Answer
$P(shut~down)=P(x ̅\lt1.98$ or $x ̅\gt2.02)=0.0074$
Work Step by Step
$μ_{x ̅}=2.0$
$σ_{x ̅}=0.00745$
Let's find the z-scores for 1.98 and 2.02:
$z=\frac{X-μ_{x ̅}}{σ_{x ̅}}=\frac{1.98-2.0}{0.00745}=-2.68$
$z=\frac{X-μ_{x ̅}}{σ_{x ̅}}=\frac{2.02-2.0}{0.00745}=2.68$
According to Table V, the area of the standard normal curve to the left of z-score equal to -2.68 is 0.0037.
According to Table V, the area of the standard normal curve to the left of z-score equal to 2.68 is 0.9963.
The area of the standard normal curve between the z-scores -2.68 and 2.68 is the difference between the area of the standard normal curve to the left of z-score equal to 2.68 and the area of the standard normal curve to the left of z-score equal to -2.68:
$P(1.98\leq x ̅\leq2.02)=0.9963-0.0037=0.9926$
The event "$x ̅\lt1.98$ or $x ̅\gt2.02$" is the complement of the event "$1.98\leq x ̅\leq2.02$".
Using the Complement Rule (see page 275):
$P(x ̅\lt1.98$ or $x ̅\gt2.02)=1-P(1.98\leq x ̅\leq2.02)=1-0.9926=0.0074$
