The Good Chocolate Company makes a variety of chocolate candies, including a 12-

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 328 grams to 352 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 .84 gram. The filling machine is set to deliver an average of 1.03 ounces per bar. Specifications for the six-bar box are 155 to 185 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

Explanation / Answer

Answer to question a :

Upper Specification limit for 12 ounce bar = USL = 352 grams

Lower Specification Limit for 12 ounce bar = LSL = 328 grams

Average fill of the process = m = 340

For the process to be considered capable ,the difference between respective specification limit and average fill must be at least 3 times the standard deviation.

That means :

USL - m > = 3 x Standard deviation

And

M – LSL > = 3 x Standard deviation

USL – m > = 3 x Standard deviation

Or , (352 -340 ) > = 3 x Standard deviation

OR, Standard deviation < = 4

M - LSL > = 3 x Standard deviation

Or, ( 340 – 328 ) > = 3 x standard deviation

Standard deviation < = 4

Therefore, the largest standard deviation permissible for the machine to be capable = 4 grams

LARGET STANDARD DEVIATION = 4 GRAMS

Answer to question b :

Average weight per bar = 1.03 Ounce

Therefore average weight of 6 bar box = m = 1.03 x 6 = 6.18 Ounce = 340/12 x 6.18 gram = 175.10 ( rounded to 2 decimal places )

Standard deviation of weight of 1 bar as given = 0.84 grams

Therefore, standard deviation of weight of box of 6 ounce bar = Sd = 0.84 x Square root ( 6) = 2.057 gram

Specification for six bar box us 155 to 185 grams,

I.e , Upper Specification Limit = USL = 185

Lower Specification Limit = LSL = 155

Therefore ,

Process Capability Index, Cpk

= Minimum ( ( USL – m)/3xSd , ( m – LSL)/3xSd)

= Minimum ( ( 185 – 175.10) / 3 x 2.057, ( 175.10 – 155)/ 3 x 2.057)

= Minimum ( 1.604, 3.257)

= 1.604

Since Cpk > 1 , therefore the process is capable

THE PROCESS IS CAPABLE

LARGET STANDARD DEVIATION = 4 GRAMS

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