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At a fast-food restaurant, the time spent by the customers is a normally distributed variable. After randomly recording the habits of 200 custowers, the average amount of time spent was found to be 26 minutes with a standard deviation of 2.8 minutes. What proportion of customers spent more than 30 minutes in the restaurant? If a customer arrives at 11:05 a.m., what is the probability that the customer will leave at 11:30 a.m. or earlier?

2007-10-14 14:16:48 · 2 answers · asked by Anonymous in Education & Reference Homework Help

2 answers

26 +2.8 = 28.8 26 - 2.8 = 23.2 1 1 standard Deviation from the mean.
2.8 x 2 = 5.6 and has the range between
20.4 to 31.6 2 standard deviations form the mean.
30 minutes is almost 2 standard deviations from the mean; approximately 95 percent for a normal distribution. So there is about a 5% of the customers will spend 30 minutes or more.
A customer arriving at 11:05 and leaving at 11:30 spends 25 min which has a 68% chance or probability. This part is encompassing approximately 68% of the population.

2007-10-14 15:06:08 · answer #1 · answered by TicToc.... 7 · 1 0

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2016-12-18 07:46:09 · answer #2 · answered by ? 4 · 0 0

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