91Ó°ÊÓ

The calculation is given below:

$\text{χ}~\text{N}(\mathrm{μ},{\mathrm{σ}}^2)$

Given,

$\begin{array}{l}\mathrm{μ}=50\\ \mathrm{σ}=3\end{array}$

$\begin{array}{c}\text{P}(\text{χ}≤{\text{χ}}_{\text{0}})=0.8413\\ \text{P}(\text{χ}≤{\text{χ}}_{\text{0}})=\text{P}(\text{z}≤{\text{z}}_{\text{0}})=0.8413\\ {\text{z}}_{\text{0}}=0.999815=\frac{{\text{χ}}_{\text{0}}−\mathrm{μ}}{\mathrm{σ}}\\ {\text{χ}}_{\text{0}}=3\left(0.999815\right)+50≈53\end{array}$

The calculation is given below:

$$\begin{gathered} \text{P}(\text{χ}â‰\yen {\text{χ}}_{\text{0}})=0.25 \\ \text{P}(\text{χ}≤{\text{χ}}_{\text{0}})=1−\text{P}(\text{χ}â‰\yen {\text{χ}}_{\text{0}})=1−0.25=0.75 \\ \text{P}(\text{χ}≤{\text{χ}}_{\text{0}})=\text{P}(\text{z}≤{\text{z}}_{\text{0}})=0.75 \\ {\text{z}}_{\text{0}}=0.674490=\frac{{\text{χ}}_{\text{0}}−\mathrm{μ}}{\mathrm{σ}} \\ {\text{χ}}_{\text{0}}=3\left(0.674490\right)+50≈52 \end{gathered}$$

The calculation is given below:

$$\begin{gathered} \text{P}(\text{χ}â‰\yen {\text{χ}}_{\text{0}})=0.95 \\ \text{P}(\text{χ}≤{\text{χ}}_{\text{0}})=1−\text{P}(\text{χ}â‰\yen {\text{χ}}_{\text{0}})=1−0.95=0.05 \\ \text{P}(\text{χ}≤{\text{χ}}_{\text{0}})=\text{P}(\text{z}≤{\text{z}}_{\text{0}})=0.05 \\ {\text{z}}_{\text{0}}=−1.6745=\frac{{\text{χ}}_{\text{0}}−\mathrm{μ}}{\mathrm{σ}} \end{gathered}$$

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

The calculation is given below:

$$\begin{gathered} \text{P}(\text{41}≤\text{χ}≤{\text{χ}}_{\text{0}})=\text{P}(\text{χ}≤{\text{χ}}_{\text{0}})−\text{P}(\text{χ}≤\text{41})=0.8630 \\ \text{=P}(\text{z}≤{\text{z}}_{\text{0}})=−\text{P}(\text{z}≤\frac{\text{41}−\text{50}}{3})=0.8630 \\ \text{=P}(\text{z}≤{\text{z}}_{\text{0}})=−\text{P}(\text{z}≤−3)=0.8630 \\ \text{P}(\text{z}≤{\text{z}}_{\text{0}})−0.00135=0.86435 \\ \text{P}(\text{z}≤{\text{z}}_{\text{0}})=0.86435 \\ {\text{z}}_{\text{0}}=1.1=\frac{{\text{χ}}_{\text{0}}−\mathrm{μ}}{\mathrm{σ}} \\ {\text{χ}}_{\text{0}}=3\left(1.1\right)+50≈53.3 \end{gathered}$$

The calculation is given below:

$$\begin{gathered} \text{P}(\text{χ}<{\text{χ}}_{\text{0}})=0.1 \\ \text{P}(\text{χ}≤{\text{χ}}_{\text{0}})=\text{P}(\text{z}â‰\yen {\text{z}}_{\text{0}})=0.1 \\ {\text{z}}_{\text{0}}=−1.28155=\frac{{\text{χ}}_{\text{0}}−\mathrm{μ}}{\mathrm{σ}} \\ {\text{χ}}_{\text{0}}=3(−1.28155)+50≈46.155 \end{gathered}$$

The calculation is given below:

$$\begin{gathered} \text{P}(\text{χ}>{\text{χ}}_{\text{0}})=0.01 \\ \text{P}(\text{χ}≤{\text{χ}}_{\text{0}})=1−\text{P}(\text{χ}>{\text{χ}}_{\text{0}})=1−0.1=0.99 \\ \text{P}(\text{χ}≤{\text{χ}}_{\text{0}})=\text{P}(\text{z}≤{\text{z}}_{\text{0}})=0.99 \\ {\text{z}}_{\text{0}}=2.326=\frac{{\text{χ}}_{\text{0}}−\mathrm{μ}}{\mathrm{σ}} \end{gathered}$$

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