Question

 A farmer focusing on the production of eco-friendly chicken eggs collects the following data about his output. In a sample of 50 eggs, the farmer finds the average egg to weigh 47 grams. The standard deviation of the egg weight is 2 grams and the distribution of weights resembles a normal distribution reasonably closely.

 The farmer can sell the eggs to a local distributor. However, they have to be in the interval between 44 grams and 50 grams (i.e., the lower specification limit is 44 grams and the upper specification limit is 50 grams).

 a. What is the capability score of the eco-friendly chicken egg operation?

 b. What percentage of the produced eggs fall within the specification limits provided by the local distributor?

 c. By how much would the farmer have to reduce the standard deviation of the operation if his goal were to obtain a capability score of Cp=2/3 (i.e., get 4.5% defects)?

Answer 1

UCL = 50

LCL = 44

Mean = 47

standard deviation (s) = 2

As mean is centered, Cp will be used for capability score

Cp = (UCL - LCL)/(6*s) = (50-44)/(6*2) = 0.5

z value (

z value for less than LCL of 44 is = (44 - 47)/2 = -1.5

Probability of values less than 44 is 0.0668 (Using z table)

z value for less than UCL of 50 is = (50 - 47)/2 = 1.5

Probability of values less than 50 is 0.9332 (Using z table)

So, probability of values within specification is = 0.9332 - 0.0668 = 0.8664 or 86.64%

For Cp = 2/3 = (UCL-LCL)/(6*s)

2/3 = (50-44)/(6*s)

s = 1.5

So, standard deviation needs to be reduced from 2 to 1.5