Answer
$P(machine~needs~an~adjustment)=0.0062$
Work Step by Step
The z-score for 0.748:
$z=\frac{x̄ −μ}{\frac{σ}{\sqrt n}}=\frac{0.748-0.75}{\frac{0.004}{\sqrt {30}}}=-2.74$
The z-score for 0.752:
$z=\frac{x̄ −μ}{\frac{σ}{\sqrt n}}=\frac{0.752-0.75}{\frac{0.004}{\sqrt {30}}}=2.74$
$P(x̄\lt0.748~or~x̄\gt0.752)=P(x̄\lt0.748)+P(x̄\gt0.752)=P(z\lt-2.74)+P(z\gt2.74)$
Given the symmetry of the normal distribution we have that: $P(z\lt-2.74)=P(z\gt2.74)$. So:
$P(x̄\lt0.748~or~x̄\gt0.752)=2P(z\lt-2.74)=2\times0.0031=0.0062$
