Answer:
Hence the probability that a randomly selected wire from company A's production will meet the specifications is 0.95455.
Step-by-step explanation:
a)[tex]P(0.14 < x < 0.16 ) = P[(0.14 - 0.15)/ 0.005) < (x - \mu) /\sigma < (0.16 - 0.15) / 0.005) ][/tex]
[tex]=P(0.14 < x < 0.16 ) = P[(0.14 - 0.15)/ 0.005)< ((x - 0.15) /0.005) < (0.16 - 0.15) / 0.005) ][/tex]
[tex]=P(z<\frac{0.01}{0.005} )- P(z<-\frac{0.01}{0.005})[/tex]
Using z table,
= 0.9773 - 0.02275
= 0.95455.