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Business Statistics

Report

Acceptable Sampling of Pins

Summary:

A company supplies bulk amount of pins to a customer.

Length of the pins made by the machine is normally distributed with a mean of 1.008inch and a standard deviation of 0.045inch.

A random sample of 50 pins is taken from the batch. The pins are acceptable in a mean interval of 1.000inch (+/-) 0.010inch.

Standard Deviation = 0.045 inch

Mean = 1.008 inch

Tolerance = .010 inch

Acceptable length of pins = 1.00 +.010 (x1)

and 1.00 - .010 (x2)

Question 1: What percentage of the pins will be acceptable to the customer? Is the probability large enough to be an acceptable level of performance?

Answer:

At x = 0.99

z1 = 0.99-1.008/(.045)/√50

= -2.828

* At x = 1.01

z2 = 1.01-1.008/0.045/√50

= 0.3142

Probability for accepted pins = P (z1) + P (z2)

= 0.4976 + 0.1217 = 0.6193

So, The Probability that a batch will be acceptable to the customer is 61.93% which is very low.

Question 2: If the lathe can be adjusted to have the mean of the lengths to any desired value, what should it be adjusted to? Why?

Answer:

To improve the acceptance probability of pins, the required mean should be in the middle of 0.99 and 1.01 so as to cover maximum area.

Thus mean should be adjusted to the value 1.000 inch.

At x = .099

z1 = .99-1.0/(.045/√50)

= -1.57

At x= 1.01

z2= 1.01-1.0/ (.045/√50) = 1.57

Probability of accepted pins = P(z1) + P(z2)

= 0.4418+0.4418

= 0.8836

So percentage of accepted pins = 88.36%

Question 3: Suppose the mean cannot be adjusted, but the standard deviation can be reduced. What maximum value of the...