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Let $X=N(-2,4)$ be a normal random variable.

a) Find $P(-3.3

b) Find $P(X^2>49)$

c)What is the type (and parameters) of the random variable $Y=3(2-x)-15$?

For (a) The answer is supposed to be.742, but I keep getting .9958. I know that $P\{y_1 \leq Y \leq y_2\} = P\{Y \leq y_2\} -P\{Y \leq y_1\} = \Phi\left(\frac{y_2-\mu}{\sigma}\right)- \Phi\left(\frac{y_1-\mu}{\sigma}\right)$, but I dont know if I am just making calculation errors. I get $\Phi(\frac{5.2+2}{2})-1+\Phi(\frac{3.3+2}{2})=.9958$. Not sure what I am doing wrong.

For (b) I am a little unsure of what to do. I don't suppose we could take the square root and find $P(X>7)$?

(c) I do not understand. I know its normal with $\mu=-3$ and $\sigma=6$, but how do you get that? I made it so $Y=6-3X-15$. Do you just plug in the values fro $\mu$ and $\sigma$ we are given for $X$ in the beginning and solve it from there?

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Thanks for explaining what you've done so far!

For (a), your error is in turning $\Phi\left(\frac{-3.3+2}{2}\right)$ into $1-\Phi\left(\frac{3.3+2}{2}\right)$. You need to be careful when you switch the minus sign.

Your approach to (b) is almost right, but you need to consider the negative square root as well. $X^2 > 49$ when $X > 7$ OR $X < -7$.

You're on the right track with (c). If $X \sim \operatorname{Normal}(\mu,\sigma^2)$, what is the distribution of $Y = aX + b$?

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    I plugged what is given for$X$into Y and got $Y=-3-6Z$. So it would be normal(-3,36).2012-11-02