Answer
$0.09558$
Work Step by Step
Write the equation for standard normal curve.
$P(a \leq Z\leq b)=\int_a^b \exp(\dfrac{-t^2}{2}) \ dt$
where, $Z=\dfrac{x-\mu}{\sigma}$
$\text{P(away from its mean) = 1-P(within its mean)}$
Plug in the above equation the given values to obtain:
$1-P(-\dfrac{5}{3} \leq Z \leq \dfrac{5}{3} )=1-\dfrac{1}{\sqrt 2 \pi}\int_{-\frac{5}{3}}^{\frac{5}{3}} \exp(\dfrac{-t^2}{2}) \ dt$
Therefore, $\text{P(Away from its mean)}=0.09558$
