Homework SourseAutomated manufacturing operations are quite precise but sti
Automated manufacturing operations are quite precise but still vary, often distributions that are close to normal to Normal. The widths in inches of slots cut by a milling machine are approximately normal, ie. N(0.8750, 0.0012). The specification allow slot widths between 0.8720 and 0.8780 inches.
a) What proportion of slots meets these specifications?
b) The GM realizes that it would cost a lot of money to check ever slot. So she decides to look at the average slot size. She takes a sample of 10 slots. What is the probability that the mean slot size will be larger than 0.877?
c) If we increase the sample size to 25, how does the probability change?
d) Using n= 25, what is the probability that the mean is between 0.8743 and 0.8759?
Solution
a) What proportion of slots meets these specifications?
P(0.8720 <X< 0.8780)
= P((0.872-0.875)/0.0012 <(X-mean)/s <(0.8780-0.875)/0.0012)
=P(-2.5<Z<2.5) =0.9876 (from standard normal table)
------------------------------------------------------------------------------------------------------------------------------
b) The GM realizes that it would cost a lot of money to check ever slot. So she decides to look at the average slot size. She takes a sample of 10 slots. What is the probability that the mean slot size will be larger than 0.877?
P(xbar>0.877) = P((xbar-mean)/(s/vn) >(0.877-0.875)/(0.0012/sqrt(10)))
=P(Z>5.27) =0 (from standard normal table)
------------------------------------------------------------------------------------------------------------------------------
c) If we increase the sample size to 25, how does the probability change?
P(xbar>0.877) = P((xbar-mean)/(s/vn) >(0.877-0.875)/(0.0012/sqrt(25)))
=P(Z>8.33) =0 (from standard normal table)
The probability becomes smaller
------------------------------------------------------------------------------------------------------------------------------
d) Using n= 25, what is the probability that the mean is between 0.8743 and 0.8759?
P(0.8743 <xbar< 0.8759)
=P((0.8743-0.875)/(0.0012/sqrt(25)) <Z< (0.8759-0.875)/(0.0012/sqrt(25)))
=P(-2.92<Z<3.75)
=0.9982 (from standard normal table)
