Answer:
The answer is "15.85%"
Step-by-step explanation:
The complete question is defined in the attachment file please find it.
Given:
[tex]\mu = 1\\\\\sigma = 0.001 \\\\[/tex]
Converting the Standard Normal:
[tex]\to P(X < x) = P( Z < \frac{( X - \mu )}{\sigma} )\\\\\to P ( 0.9998 < X < 1.0002 ) = P ( Z < \frac{( 1.0002 - 1 )}{0.001}) - P ( Z < \frac{( 0.9998 - 1 )}{0.001})\\\\[/tex]
[tex]= P ( Z < 0.2) - P ( Z < -0.2 )\\\\= 0.5793 - 0.4207\\\\= 0.1585\\\\= 15.85\%\\\\[/tex]
