the percentage of defectives we can expect if we set the tolerance limits at 2 ± 0.02 in is
We need to calculate the percentage of defective productions. That means if rods are of length > $2.02$ or < $1.98$
$Z = \frac{X - \lambda }{\sigma}$
$P(1.98 \leq X \leq 2.02) = P(X \leq 2.02) - P( X \leq 1.98)$
$= P(Z \leq \frac{2.02 - 2}{0.008}) - P( Z \leq \frac{1.98 - 2}{0.008})$
$= P(Z \leq 2.5) - P( Z \leq -2.5)$
$= P(Z \leq 2.5) - (1 - P( Z \leq 2.5))$
So the question contains an error "ϕ(0.25) = 0.99"
but $\phi(2.5) = 0.99$
$= P(Z \leq 2.5) - (1 - P( Z \leq 2.5))$
$= 0.99 - (1 - 0.99)$
$P(1.98 \leq X \leq 2.02) = 0.98$
Hence the percentage of defectives we can expect is 100% - 98% which is 2%