If $X$ is normally distributed with $\mu=2$ , $\sigma=2$., find a number $x_{0}$ such that $f\left(x\right)$
a) $$P\left(X>x_{0}\right)=.10$$ My try: $$Z=\frac{X-\mu}{\sigma}$$ $$P\left(Z>\frac{x_{0}-2}{2}\right)=1-X=.10$$ Where $X=\frac{x_{0}-2}{2}$, solving for $X$,$$X=0.9$$ $$\frac{x_{0}-2}{2}=0.9$$ Checking the table values for the normal distribution $$x_{0}=3.8$$ b) $$P\left(X>-x_{0}\right)=.20$$ My try: $$\frac{-x_{0}-2}{2}=0.8$$ $$x_{0}=-3.6$$ $$P\left(Z>\frac{-3.6-2}{2}\right)=1-2.8=-1.85=0.08851$$
Can you please provide me some feedback, for both solutions, I tried to check if the values are correct but I don't get the same value for b)?
You have to apply the normal distribution function. The equation is $P(X>x_0)=1-P(X<x_0)$ (converse probability).
This is the same as $P(X>x_0)=1-\Phi\left(\frac{x_0-2}{2} \right)=0.1$. Here $\Phi\left(z\right)$ is the cdf of the normal distribution. Solving the equation for $x$. Adding $\Phi\left(\frac{x_0-2}{2} \right)$
$1=0.1+\Phi\left(\frac{x_0-2}{2} \right) \quad |-0.1$
$1-0.1=\Phi\left(\frac{x_0-2}{2} \right) \quad $
$0.9=\Phi\left(\frac{x_0-2}{2} \right) \quad $
Now we have to use the inverse function in order to isolate $x_0$.
$\Phi^{-1}\left(0.9\right) =\frac{x_0-2}{2} \quad $
Now you look at a table and look where the function is approximately $0.9$. This is at $\Phi\left(1.28\right)= .89973\approx 0.9$. So $\Phi^{-1}\left(0.9\right)=1.28$. The equation becomes
$$1.28=\frac{x_0-2}{2} \quad $$
Is it comprehensible up to here? If yes, then I think you can go on by yourself.