Question

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 330 grams to 350 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 .82 gram. The filling machine is set to deliver an average of 1.01 ounces per bar. Specifications for the six-bar box are 157 to 183 grams. Is the process capable? Hint: The variance for the box is equal to six times the bar variance.
  

• Yes

• 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              ounces

Answer 1

Ans a)

USL (Upper Specification Limit) =350 gms
LSL (Lower Specification Limit) =330 gms
SD (Standard Deviation) =?
Mean = 340 gms

Formula for finding process Capability (C) is given by,

C= (USL-LSL) / 6

Thus,. C= (350-330) / 6SD

Let Cp be 1.33 (assumption) and find σ for the given formula:

Therefore, 1.33= (350-330)/6SD

1.33 * (6SD)= 20

(1.33) * 6SD= 20

7.98 * SD= 20

SD = 20/7.98

SD= 2.506

Thus, the largest standard deviation (in grams) that the machine that fills the bar molds can have and still be considered capable is 2.50

Ans b): Lets check whether the process is capable or not. Ideally the process is capable is Cpk>=1.33.

1 ounce = 28.33 grams (assumption which is practically true)

Bar variance = (.82) ^2 = 0.672

Box variance = 6 * 0.672 = 4.032

Box standard deviation = root (4.032) = 2.007

Average box weight = 6 *1.01 = 6.06 ounces * 28.33 = 171.68 grams for the six-bar box.

Now Cpk is calculated by the formula, Cpk= min (USL−mean)/3σ,(mean−LSL)/3σ)

Thus, Cpk = (171.68-157)/ (3*2.007) = 2.3, Cpk = (183-171.68)/ (3*2.007) = 1.88

Cpk (min) = 1.88

Because 1.88 ≥ 1.33, this process is capable.

Ans C - Lowest setting in ounces LSL can be calculated by formula Cpk= (mean−LSL)/3σ)

Thus, 1.88 = (183 - LSL) / (3*2.007)

Thus, 183 - LSL = 1.88 * 6.02 = 11.31 gms

Converting gms into ounces = 11.31/28.33 = 0.4 ounce