I must resolve a quiz about normal random variable.
"Let X be a normal random variable of mean 3 and variance 4. Prove that $P((X-2)^2 > 4) =0.3753$
So, I made: $Z = Norm(1,4)$
Using formula $K = \alpha X + \beta$ with $X = Norm(\mu,(\omega)^2)$ So $K = Norm(\alpha \mu + \beta, (\alpha \omega)^2$
So the problem become:
$P(K^2 > 4) = P(K>2) + P(K<-2)$
And denoting normalized $K$ with $K'$
$P(K'>1/2) + P(K'<-1/2) = 2(1 - \phi(1/2)) = 0.617$
I don't understand where I make the mistake
I see you're using $Z$ and $K$. I think they're the same.
Anyway, if $K=X-2$, then it is $K$ that has $N(1,4)$ distribution, and so, when you have—for instance— $$P(K<-2)$$ you have to substract one, and the region you get is not symmetric around $0$ but around $1$.
That is, $P(K>2)=P\left(K'>\frac12\right)$, but $$P(K<-2)=P\left(K'<\frac{-2-1}2\right)=P\left(K'<-\frac32\right).$$
Also, you could have done this: $$(X-2)^2>4 \quad\iff\quad |X-2|>2 \quad \iff$$ $$\iff\quad X-2>2 \quad\vee\quad X-2<-2 \quad\iff$$ $$\iff\quad X>4 \quad\vee\quad X<0,$$ and having in mind that $X\sim N(3,4)$, if we call $Z=\frac{X-3}2$ (which is standard normal), $$(X-2)^2>4 \quad\iff\quad Z>\frac{4-3}2=\frac12 \;\vee\; Z<\frac{0-3}2=-\frac32.$$
So $$P\big((X-2)^2>4\big)=P\left(Z<\frac12\right)+P\left(Z>-\frac32\right).$$
Yes, $X^2>$4 means $X< -2$ or $X> 2$. With mean $\mu= 3$ and variance 4, so standard deviation $\sigma= \sqrt{4}= 2$, The "standard normal variable" is $z= \frac{x- \mu}{\sigma}= \frac{x- 3}{2}$. With x= 2, $z= \frac{2- 3}{2}= -\frac{1}{2}$ and with x= -2, $z= \frac{-2-3}{2}= -\frac{5}{2}$.
Looking at the standard normal distribution, P(z> -0.5)= 0.6915 and P(z< -2.5)= 0.0062. So the probability you want is 0.6915+ 0.0062= 0.6977.
