91Ó°ÊÓ

The calculation is given below:

Given,

$\mathrm{χ}$is normally distributed

$$\begin{gathered} \mathrm{μ}=11 \\ \mathrm{σ}=2 \end{gathered}$$

localid="1652160102930" $$\begin{gathered} \text{P}(10≤\text{χ}≤12)=\text{P}\left(\frac{{\displaystyle 10−\mathrm{μ}}}{{\displaystyle \mathrm{σ}}}≤\frac{{\displaystyle \text{χ}−\mathrm{μ}}}{{\displaystyle \mathrm{σ}}}≤\frac{{\displaystyle \text{12}−\mathrm{μ}}}{{\displaystyle \mathrm{σ}}}\right) \\ =\text{P}\left(\frac{{\displaystyle 10−11}}{{\displaystyle 2}}≤\text{z}≤\frac{{\displaystyle \text{12}−11}}{{\displaystyle 2}}\right) \\ =\text{P}(−0.5≤\text{z}≤0.5) \\ =\text{P}(\text{z}≤0.5)−\text{P}(\text{z}≤−0.5) \\ =\text{P}(\text{z}≤0.5)−(1−\text{P}(\text{z}≤−0.5)) \\ =0.6915−(1−0.6915) \\ \text{P}(10≤\text{χ}≤12)=0.383 \end{gathered}$$

The calculation is given below:

$$\begin{gathered} \text{P}(6≤\text{χ}≤\text{10})=\text{P}\left(\frac{{\displaystyle 6−\mathrm{μ}}}{{\displaystyle \mathrm{σ}}}≤\text{z}≤\frac{{\displaystyle \text{10}−\mathrm{μ}}}{{\displaystyle \mathrm{σ}}}\right) \\ =\text{P}\left(\frac{{\displaystyle 6−11}}{{\displaystyle 2}}≤\text{z}≤\frac{{\displaystyle \text{10}−11}}{{\displaystyle 2}}\right) \\ =\text{P}(−2.5≤\text{z}≤−0.5) \\ =\text{P}(\text{z}≤−0.5)−\text{P}(\text{z}≤−2.5) \\ =(\text{1}−\text{P}(\text{z}≤0.5))−(1−\text{P}(\text{z}≤2.5)) \\ =(1−0.6915)−(1−0.9938) \\ =0.3023 \end{gathered}$$

The calculation is given below:

$$\begin{gathered} \text{P}(\text{13}≤\text{χ}≤\text{16})=\text{P}\left(\frac{{\displaystyle 13−11}}{{\displaystyle 2}}≤\text{z}≤\frac{{\displaystyle 16−11}}{{\displaystyle 2}}\right) \\ =\text{P}(1≤\text{z}≤2.5) \\ =\text{P}(\text{z}≤2.5)−\text{P}(\text{z}≤1) \\ =0.9938−0.8413 \\ =0.1525 \end{gathered}$$

The calculation is given below:

$$\begin{gathered} \text{P}(\text{7.8}≤\text{χ}≤\text{12.6})=\text{P}\left(\frac{{\displaystyle 7.8−11}}{{\displaystyle 2}}≤\text{z}≤\frac{{\displaystyle \text{12.6}−11}}{{\displaystyle 2}}\right) \\ =\text{P}(−1.6≤\text{z}≤0.8) \\ =\text{P}(\text{z}≤0.8)−\text{P}(\text{z}≤−1.6) \\ =\text{P}(\text{z}≤−0.8)−(1−(\text{z}≤−1.6)) \\ =0.7881−(1−0.9452) \\ =0.7333 \end{gathered}$$

The calculation is given below:

$$\begin{gathered} \text{P}(\text{χ}â‰\yen \text{13.24})=\text{P}\left(\frac{{\displaystyle \text{χ}−\mathrm{μ}}}{{\displaystyle \mathrm{σ}}}â‰\yen \frac{{\displaystyle \text{13.24}−\mathrm{μ}}}{{\displaystyle \mathrm{σ}}}\right) \\ =\text{P}\left((\text{z}â‰\yen \frac{{\displaystyle \text{13.24}−11}}{{\displaystyle 2}}\right) \\ =\text{P}(\text{z}â‰\yen 1.12) \\ =1−\text{P}(\text{z}≤1.12) \\ =1−0.8686 \\ =0.1314 \end{gathered}$$

The calculation is given below:

$$\begin{gathered} \text{P}(\text{χ}â‰\yen \text{7.62})=\text{P}\left(\frac{{\displaystyle \text{χ}−\mathrm{μ}}}{{\displaystyle \mathrm{σ}}}â‰\yen \frac{{\displaystyle \text{7.62}−\mathrm{μ}}}{{\displaystyle \mathrm{σ}}}\right) \\ =\text{P}\left(\text{z}â‰\yen \frac{{\displaystyle \text{7.62}−11}}{{\displaystyle 2}}\right) \\ =\text{P}(\text{z}â‰\yen −1.69) \\ =\text{P}(\text{z}≤1.69) \\ =0.9545 \end{gathered}$$