Suppose x is a normally distributed random variable with $\mathrm{\mu }$= 11 and $\mathrm{\sigma }$= 2. Find each of the following:

a)$\mathbf{P}(\mathbf{10}\le \mathbf{\chi }\le \mathbf{12})$

b) $\mathbf{P}(\mathbf{6}\le \mathbf{\chi }\le \mathbf{10})$

c)$\mathbf{P}(\mathbf{13}\le \mathbf{\chi }\le \mathbf{16})$

d)$\mathbf{P}(\mathbf{7}\mathbf{.}\mathbf{8}\le \mathbf{\chi }\le \mathbf{12}\mathbf{.}\mathbf{6})$

e)$\mathbf{P}(\mathbf{\chi }\ge \mathbf{13}\mathbf{.}\mathbf{24})$

f)$\mathbf{P}(\mathbf{\chi }\ge \mathbf{7}\mathbf{.}\mathbf{62})$

The calculation is given below:

Given,

$\mathrm{\chi }$is normally distributed

$$\begin{gathered} \mathrm{\mu }=11 \\ \mathrm{\sigma }=2 \end{gathered}$$

localid="1652160102930" $$\begin{gathered} \text{P}(10\le \text{χ}\le 12)=\text{P}\left(\frac{{\displaystyle 10-\mu }}{{\displaystyle \sigma }}\le \frac{{\displaystyle \text{χ}-\mu }}{{\displaystyle \sigma }}\le \frac{{\displaystyle \text{12}-\mu }}{{\displaystyle \sigma }}\right) \\ =\text{P}\left(\frac{{\displaystyle 10-11}}{{\displaystyle 2}}\le \text{z}\le \frac{{\displaystyle \text{12}-11}}{{\displaystyle 2}}\right) \\ =\text{P}(-0.5\le \text{z}\le 0.5) \\ =\text{P}(\text{z}\le 0.5)-\text{P}(\text{z}\le -0.5) \\ =\text{P}(\text{z}\le 0.5)-(1-\text{P}(\text{z}\le -0.5)) \\ =0.6915-(1-0.6915) \\ \text{P}(10\le \text{χ}\le 12)=0.383 \end{gathered}$$

The calculation is given below:

$$\begin{gathered} \text{P}(6\le \text{χ}\le \text{10})=\text{P}\left(\frac{{\displaystyle 6-\mu }}{{\displaystyle \sigma }}\le \text{z}\le \frac{{\displaystyle \text{10}-\mu }}{{\displaystyle \sigma }}\right) \\ =\text{P}\left(\frac{{\displaystyle 6-11}}{{\displaystyle 2}}\le \text{z}\le \frac{{\displaystyle \text{10}-11}}{{\displaystyle 2}}\right) \\ =\text{P}(-2.5\le \text{z}\le -0.5) \\ =\text{P}(\text{z}\le -0.5)-\text{P}(\text{z}\le -2.5) \\ =(\text{1}-\text{P}(\text{z}\le 0.5))-(1-\text{P}(\text{z}\le 2.5)) \\ =(1-0.6915)-(1-0.9938) \\ =0.3023 \end{gathered}$$

The calculation is given below:

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

The calculation is given below:

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

The calculation is given below:

$$\begin{gathered} \text{P}(\text{χ}\ge \text{13.24})=\text{P}\left(\frac{{\displaystyle \text{χ}-\mu }}{{\displaystyle \sigma }}\ge \frac{{\displaystyle \text{13.24}-\mu }}{{\displaystyle \sigma }}\right) \\ =\text{P}\left((\text{z}\ge \frac{{\displaystyle \text{13.24}-11}}{{\displaystyle 2}}\right) \\ =\text{P}(\text{z}\ge 1.12) \\ =1-\text{P}(\text{z}\le 1.12) \\ =1-0.8686 \\ =0.1314 \end{gathered}$$

The calculation is given below:

$$\begin{gathered} \text{P}(\text{χ}\ge \text{7.62})=\text{P}\left(\frac{{\displaystyle \text{χ}-\mu }}{{\displaystyle \sigma }}\ge \frac{{\displaystyle \text{7.62}-\mu }}{{\displaystyle \sigma }}\right) \\ =\text{P}\left(\text{z}\ge \frac{{\displaystyle \text{7.62}-11}}{{\displaystyle 2}}\right) \\ =\text{P}(\text{z}\ge -1.69) \\ =\text{P}(\text{z}\le 1.69) \\ =0.9545 \end{gathered}$$