# James' question about Normal Distribution

#### Prove It

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MHB Math Helper

(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*}