Answer
$P(x ̅\gt0)=0.4522$
Work Step by Step
$μ_ {x ̅}=-0.05$
$σ_{x ̅}=\frac{σ_X}{\sqrt n}=\frac{5.76}{\sqrt {200}}=0.407$
Let's fint the z-score for 0:
$z=\frac{X-μ_ {x ̅}}{σ_{x ̅}}=\frac{0-(-0.05)}{0.407}=0.12$
According to Table V, the area of the standard normal curve to the left of z-score equal to 0.12 is 0.5478.
But, we want the area of the standard normal curve to the right (greater than 0) of z-score equal to 0.12:
$1-0.5478=0.4522$
