Vtyjkyuk

I have a problem solving this exercise. I have this:

1. $P(0 le Z le z_2) = 0.3$




2. $P(Z le z_1) = 0.3$




3. $P(z_1 le Z le z_2) = 0.8$

I need to find the $z$ values for each given probability. I already solved the first and the second like this:

1. I calculated the inverse standard normal distribution (with LibreOffice Calc) and I found that $z_2$ is $0.841$




2. I calculated the inverse standard normal distribution (with LibreOffice Calc) and I found that $z_1$ is $-0.524$

How can I find $z_1$ and $z_2$ of the third point of the exercise?

By using 1. and 2. , there is no solution
$$P(z_1 le Z le z_2)=P( Z le z_2)-P( Z le z_1)+P( Z le 0)-P( Z le 0)$$
$$= [P( Z le z_2)-P( Z le 0)]-P( Z le z_1)+P( Z le 0)$$
$$= P(0 le Z le z_2)-P( Z le z_1)+P( Z le 0)$$

We have that $P( Z le 0)=0.5$,$P(0 le Z le z_2)=0.3$ and $P(z_1 le Z le z_2)=0.8$, but we do not have $$0.8 ne 0.3-0.3+0.5$$

If we solve independently of question 1. and 2.

Let $(z_1,z_2) in mathbbR^2$ , we have $z_1 < z_2$ and

$$P(z_1 le Z le z_2) = Phi(z_2)-Phi(z_1)=0.8.$$

Obviously, $Phi(z_1)<0.2$ because $Phi(z_2)<1$, which imposes that $z_1<Phi^-1(0.2) approx-0.81462$

Then, we have $$z_2=Phi^-1(0.8+Phi(z_1)).$$

Finally , the space of solutions is $$z_1<Phi^-1(0.2),z_2=Phi^-1(0.8+Phi(z_1)) $$