Suppose $\mathbf{\chi }$ is a normally distributed random variable with $\mathbf{\mu }\mathbf{=}\mathbf{50}$ and $\mathbf{\sigma }\mathbf{=}\mathbf{3}$ . Find a value of the random variable, call it ${\mathbf{\chi }}_{\mathbf{0}}$ , such that

a)$\mathbf{P}(\mathbf{\chi }\le {\mathbf{\chi }}_0)\mathbf{=}\mathbf{.}\mathbf{8413}$

b)$\mathbf{P}(\mathbf{\chi }>{\mathbf{\chi }}_0)\mathbf{=}\mathbf{.}\mathbf{025}$

c)$\mathbf{P}(\mathbf{\chi }>{\mathbf{\chi }}_0)\mathbf{=}\mathbf{.}\mathbf{95}$

d)$\mathbf{P}(\mathbf{41}\le \mathbf{\chi }\le {\mathbf{\chi }}_0)\mathbf{=}\mathbf{.}\mathbf{8630}$

e) 10% of the values of role="math" localid="1652160513072" $\mathbf{\chi }$are less thanrole="math" localid="1652160519976" ${\mathbf{\chi }}_{\mathbf{0}}$

f)1% of the values of$\mathit{\chi }$ are greater than${\mathbf{\chi }}_{\mathbf{0}}$

The calculation is given below:

$\text{χ}~\text{N}(\mathrm{\mu },{\mathrm{\sigma }}^2)$

Given,

$\begin{array}{l}\mu =50\\ \sigma =3\end{array}$

$\begin{array}{c}\text{P}(\text{χ}\le {\text{χ}}_{\text{0}})=0.8413\\ \text{P}(\text{χ}\le {\text{χ}}_{\text{0}})=\text{P}(\text{z}\le {\text{z}}_{\text{0}})=0.8413\\ {\text{z}}_{\text{0}}=0.999815=\frac{{\text{χ}}_{\text{0}}-\mu }{\sigma }\\ {\text{χ}}_{\text{0}}=3\left(0.999815\right)+50\approx 53\end{array}$

The calculation is given below:

$$\begin{gathered} \text{P}(\text{χ}\ge {\text{χ}}_{\text{0}})=0.25 \\ \text{P}(\text{χ}\le {\text{χ}}_{\text{0}})=1-\text{P}(\text{χ}\ge {\text{χ}}_{\text{0}})=1-0.25=0.75 \\ \text{P}(\text{χ}\le {\text{χ}}_{\text{0}})=\text{P}(\text{z}\le {\text{z}}_{\text{0}})=0.75 \\ {\text{z}}_{\text{0}}=0.674490=\frac{{\text{χ}}_{\text{0}}-\mu }{\sigma } \\ {\text{χ}}_{\text{0}}=3\left(0.674490\right)+50\approx 52 \end{gathered}$$

The calculation is given below:

$$\begin{gathered} \text{P}(\text{χ}\ge {\text{χ}}_{\text{0}})=0.95 \\ \text{P}(\text{χ}\le {\text{χ}}_{\text{0}})=1-\text{P}(\text{χ}\ge {\text{χ}}_{\text{0}})=1-0.95=0.05 \\ \text{P}(\text{χ}\le {\text{χ}}_{\text{0}})=\text{P}(\text{z}\le {\text{z}}_{\text{0}})=0.05 \\ {\text{z}}_{\text{0}}=-1.6745=\frac{{\text{χ}}_{\text{0}}-\mu }{\sigma } \end{gathered}$$

${\text{χ}}_{\text{0}}=3(-1.645)+50\approx 45$

The calculation is given below:

$$\begin{gathered} \text{P}(\text{41}\le \text{χ}\le {\text{χ}}_{\text{0}})=\text{P}(\text{χ}\le {\text{χ}}_{\text{0}})-\text{P}(\text{χ}\le \text{41})=0.8630 \\ \text{=P}(\text{z}\le {\text{z}}_{\text{0}})=-\text{P}(\text{z}\le \frac{\text{41}-\text{50}}{3})=0.8630 \\ \text{=P}(\text{z}\le {\text{z}}_{\text{0}})=-\text{P}(\text{z}\le -3)=0.8630 \\ \text{P}(\text{z}\le {\text{z}}_{\text{0}})-0.00135=0.86435 \\ \text{P}(\text{z}\le {\text{z}}_{\text{0}})=0.86435 \\ {\text{z}}_{\text{0}}=1.1=\frac{{\text{χ}}_{\text{0}}-\mu }{\sigma } \\ {\text{χ}}_{\text{0}}=3\left(1.1\right)+50\approx 53.3 \end{gathered}$$

The calculation is given below:

$$\begin{gathered} \text{P}(\text{χ}<{\text{χ}}_{\text{0}})=0.1 \\ \text{P}(\text{χ}\le {\text{χ}}_{\text{0}})=\text{P}(\text{z}\ge {\text{z}}_{\text{0}})=0.1 \\ {\text{z}}_{\text{0}}=-1.28155=\frac{{\text{χ}}_{\text{0}}-\mu }{\sigma } \\ {\text{χ}}_{\text{0}}=3(-1.28155)+50\approx 46.155 \end{gathered}$$

The calculation is given below:

$$\begin{gathered} \text{P}(\text{χ}>{\text{χ}}_{\text{0}})=0.01 \\ \text{P}(\text{χ}\le {\text{χ}}_{\text{0}})=1-\text{P}(\text{χ}>{\text{χ}}_{\text{0}})=1-0.1=0.99 \\ \text{P}(\text{χ}\le {\text{χ}}_{\text{0}})=\text{P}(\text{z}\le {\text{z}}_{\text{0}})=0.99 \\ {\text{z}}_{\text{0}}=2.326=\frac{{\text{χ}}_{\text{0}}-\mu }{\sigma } \end{gathered}$$

${\text{χ}}_{\text{0}}=3\left(2.326\right)+50\approx 56.978$