Customers Make Purchases At A Convenience Store On Average

customers Make Purchases At A Convenience Store On Average Every F

Customers make purchases at a convenience store, on average, every fourteen minutes. It is fair to assume that the time between customer purchases is exponentially distributed. Jack operates the cash register at this store.

a-1. What is the rate parameter λ? (Round your answer to 4 decimal places.)

a-2. What is the standard deviation of this distribution? (Round your answer to 1 decimal place.)

b. Jack wants to take a ten-minute break. He believes that if he goes right after he has serviced a customer, he will lower the probability of someone showing up during his ten-minute break. Is he right in this belief? Yes No

c. What is the probability that a customer will show up in less than ten minutes? (Round intermediate calculations to 4 decimal places and final answer to 4 decimal places.)

d. What is the probability that nobody shows up for over forty-five minutes? (Round intermediate calculations to 4 decimal places and final answer to 4 decimal places.)

Sample Paper For Above instruction

The question involves understanding the exponential distribution in the context of customer arrivals at a convenience store. The key is to determine the rate parameter λ, recognize the properties of the exponential distribution, and compute associated probabilities.

Given that customers arrive every 14 minutes on average, the exponential distribution parameter λ is the reciprocal of the mean inter-arrival time. Mathematically, λ = 1 / μ, where μ is the mean time between arrivals. Here, μ = 14 minutes, so λ = 1 / 14 ≈ 0.0714. Rounded to four decimal places, λ = 0.0714.

The standard deviation for an exponential distribution is also equal to the mean, so it is 14 minutes. However, noting the standard deviation, it specifically is equal to the mean, which confirms the properties of the exponential distribution used here.

The probability that a customer arrives within a specific time frame t in an exponential distribution is given by P(T

To find the probability that no customer arrives in over 45 minutes, we compute P(T > 45) = e^{-λ45} ≈ e^{-0.071445} ≈ e^{-3.213} ≈ 0.0400. Therefore, the probability that nobody shows up for over forty-five minutes is approximately 4.00%.

Regarding Jack's belief about taking a break right after servicing a customer, since the exponential distribution models the time between arrivals as memoryless, the probability of a customer arriving in the next ten minutes does not depend on when the last customer arrived. Therefore, going immediately after servicing a customer does not reduce the probability of customer arrivals in the next ten minutes. Jack's belief that this action lowers the probability is incorrect. The probability remains the same regardless of when he takes his break, aligned with the memoryless property of the exponential distribution.

References

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