# 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 326 grams to 354 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.) St

###### Question:

a.

Specifications for the 12-ounce bar are 326 grams to 354 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 ? in grams

b.

The machine that fills the bar molds for the 1-ounce bars has a standard deviation of .86 gram. The filling machine is set to deliver an average of 1.03 ounces per bar. Specifications for the six-bar box are 153 to 187 grams. Is the process capable? Hint: The variance for the box is equal to six times the bar variance.

Yes?

or

No?

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 ? in ounces

## Answers

Answer:A) ≈ 3.509 grams B) The process is not capable because the lowest Cpk which is 1.89 ≈ 1.9 is far from the ideal 1.33 capability index number ( NO )c) 1.051 ouncesExplanation:A) The largest standard deviation ( in grams) This can be calculated applying the capability index formula and according to the capability index formula the value at which a process is capable is at : 1.33hence the largest standard deviation = upper limit - lower limit / 6 * 1.33 ( 354 - 326 ) / 7.98 = 3.5087 ≈ 3.509 grams B) we first calculate the process mean and std of the box standard deviation = [tex]\sqrt{variance }[/tex]std of the six-bar box = [tex]\sqrt{6*0.86^2}[/tex] = 2.106process mean = 6 * ( 1.03 * 28.33) = 175.079the mean is not centered between the upper specification and the lower specification hence we will apply the formulae used in calculating the capability index for an uncentered processCpk = (upper value - process mean) / (3 * std of box), = (187 - 175.079) / ( 3 * 2.106) = 11.921 / 6.318 = 1.89Cpk = (process mean - lower value ) / (3*std of box) = ( 175.079 - 153 ) / (3 * 2.106) = 22.079 / 6.318 = 3.49The process is not capable because the lowest Cpk which is 1.89 ≈ 1.9 is far from the ideal 1.33 capability index number ( N0 )c ) lowest setting calculate the value of the mean using capability index of 1.33 Cpk = (upper value - mean ) / 3 * std mean = 187 - 1.33 * 3 * 2.106 = 187 - 8.40 = 178.60 grams Cpk = ( process mean - lower value ) / 3 * std mean = 1.33 * 3 * 2.106 + 153 = 8.40 + 153 = 161.40 grams the lowest setting in ounces = 178.60 / ( 6 bar * 28 .33) = 178.60 / 169.98 = 1.051 ounces