ezto.mheducati A teller at a drive-up window at a bank had the following service

Question

ezto.mheducati A teller at a drive-up window at a bank had the following service times (in minutes) for 20 randomly selectecd 4.5 4.2 4.2 4.3 4.3 4.6 4.5 4.4 4.7 4.3 4.5 4.6 4.4 4.4 4.6 4.7 4.6 4.8 4.5 4.9 a. Determine the mean of each sample. (Round your answers to 1 decimal place.) b. If the process para value to 1 decimal place and "Standard deviation" value to 3 decimal places.) meters are unknown, estimate its mean and standard deviation. (Round "Mean" Mean Standard deviation c. Estimate the mean and standard deviation of the sampling distribution. (Round "Mean" value to 1 decimal place and "Standard deviation" value to 3 decimal places.) Mean Standard deviation 2

Explanation / Answer

(a)

(b)

Process mean (estimated by the mean of the sampling means) = 4.5
Process standard deviation = 0.192

(c)

Mean of the sampling distribution = 4.5
Std. deviation of the sampling distribution = 0.192 / SQRT(5) = 0.086

(d)

UCL = 4.5 + 3*0.086 = 4.758
LCL = 4.5 - 3*0.086 = 4.242

Risk = 2*(1 - NORM.S.DIST(3,TRUE)) = 0.0026

(e)

UCL = 4.86

So, 4.86 = 4.5 + Z*0.086
or, Z = (4.86 - 4.50) / 0.086 = 4.186

Risk = 2*(1 - NORM.S.DIST(4.186,TRUE)) = 0.00

(f)

No. All sample means are within the range [4.14, 4.86]

Sample 1 2 3 4 Avg. x1 4.5 4.6 4.5 4.7 x2 4.2 4.5 4.6 4.6 x3 4.2 4.4 4.4 4.8 x4 4.3 4.7 4.4 4.5 x5 4.3 4.3 4.6 4.9 Sample means 4.3 4.5 4.5 4.7 4.5 STDEV() 0.192

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