Question

Many of a bank’s customers use its automated banking machine (ABM). During the early evening hours in the summer months, customers arrive at the ABM at the rate of one every other minute (assume Poisson). Each customer spends an average of 96 seconds completing the transaction. Transaction times are exponentially distributed. Assume that the length of the queue is not a constraint. Determine the probability that a customer waits five minutes or more in line.

Answer 1

The probability that a customer waits five minutes or more in line is 0.918.

Step-by-step explanation:

Given that customers arrive at the ABM at a rate of one every other minute (assume Poisson) and that each customer spends an average of 96 seconds completing the transaction, we can calculate the average time between customer arrivals using the exponential distribution.

The exponential distribution is defined by the parameter lambda, which is equal to the reciprocal of the mean. In this case, lambda = 1/(2 minutes) = 0.5.

To find the probability that a customer waits five minutes or more in line, we need to calculate the cumulative distribution function (CDF) of the exponential distribution at t = 5 minutes.

P(X ≥ 5) = 1 - P(X < 5) = 1 - e^(-λt) = 1 - e^(-0.5 * 5) ≈ 1 - e^(-2.5) ≈ 1 - 0.082 = 0.918

Answer 2

To find the probability that a customer waits five minutes or more in line at a bank's ABM, calculate the cumulative distribution function of the exponential distribution. Use the formula P(X < t) = 1 - e^(-λt), where X is the exponential random variable and t is the waiting time. Subtract the probability of waiting less than five minutes from 1 to find the probability of waiting five minutes or more.

Step-by-step explanation:

To find the probability that a customer waits five minutes or more in line, we need to calculate the cumulative distribution function (CDF) of the exponential distribution.

First, let's calculate the rate parameter (λ) of the exponential distribution. Since customers arrive at a rate of one every other minute, λ = 1/2.

Next, we need to find the probability of waiting less than five minutes. We can use the formula P(X < t) = 1 - e^(-λt), where X is the exponential random variable and t is the waiting time.

So, the probability of waiting less than five minutes is P(X < 5) = 1 - e^(-1/2 × 5) = 1 - e^(-2.5).