Question

Find a$\mathbf{z}$-score, call it${\mathbf{z}}_0$, such that

a.$\mathbf{P}(\mathbf{z}\le {\mathbf{z}}_0)\mathbf{=}\mathbf{.}\mathbf{5080}$

b.$\mathbf{P}(\mathbf{z}\ge {\mathbf{z}}_0)\mathbf{=}\mathbf{.}\mathbf{5517}$

c.$\mathbf{P}(\mathbf{z}\ge {\mathbf{z}}_0)\mathbf{=}\mathbf{.}\mathbf{1492}$

d.$\mathbf{P}({\mathbf{z}}_0\le \mathbf{z}\le \mathbf{.}\mathbf{59})\mathbf{=}\mathbf{.}\mathbf{4773}$

Short answer

a.$z_0=\text{0.02}$

b.$z_0=\text{-0.129}$

c. $z_0=\text{1.039}$

d. $z_0=\text{-0.689}$

Step-by-step solution

Given information

z-score is denoted by.$z_0$

CalculateP(z≤z0)=.5080

$\begin{array}{c}P(z\le z_0)=.5080\\ \Phi \left(z_0\right)=.5080\\ z_0={\Phi }^{-1}\left(.5080\right)\\ z_0=\text{0.02}\end{array}$

Hence.$z_0=\text{0.02}$

CalculateP(z≥z0)=.5517

$\begin{array}{c}P(z\ge z_0)=.5517\\ 1-P(z\le z_0)=.5517\\ 1-\Phi \left(z_0\right)=.5517\\ \Phi \left(z_0\right)=1-.5517\\ \Phi \left(z_0\right)=\text{0.4483}\\ z_0={\Phi }^{-1}\left(\text{0.4483}\right)\\ z_0=\text{-0.129}\end{array}$

Hence$z_0=\text{-0.129}$.

CalculateP(z≥z0)=.1492

$\begin{array}{c}P(z\ge z_0)=.1492\\ 1-P(z\le z_0)=.1492\\ 1-\Phi \left(z_0\right)=.1492\\ \Phi \left(z_0\right)=1-.1492\\ \Phi \left(z_0\right)=\text{0.8508}\\ z_0={\Phi }^{-1}\left(\text{0.8508}\right)\\ z_0=\text{1.039}\end{array}$

Hence.$z_0=\text{1.039}$

CalculateP(z0≤z≤.59)=.4773

$\begin{array}{c}P(z_0\le z\le .59)=.4773\\ P(z\le .59)-P(z\le z_0)=.4773\\ P(z\le z_0)=P(z\le .59)-.4773\\ \Phi \left(z_0\right)=0.7224-0.4773\\ \Phi \left(z_0\right)=\text{0.2451}\\ z_0={\Phi }^{-1}\left(\text{0.2451}\right)\\ z_0=\text{-0.689}\end{array}$

Hence $z_0=-0.689$.