MHB James' question about Normal Distribution

[Prove It]

(a) We are told $\displaystyle \begin{align*} \textrm{Pr}\,\left( X < 3 \right) = \textrm{Pr}\,\left( Z < a \right) \end{align*}$, so if $\displaystyle \begin{align*} x = 3 \end{align*}$ and $\displaystyle \begin{align*} z = a \end{align*}$ then we have

$\displaystyle \begin{align*} z &= \frac{x - \mu}{\sigma} \\ a &= \frac{3 - 5}{2} \\ a &= \frac{-2}{\phantom{-}2} \\ a &= -1 \end{align*}$(b) We are told $\displaystyle \begin{align*} \textrm{Pr}\,\left( X > 8 \right) = \textrm{Pr}\,\left( Z > b \right) \end{align*}$, so if $\displaystyle \begin{align*} x = 8 \end{align*}$ then we have

$\displaystyle \begin{align*} z &= \frac{x - \mu}{\sigma} \\ b &= \frac{8 - 5}{2} \\ b &= \frac{3}{2} \\ b &= 1.5 \end{align*}$(c) We are told $\displaystyle \begin{align*} \textrm{Pr}\,\left( X > 6 \right) = \textrm{Pr}\,\left( Z < c \right) \end{align*}$, so by symmetry, $\displaystyle \begin{align*} \textrm{Pr}\,\left( X > 6 \right) = \textrm{Pr}\,\left( Z > -c \right) \end{align*}$, and thus if $\displaystyle \begin{align*} x = 6 \end{align*}$ then

$\displaystyle \begin{align*} z &= \frac{x - \mu}{\sigma} \\ -c &= \frac{6 - 5}{2} \\ -c &= \frac{1}{2} \\ c &= -\frac{1}{2} \end{align*}$

 

[Mark44]

View attachment 309785
(a) We are told $\displaystyle \begin{align*} \textrm{Pr}\,\left( X < 3 \right) = \textrm{Pr}\,\left( Z < a \right) \end{align*}$, so if $\displaystyle \begin{align*} x = 3 \end{align*}$ and $\displaystyle \begin{align*} z = a \end{align*}$ then we have

$\displaystyle \begin{align*} z &= \frac{x - \mu}{\sigma} \\ a &= \frac{3 - 5}{2} \\ a &= \frac{-2}{\phantom{-}2} \\ a &= -1 \end{align*}$(b) We are told $\displaystyle \begin{align*} \textrm{Pr}\,\left( X > 8 \right) = \textrm{Pr}\,\left( Z > b \right) \end{align*}$, so if $\displaystyle \begin{align*} x = 8 \end{align*}$ then we have

$\displaystyle \begin{align*} z &= \frac{x - \mu}{\sigma} \\ b &= \frac{8 - 5}{2} \\ b &= \frac{3}{2} \\ b &= 1.5 \end{align*}$(c) We are told $\displaystyle \begin{align*} \textrm{Pr}\,\left( X > 6 \right) = \textrm{Pr}\,\left( Z < c \right) \end{align*}$, so by symmetry, $\displaystyle \begin{align*} \textrm{Pr}\,\left( X > 6 \right) = \textrm{Pr}\,\left( Z > -c \right) \end{align*}$, and thus if $\displaystyle \begin{align*} x = 6 \end{align*}$ then

$\displaystyle \begin{align*} z &= \frac{x - \mu}{\sigma} \\ -c &= \frac{6 - 5}{2} \\ -c &= \frac{1}{2} \\ c &= -\frac{1}{2} \end{align*}$

Your answers look fine but I wouldn't have done things as you did.
You're given that $\mu = 5$ and $\sigma = 2$, so $Z = \frac{X - \mu}\sigma$. Substituting for the given parameters, we have $Z = \frac{X - 5}2 \Rightarrow X = 2Z + 5$.

For part a, $Pr(X < 3) = Pr(2Z + 5 < 3) = Pr(2z < -2) = Pr(Z < -1)$, so $a = -1$, same answer that you gave, but probably cleaner in its derivation. The other two parts are done similarly.