Let x be a random variable that represents checkout time (time spent in the actual checkout process) in minutes in the express l
ane of a large grocery. Based on a consumer survey, the mean of the x distribution is about μ = 2.7 minutes, with standard deviation σ = 0.6 minute. Assume that the express lane always has customers waiting to be checked out and that the distribution of x values is more or less symmetric and mound-shaped. What is the probability that the total checkout time for the next 30 customers is less than 90 minutes? Let us solve this problem in steps.
1 answer:
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Answer:
x=2.4650
Step-by-step explanation:
I assume that you mean
so let s use the ln() function
so x = ln(15)/ln(3)
x = 2.4650