<12$ with something logically equivalent. (Changing means and standard deviations is irrelevant here, you are working with the distribution $X$.)

In this case, following your agenda, you would get $-\sqrt{12}<X-10<\sqrt{12}$, and then $-\sqrt{12}+10<X<\sqrt{12}+10$.

So, now I am trusting that you can compute this (normal) probability with the expression inside modified to this equivalent form: $P(-\sqrt{12}+10<X<\sqrt{12}+10)=$ ?

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