Question

Question # 2 The diameter of a shaft in an optical storage drive is normally distributed with mean 0.2508 inch and standard deviation 0.0005 inch. The specifications on the shaft are 0.2500 ± 0.0015 inch. What proportion of shafts conforms to specifications? he effective life of a component used in a jet-turbine aircraft engine is a random variable f effective life is fairly close to a normal distribution. The engine manufacturer introduces an improvement 050 hours and decreases the standard deviation to 30 hours. Suppose that a random sample of ni = 16 components is selected from the "old" process and a random sample of n2 25 components is selected from the "improved" process. What is the probability that the difference in the two sample means X2-X is at least 25 hours? Assume that the old and improved processes can with mean 5000 hours and standard deviation 40 hours. The distribution o into the manufacturing process for this component that increases the mean life to 5 be regarded as independent populations.

Answer 1

2)

i)

Z = (X - mean)/sd

=(X - 0.2508)/0.0005

P ( 0.2485<X<0.2515 )=P ( −4.6<Z<1.4 )

= 0.9192

ii)

mu1 - mu2 = 5000 - 5050 = -50
sqrt(40^2 /16 + 30^2 /25) = 11.66190

Z = ((Xbar1 - Xbar2)- (-50) )/11.66190

P(|Xbar1 - Xbar2| > 25)
= 0.984