A maufacturing process produces machine parts with a mean di

A maufacturing process produces machine parts with a mean diameter of 3,60 cm and a standard deviation of 0.02 cm.

Solution

X bar is normal with 3.6 and 0.02 cm

a) For percent first let us compute probability

P(3.57<x<3.63) = P(-0.03<x-mu<0.03)

= P(-1.5<z<1.5) = 0.8664

Hence 86.64% are within this values

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b) If sample size = 5

std error = 0.02/rt 5 = 0.0089

P(|x-mu|<0.03) = P(|z|<0.03/0.0089)

= P(|z|<3.354)

=1.00(almost definite)

Hence 100% of samples