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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?

  • 0
    If $z_1$ and $z_2$ in the third line are supposed to be different from $z_1$ and $z_2$ in the first two lines, then you don't have enough information to find them. – 2017-01-30

1 Answers 1

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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 $(z_1,z_2) \in \mathbb{R^2}$ , 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 $$\{(z_1,z_2)\in \mathbb{R^2}|z_1<\Phi^{-1}({0.2}),z_2=\Phi^{-1}(0.8+\Phi(z_1)) \}$$