Question

Find a value of the standard normal random variable z, call it ${\mathbf{z}}_{\mathbf{0}}$, such that

$$\begin{gathered} \mathbf{a}\mathbf{.}\mathbf{}\mathbf{P}\left(z\le z_0\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2090} \\ \mathbf{b}\mathbf{.}\mathbf{}\mathbf{P}\left(z\le z_0\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{7090} \\ \mathbf{c}\mathbf{.}\mathbf{}\mathbf{P}\left(-z_0\le z<z_0\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8472} \\ \mathbf{d}\mathbf{.}\mathbf{}\mathbf{P}\left(-z_0\le z<z_0\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1664} \\ \mathbf{e}\mathbf{.}\mathbf{}\mathbf{P}\left(z_0\le z\le z_0\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4798} \\ \mathbf{f}\mathbf{.}\mathbf{}\mathbf{P}\left(-1<z<z_0\right) \end{gathered}$$

Short answer

$$\begin{gathered} \mathrm{a}.{\mathrm{z}}_0=-0.81 \\ \mathrm{b}.{\mathrm{z}}_0=0.55 \\ \mathrm{c}.{\mathrm{z}}_0=1.43 \\ \mathrm{d}.{\mathrm{z}}_0=0.21 \\ \mathrm{e}.{\mathrm{z}}_0=-2.05 \\ \mathrm{f}.\mathrm{P}\left(-1<\mathrm{z}<{\mathrm{z}}_0\right)=\mathrm{\Phi }\left({\mathrm{z}}_0\right)-0.1587 \end{gathered}$$

Step-by-step solution

Given information

z is a standard normal variable which is called${\mathrm{z}}_0$

Finding the value of z0 when P(z≤z0)=0.2090

$$\begin{gathered} a. \\ P\left(z\le {\mathrm{z}}_0\right)=0.2090 \\ \Phi \left({\mathrm{z}}_0\right)=0.2090 \\ {\mathrm{z}}_0={\mathrm{\Phi }}^{-1}\left(0.2090\right) \\ {\mathrm{z}}_0=-0.81 \end{gathered}$$

Therefore, the${\mathrm{z}}_0$ score is -0.81.

Finding the value of z0 when P(z≤z0)=0.7090

$$\begin{gathered} \mathrm{b}. \\ \mathrm{P}(\mathrm{z}\le {\mathrm{z}}_0)=0.7090 \\ \mathrm{\Phi }\left({\mathrm{z}}_0\right)=0.7090 \\ {\mathrm{z}}_0={\mathrm{\Phi }}^{-1}\left(0.7090\right) \\ {\mathrm{z}}_0=0.55 \end{gathered}$$

Therefore, the${\mathrm{z}}_0$ score is 0.55.

Finding the value z0 of when P(- z0≤z< z0)=0.8472

$$\begin{gathered} \mathrm{c}. \\ \mathrm{P}\left(-{\mathrm{z}}_0\le \mathrm{z}<{\mathrm{z}}_0\right)=0.8472 \\ 2\mathrm{P}\left(0\le \mathrm{z}<{\mathrm{z}}_0\right)=0.8472 \\ \mathrm{P}\left(0\le \mathrm{z}<{\mathrm{z}}_0\right)=0.4236 \\ \mathrm{P}\left(\mathrm{z}<{\mathrm{z}}_0\right)-\mathrm{P}\left(\mathrm{z}<0\right)=0.4236 \\ \mathrm{P}\left(\mathrm{z}<{\mathrm{z}}_0\right)-0.50=0.4236 \\ \mathrm{\Phi }\left({\mathrm{z}}_0\right)=0.9236 \\ {\mathrm{z}}_0={\mathrm{\Phi }}^{-1}\left(0.9236\right) \\ {\mathrm{z}}_0=1.43 \end{gathered}$$

Therefore, the${\mathrm{z}}_0$ score is 1.43.

Finding the value of z0 when P(- z0≤z< z0)=0.1664

$$\begin{gathered} \mathrm{d}. \\ \mathrm{P}\left(-{\mathrm{z}}_0\le \mathrm{z}<{\mathrm{z}}_0\right)=0.1664 \\ 2\mathrm{P}\left(0\le \mathrm{z}<{\mathrm{z}}_0\right)=0.1664 \\ \mathrm{P}\left(0\le \mathrm{z}<{\mathrm{z}}_0\right)=0.0832 \\ \mathrm{P}\left(\mathrm{z}<{\mathrm{z}}_0\right)-\mathrm{P}\left(\mathrm{z}<0\right)=0.0832 \\ \mathrm{P}\left(\mathrm{z}<{\mathrm{z}}_0\right)-0.50=0.832 \\ \mathrm{P}\left(\mathrm{z}<{\mathrm{z}}_0\right)=0.5832 \\ \mathrm{\Phi }\left({\mathrm{z}}_0\right)=0.5832 \\ {\mathrm{z}}_0={\mathrm{\Phi }}^{-1}\left(0.5832\right) \\ {\mathrm{z}}_0=0.21 \end{gathered}$$

Therefore, the$z_0$ score is 0.21.

Finding the value of z0 when P(-1<z<z0)

$$\begin{gathered} e. \\ \mathrm{P}\left({\mathrm{z}}_0\le \mathrm{z}<0\right)=0.4798 \\ \mathrm{P}\left(z<0)-P(\mathrm{z}<{\mathrm{z}}_0\right)=0.4798 \\ 0.50-P\mathrm{P}\left(\mathrm{z}<{\mathrm{z}}_0\right)=0.4798 \\ \mathrm{P}\left(\mathrm{z}<0\right)=0.020 \\ \mathrm{\Phi }\left({\mathrm{z}}_0\right)=0.0202 \\ {\mathrm{z}}_0=-2.05 \end{gathered}$$

Therefore, the${\mathrm{z}}_0$ score is -2.05.

Finding the value of z0 when P(-1<z<z0)

$$\begin{gathered} \mathrm{f}. \\ \mathrm{P}\left(-1<\mathrm{z}<{\mathrm{z}}_0\right)=P\left(z<z_0\right)-P\left(z\le -1\right) \\ =\mathrm{\Phi }\left({\mathrm{z}}_0\right)-\mathrm{\Phi }\left(-1\right) \\ =\mathrm{\Phi }\left({\mathrm{z}}_0\right)-\mathrm{\Phi }\left(-1\right) \\ =\mathrm{\Phi }\left({\mathrm{z}}_0\right)-0.1587 \end{gathered}$$

Therefore, the${\mathrm{z}}_0$ score is -0.1587.