Question

Problem 10-27 The Good Chocolate Company makes a variety of chocolate candies, including a 12-ounce chocolate bar (340 grams) and a box of six 1-ounce chocolate bars (170 grams). a.Specifications for the 12-ounce bar are 324 grams to 356 grams. What is the largest standard deviation (in grams) that the machine that fills the bar molds can have and still be considered capable if the average fill is 340 grams? (Round your intermediate calculations to 2 decimal places and final answer to 3 decimal places.) Standard deviation grams b.The machine that fills the bar molds for the 1-ounce bars has a standard deviation of .78 gram. The filling machine is set to deliver an average of 1.02 ounces per bar. Specifications for the six-bar box are 152 to 188 grams. Is the process capable? Hint: The variance for the box is equal to six times the bar variance. O Yes Ο Νο c.What is the lowest setting in ounces for the filling machine that will provide capability in terms of the six-bar box? (Round your intermediate calculations to 2 decimal places and final answer to 3 decimal places.) Lowest setting ounces

Answer 1

a.

For a capable process, Cp = 1.33

So, 1.33=(UCL-LCL)/(6*standard deviation)

or, 1.33 =(356-324)/(6*standard deviation)

or, standard deviation = (356-324)/(6*1.33) = 4.010025063 = 4.01 (Rounded to 3 decimal places)

b.

standard deviation = sqrt(6*0.78^2)= 1.910601999

Mean in grams = 6*1.02*28.33 = 173.3796

Cpk = min((UCL-mean)/(3*standard deviation),(mean-LCL)/(3*standard deviation))

=min((188-173.3796)/(3*1.910601999),(173.3796-152)/(3*1.910601999)) = 2.550749277

The process is capable as Cpk is greater than 1.33

c.

Cpk(upper) = 188–(1.33*3*1.91) = 180.3791 = 180.38(Rounded to 2 decimal places)
Cpk(lower) = 152+(1.33*3*1.91) = 159.6209= 159.62(Rounded to 2 decimal places)

Mean = 159.62/(6*28.33) = 0.939051653 = 0.939 ounces (Rounded to 3 decimal places) The lowest setting is 0.939 ounces