Answer :
The probability that the interarrival time will be less than 0.5 minutes between customers is 0.5987.Answer:Thus, the probability the interarrival time will be less than 30 seconds (0.5 minutes) between customers is 0.5987.
In a busy day, 500 customers enter a store.Mean interarrival time is 0.4 minutes.Standard deviation is 0.4 minutes.The probability that the inter-arrival time will be less than 30 seconds between customers will be calculated as follows:Solution:To solve the problem, we can use the z-score formula which is:where, x = raw score, μ = population mean and σ = standard deviationThe given mean inter-arrival time = 0.4 minutes.To find the probability of inter-arrival time being less than 0.5 minutes, we convert it into a z-score using the formula as below. z=(x - μ) / σThe formula can be re-written as:x = μ + z × σFrom the problem,x = 0.5 minutes.μ = 0.4 minutes.σ = 0.4 minutes.z = (0.5 - 0.4) / 0.4z = 0.25The z-table shows that the probability corresponding to z-score 0.25 is 0.5987.
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