The quality-control manager of a local fast food restaurant wishes to analyze th

Question

The quality-control manager of a local fast food restaurant wishes to analyze the length of time that a car spends at the drive-through window waiting for an order. According to records obtained from the restaurant, it is determined that the mean time spent at the window is 58.9 seconds with a standard deviation of 13.1 seconds. If a random sample of 55 cars is to be chosen, what are the mean and standard deviation of the sampling distribution for the mean time spent at the window? According to the 68-95-99.7 (Empirical) Rule, approximately 68% of all possible sample means arc between and seconds approximately 95% of all possible sample means are between and seconds approximately 99.7% of all possible sample means arc between and seconds

Explanation / Answer

If 55 cars are chosen then the mean will remain same as population mean mu=58.9 and standard deviation is 13.1/root over 55=1.76

According to the empirical rule 68% of all possible sample means lie between mu-sigma and mu+sigma that is 58.9-1.76=57.14 and 58.9+1.76=60.66

According to the empirical rule 95% of all possible sample means lie between mu-2sigma and mu+2sigma that is 58.9-2*1.76=55.38 and 58.9+2*1.76=62.42.

According to the empirical rule 99.7% of all possible sample means lie between mu-2sigma and mu+2sigma that is 58.9-3*1.76=53.62 and 58.9+3*1.76=64.18.

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