Question

Assume $X$ is normally distributed with a mean of $12$ and a standard deviation of $2$.

Determine the value for $x$ that solves each of the following.

Round the answers to $2$ decimal places.

$\left(a\right)$ $\text{P}\left(X>x\right)=0.5$

$x=?$

$\left(b\right)$ $\text{P}\left(X>x\right)=0.95$

$x=?$

$\left(c\right)$ $\text{P}\left(x<X<12\right)=0.2$

$x=?$

$\left(d\right)$ $\text{P}\left(-x<X-12<x\right)=0.95$

$x=?$

$\left(e\right)$ $\text{P}\left(-x<X-12<x\right)=0.99$

$x=?$

Answer 1

Step-1: Calculate the value of $x$ for $\text{P}\left(X>x\right)=0.5$

$x=\text{Mean}+\left(z\times SD\right)$

$\text{P}\left(X>x\right)=0.5$

Here, $z=0$

$\therefore x=12+\left(0\times 2\right)=12$

Step-2: Calculate the value of $x$ for $\text{P}\left(X>x\right)=0.95$

$\text{P}\left(X>x\right)=0.95$

Here, $z=-1.645$

$\therefore x=12+\left(-1.645\times 2\right)=8.71$

Step-3: Calculate the value of $x$ for $\text{P}\left(x<X<12\right)=0.2$

$\text{P}\left(x<X<12\right)=0.2$

Here, $z=-0.525$

$\therefore x=12+\left(-0.525\times 2\right)=10.95$

Step-4: Calculate the value of $x$ for $\text{P}\left(-x<X-12<x\right)=0.95$

$\text{P}\left(-x<X-12<x\right)=0.95$

Here, $\text{P}\left(-1.96<z<1.96\right)=0.9500$

$\therefore x=1.96$

Step-5: Calculate the value of $x$ for $\text{P}\left(-x<X-12<x\right)=0.99$

$\text{P}\left(-x<X-12<x\right)=0.99$

Here, $\text{P}\left(-2.575<z<2.575\right)=0.99$

$\therefore x=0.99$

Hence, the values of $x$ are,

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{}\mathit{x}\mathbf{=}\mathbf{12} \\ \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{}\mathit{x}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{71} \\ \mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{}\mathbf{10}\mathbf{.}\mathbf{95} \\ \mathbf{\left(}\mathit{d}\mathbf{\right)}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{96} \\ \mathbf{\left(}\mathit{e}\mathbf{\right)}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{99} \end{gathered}$$