6-) A big is a full of candies and the following data is provided The lower spec

Question

6-) A big is a full of candies and the following data is provided The lower specification limit is 48 pound and upper specification limit is 56 pound, the population mean weight of the candy bag is 50 pound with a standard deviation of 1.75 pound. (normal distribution exist) A) What percentage of the bag contain less than 48 pound? B) What would be average weight of the bag if we desire 4% of the bag to be below 48 pound Multiple Choice--True/False--Essay type of question- Short Problem Solving 1-) Please explain the difference between special cause variation and normal cause variation. Please explain with details and give an example 2-) What of the following charts uses past information to predict what the next process outcome would be A) MR chart B) U chart C) Np Chart D) P Chart 3-) The diameter of a tire is 3 mm but it can get as large as 3.03 mm and as small as 2.97 mm. The management decided to take 25 samples of these tires and they discovered that these components have a grand mean of 3.01 mm and a std of 0.02 mm. The management would like to know probability of producing a bad product? 4-) What is the difference between CP and CPk. Which one of those give us a better picture in terms of analyzing capability of a process? Please explain 5-) Company SEMO tested 120 products for 60 hours each During the test, 6 breakdowns occurred. Compute the number of failures per hour and MTBF 6-) Please explain the control phase of DMAIC six sigma approach

Explanation / Answer

6)

Upper tolerance limit 56

Lower tolerance limit 48

Mean 50

Standard deviation     1.75

Find the z value at 48 = (Value - mean) / standard deviation = (48 - 50)/1.75 = -1.14286

Using formula =Normsdist(-1.14286) in excel we get the probability = 0.1265 or 12.65%

part (b)

Using formula = Normsinv(0.04) in excel we get the z value -1.75069.

we know that, z value at 48 = (Value - mean) / standard deviation

or

-1.75069 = (48 - Mean)/1.75

solving it will give

Mean = 51.0637

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