Q1. Pepsi uses six-sigma technique to evaluate whether its 15 oz Gatorade has co
ID: 356087 • Letter: Q
Question
Q1. Pepsi uses six-sigma technique to evaluate whether its 15 oz Gatorade has consistent weight. Pepsi took a sampling of 1000 15 oz Gatorade and found out that the average weight was 15.12 oz with a standard deviation of 0.3 oz. Pepsi considers a tolerance of 4% from 15 oz acceptable a). What is the percentage of 15 oz Gatorade that meets the specification? b). What are the percentages of 15 oz Gatorade that are considered "overweight" and "underweight? c). If Pepsi improves its production process, and samples another 1000 bottles and finds a average weight of 15 oz with a standard deviation of 0.25 oz. What is the % of 15 oz Gatorade that meets the specification? What quality standard does this new process achieve (in terms of number of"sigma")? d) Following c, what is the standard deviation that Pepsi needs to achieve in order to achieve the 6-sigma standard without having to change the specification? Q2. If x is normally distributed with mean and standard deviation ?-4, and given that the probability that x is less than 32 is 0.0228. Find the value of H. Q3. A lightbulb has a normally distributed light output with mean 5000 end foot-candles and standard deviation of 50 end foot-candles. Find a lower tolerance limit such that only 0.2% of the bulbs will not exceed this limit.Explanation / Answer
Mean, m = 15.12 oz
Standard deviation, s = 0.3 oz
USL = 15*(1+0.04) = 15.6 oz
LSL = 15*(1-0.04) = 14.4 oz
a) For x=USL, z = (USL-m)/s = (15.6-15.12)/0.3 = 1.6
P(z<=1.6) = NORMSDIST(1.6) = 0.9452
For x=LSL, z = (LSL-m)/s = (14.4-15.12)/0.3 = -2.4
P(z<=-2.4) = NORMSDIST(-2.4) = 0.0082
Percentage that meets specification = 0.9452-0.0082 = 0.9370
b) Percentage that are considered overweight = 1-0.9452 = 0.0548
Percentage that are considered underweight = 0.0082
Percentage that are considered overweight and underweight = 0.0548+0.0082 = 0.0630
c)
Mean, m = 15.12 oz
Standard deviation, s = 0.3 oz
a) For x=USL, z = (USL-m)/s = (15.6-15)/0.25 = 2.4
P(z<=2.4) = NORMSDIST(2.4) = 0.9918
For x=LSL, z = (LSL-m)/s = (14.4-15)/0.25 = -2.4
P(z<=-2.4) = NORMSDIST(-2.4) = 0.0082
Percentage that meets specification = 0.9918-0.0082 = 0.9836
Process is perfectly centered, as the process mean is equal to the specification mean.
Therefore, Sigma level = (USL-LSL)/s = (15.6-14.4)/0.25 = 4.8 sigma
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