Answer
0.0668 of ball bearings have a diameter greater than 5.03mm.
Work Step by Step
$\mu$ = 5, $\sigma$ = 0.02
Want to find $P( x > 5.03 )$
i) Convert 5.03 to a z-score:
z = $\frac{x - \mu}{\sigma}$
z = $\frac{5.03 - 5}{0.02}$
z= 1.5
ii) $P(x > 5.03)$
$= P(z > 1.5)$
$= 1 - P( z < 1.5)$
$= 1-0.9332$
$= 0.0668$
