The random variable x has a normal distribution with $\mathit{\mu }\mathbf{=}\mathbf{40}$and ${\mathit{\sigma }}^{\mathbf{2}}\mathbf{=}\mathbf{36}$. Find a value of x, call it${\mathit{x}}_{\mathbf{0}}$, such that

a.$\mathit{P}\left(x\ge x_0\right)\mathbf{=}\mathbf{0}\mathbf{.10}$

b.$\mathit{P}\mathbf{(}\mathit{\mu }\mathbf{\le }\mathbf{x}\mathbf{\le }{\mathbf{x}}_{\mathbf{0}}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}$

c.$\mathit{P}\mathbf{(}\mathbf{x}\mathbf{\le }{\mathbf{x}}_{\mathbf{0}}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$

d.$\mathit{P}\mathbf{(}\mathbf{x}\mathbf{\ge }{\mathbf{x}}_{\mathbf{0}}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}$

e.$\mathit{P}\left(x_0\le x<\mu \right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{45}$

Random variable x follows normal distribution with mean 40 and variance 36

That is $\begin{array}{l}{\sigma }^2=36\\ \sigma =6\end{array}$

The value of$x_0$can be calculated as

$P\left(x\ge x_0\right)=0.10$

Hence

$P\left(z\ge z_0\right)=0.10$

Consider$P\left(z\ge z_0\right)$

$$\begin{gathered} P\left(z\ge z_0\right)=0.5-P\left(0\le z\le z_0\right) \\ =0.5-A \\ =0.1 \end{gathered}$$

From the normal curve we can say

$$\begin{gathered} A=0.5-0.1 \\ =0.4 \end{gathered}$$

The $z_0$value which represents the area A is 1.28

Hence the standard normal variable is,

$$\begin{gathered} z_0=\frac{x_0-\mu }{\sigma } \\ \frac{x_0-40}{6}=1.28 \\ \frac{x_0-40}{6}=1.28 \\ {x}_0-40=1.28\times 6 \\ {x}_0=7.68+40 \\ {x}_0=47.68 \end{gathered}$$

Hence the value of$\mathrm{P}(x\ge x_0)=0.10$is 47.68

b.

The value of$x_0$can be calculated as:

$$\begin{gathered} z_1=\frac{\mu -\mu }{\sigma } \\ =0 \end{gathered}$$

$P\left(\mu \le x\le x_0\right)=0.40$

Hence

$\begin{array}{l}P\left(z_1\le x\le z_0\right)=0.40\\ P\left(0\le x\le z_0\right)=0.40\end{array}$

The $z_0$value which represents the area A is 1.28

Hence the standard normal variable is,

$$\begin{gathered} z_0=\frac{x_0-\mu }{\sigma } \\ \frac{x_0-40}{6}=1.28 \\ {x}_0-40=1.28\times 6 \\ {x}_0-40=7.68 \\ {x}_0=7.68+40 \\ {x}_0=47.68 \end{gathered}$$

Hence the value of $P\left(\mu \le x\le x_0\right)=0.40$is 47.68

The value of$x_0$can be calculated as:

$\mathrm{P}(x\le x_0)=0.005$

$\mathrm{P}(z<{}_0)=0.05$

consider $\mathrm{P}(z>z_0)$

$$\begin{gathered} P\left(z>z_0\right)=0.5-P\left(-z_0\le z\le 0\right) \\ =0.5-A \\ =0.05 \end{gathered}$$

$$\begin{gathered} A=0.5-0.05 \\ =0.45 \end{gathered}$$

The$z_0$value which represents the area A is -1.645

Hence the standard normal variable is,

$Z_0=\frac{x_0-\mu }{\sigma }$

$\begin{array}{l}\frac{x_0-40}{6}=-1.645\\ x_0=-1.645\times 6+40\\ x_0=30.13\end{array}$

Hence the value of $P\left(x\le x_0\right)=0.05$is 30.13

The value of$x_0$can be calculated as:

$\mathrm{P}(x\ge x_0)=0.40$

Hence

$$\begin{gathered} P\left(z\ge z_0\right)=0.40 \\ \mathrm{Consider}P\left(z\ge z_0\right) \end{gathered}$$

$$\begin{gathered} \mathrm{P}\left(\mathrm{z}\ge {\mathrm{z}}_0\right)=0.5-\mathrm{P}\left(0\le \mathrm{z}\le {\mathrm{z}}_0\right) \\ =0.5-A \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}0.4 \end{gathered}$$

$$\begin{gathered} A=0.5-0.4 \\ =0.1 \end{gathered}$$

The$z_0$value which represents the area A is 0.255

$z_0=\frac{x_0-\mu }{\sigma }$

$\begin{array}{l}\frac{x_0-40}{6}=0.255\\ x_0-40=0.255\times 6\\ x_0=41.53\end{array}$

hence the value of the $\mathrm{P}\left(\mathrm{x}\ge {\mathrm{x}}_0\right)=0.40$is 41.53

The value$x_0$of can be calculated as:

$$\begin{gathered} z_1=\frac{\mu -\mu }{\sigma } \\ =0 \end{gathered}$$

$P\left(\mu \le x\le x_0\right)=0.45$

Hence

$\begin{array}{l}P\left(z_1\le x\le z_0\right)=0.45\\ P\left(0\le x\le z_0\right)=0.45\end{array}$

The $z_0$value which represents the area A is -1.645

Hence the standard normal variable is,

$Z_0=\frac{x_0-\mu }{\sigma }$

$\begin{array}{l}\frac{x_0-40}{6}=-1.645\\ x_0=-1.645\times 6+40\\ x_0=30.13\end{array}$

$P\left(x_0\le x<\mu \right)=0.45\mathrm{is}30.13$

Hence the value of