A supermarket selected a sample of 190 of its customers and measured how long th

Question

A supermarket selected a sample of 190 of its customers and measured how long they took to be served at the checkout counter. If too many customers wait too long, the supermarket intends to hire more checkout personnel. Specifically the supermarket would like at least $ 5% of its customers to be checked out in less than 8.5 minutes. From the data, the 90^th and 60^th percentiles were computed to be 8.8 minutes and 7.1 minutes, respectively. The ranges of the data was 13 minutes and the fastest anyone was checked out was 1.9 minutes. Approximately how many customers waited less than 7.1 minutes to be checked out?

Explanation / Answer

Here we have given that supermarket selected sample of 190 of its customer.

i.e. sampe size n = 190

also we have given that the 85% of its customer checked out in less than 8.5 minutes i.e p = 0.85

also we have given 90th and 60th percentile

90th percentie = 8.8 min

60th percentile = 7.1 min

Now we have to find that approximately how many customers waited less than 7.1 min to be checked out.

We know that the 60 th percentile = 7.1 min

i.e P( X<=7.1) = 0.6

To find number of customer multiply this probability by n

So number of customers waited less than 7.1 min for check out = n* P(X<= 7.1)

= 190*0.6 = 114

So number of customers waited less than 7.1 min for check out are 114

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