At a busy coffee house the amount of time it takes a customer to receive their b

Question

At a busy coffee house the amount of time it takes a customer to receive their beverage after ordering is a normally distributed random variable with a mean of   and 5 and standard deviation of 3.674 minutes..

a.Suppose random sample of 6 customers is observed. What is the mean and standard deviation of the sampling distribution of ? (did this one)

mean=5, standard deviation=1.49

a. For a sample size of 6, what is the probability that the mean time it takes to receive a beverage is less than 4 minutes?

Explanation / Answer

Mean is 5 and standard deviation is 3.674

For sample size 6, the standard error ,SE is s/sqrt(N)=3.674/sqrt(6)=1.499

z is (x-mean)/SE

thus P(xbar<4)=P(z<(4-3.674)/1.499)=P(z<0.22)

from normal distribution table we get 0.5871

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