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James' question about Normal Distribution

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
normal distribution.jpg

(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

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