Suppose that $X$ is a random variable that has a normal distribution with mean = 5 and standard deviation = 10. Evaluate the following probabilities:
$\mathrm{Prob}((X-10)^2 < 12)$
Suppose that $X$ is a random variable that has a normal distribution with mean = 5 and standard deviation = 10. Evaluate the following probabilities:
$\mathrm{Prob}((X-10)^2 < 12)$
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}
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