Question: 1. Fixing a UCL and LCL at 1 standard deviation each (take area as 68% approximately) means that the probability of making a Type 1

1.

Fixing a UCL and LCL at 1 standard deviation each (take area as 68% approximately) means that the probability of making a Type 1 error is approximately:

"16% for each tail (above or below the UCL and LCL, respectively)"

"0.30% for each tail (above or below the UCL and LCL, respectively)"

"1.60% for each tail (above or below the UCL and LCL, respectively)"

"2.5% for each tail (above or below the UCL and LCL, respectively)"

2.

We know that variability in arrivals/service rates generally increases the probability of longer lines.

Shop A has an average arrival rate of 64/hr (std dev. is 16) and an average service rate of 75hr (std dev. is 8).

Shop B has an average arrival rate of 28/hr (std deviation is 7) and an average service rate of 35/hr (std dev.is 5).

Based on variability considerations alone, which system has a greater probability of a line developing? Assume the shops are identical in all other relevant respects.

Hint: Compare Coeff of Variation (= Std. Deviation/Mean) of arrival rates as well as of service rates between the two shops.

a.

The probability of a line developing is the same for both shops

b.

Shop A

c.

Shop B

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