Question

the manager of a grocery store has taken a random sample of 100 customers the average length of time it took these 100 customers to check out was 3 minutes. it is known that the standard deviation of the population of check out times is 1 minute. with a .95 probability, the sample mean will provide a margin of error

Answer 1

with a .95 probability, the sample mean will provide a margin of error 0.196

Explanation:

margin of error (ME) from the mean can be calculated using the formula

ME=
`(z*s)/(√(N) )` where

• z is the corresponding statistic in the .95 confidence level (1.96)

• s is the population standard deviation (1 min.)

• N is the sample size (100)

ME=
`(1.96*1)/(√(100) )`=0.196

95% confidence interval for check-out time would be 3 ±0.196 min