Chapter 6 Homework Q1. Pepsi uses six-sigma technique to evaluate whether its 15

Question

Chapter 6 Homework 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. a) Find the value of . b) What is the probability that x is greater than 43? c) What is the probability that x is between 35 and 45? d) What is the x value so that only 4.01% of all possible outcomes exceeds that value. Q3. A lightbulb has a normally distributed light output with mean 5000 end foot-candles and standard deviation of 50 end foot-candles. a. Find a lower tolerance limit such that only 0.2% of the bulbs will not exceed this limit. b. What fraction of these lightbulbs will have a lower than an output of 5080 end footcandles? c. If 50000 lightbulbs are made, how many of them will have an output stronger than5080 end foot-candles? How many of them will have an output weaker than 4890 end footcandles? Please show work without excel. Thank you!

Explanation / Answer

Question 1

a). What is the percentage of 15 oz Gatorade that meets the specification?

Solution:

Specification = 15 -/+ 4% of 15 = 15 -/+ 0.6

Lower specification = 15 – 0.6 = 14.4

Upper specification = 15 + 0.6 = 15.6

We are given

Mean = 15.12

SD = 0.3

We have to find P(14.4<X<15.6)

P(14.4<X<15.6) = P(X<15.6) – P(X<14.4)

For X<15.6

Z = (15.6 – 15.12) / 0.3 = 1.6

P(Z<1.6) = 0.945201

For X<14.4

Z = (14.4 – 15.12) / 0.3 = -2.4

P(Z<-2.4) = 0.008198

P(14.4<X<15.6) = P(X<15.6) – P(X<14.4)

P(14.4<X<15.6) = 0.945201 - 0.008198

P(14.4<X<15.6) = 0.937003

Required probability = 0.937003

Required percentage = 93.70%

93.70% percentage of 15 oz Gatorade meets the specification.

b). What are the percentages of 15 oz Gatorade that are considered “overweight” and “underweight”?

Solution:

P(Gatorade are overweight) = P(X>15.6)

P(X>15.6) = 1 – P(X<15.6)

Z = (15.6 – 15.12) / 0.3 = 1.6

P(Z<1.6) = 0.945201

P(X>15.6) = 1 – P(X<15.6)

P(X>15.6) = 1 – 0.945201 = 0.054799

About 5.48% of 15 oz Gatorade that are considered “overweight”.

P(Gatorade are underweight) = P(X<14.4)

Z = (14.4 – 15.12) / 0.3 = -2.4

P(Z<-2.4) = 0.008198

About 0.82% of 15 oz Gatorade that are considered “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”)?

Solution:

Specification = 15 -/+ 4% of 15 = 15 -/+ 0.6

Lower specification = 15 – 0.6 = 14.4

Upper specification = 15 + 0.6 = 15.6

We are given

Mean = 15

SD = 0.25

We have to find P(14.4<X<15.6)

P(14.4<X<15.6) = P(X<15.6) – P(X<14.4)

For X<15.6

Z = (15.6 – 15) / 0.25 = 2.4

P(Z<2.4) = 0.991802

For X<14.4

Z = (14.4 – 15) / 0.25 = -2.4

P(Z<-2.4) = 0.008198

P(14.4<X<15.6) = P(X<15.6) – P(X<14.4)

P(14.4<X<15.6) = 0.991802 - 0.008198

P(14.4<X<15.6) = 0.983605

Required probability = 0.983605

Required percentage = 98.36%

98.36% percentage of 15 oz Gatorade meets the specification.

Now, we have *0.25 = 0.60 , = 0.60/0.25 = 2.4

2.4 quality standards the new process achieves.

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?

Solution:

We have 6* = 0.60

So, = 0.60/6 = 0.10

In order to achieve the 6-sigma standard without having to change the specification, standard deviation should be 0.10.

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