Question

The time between successive arrivals of calls to emergency service is random variable which follows exponential...

The time between successive arrivals of calls to emergency service is random variable which
follows exponential distribution. It was observed that on average calls arrive to emergency
service every four minutes (1 / λ = 4min) and average number of calls in one minute is λ = 0.25 calls/ 1 min

The probability that the time between successive calls is less than 2 minutes is ______

A call just arrived to emergency service. The probability that next call will arrive to emergency service no sooner than in 3 minutes is __________.

The probability that the time between successive calls is between 6 and 8 minutes is ______

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The time between arrivals at a toll booth follows an exponential distribution with a mean time...
The time between arrivals at a toll booth follows an exponential distribution with a mean time between arrivals of 2 minutes. What is the probability that the time between two successive arrivals will be less than 3 minutes? What is the probability that the time will be between 3 and 1 minutes?
The time between telephone calls to a cable televisiom service call center follows an exponential distribution...
The time between telephone calls to a cable televisiom service call center follows an exponential distribution with a mean of 1.5 minutes. a. What is the probability that the time between the next two calls will be 48 seconds or less? b. What is the probability that the between the next two calls will be greater than 118.5 seconds?
Let X = the time between two successive arrivals at the drive-up window of a local...
Let X = the time between two successive arrivals at the drive-up window of a local bank. If X has an exponential distribution with λ = 1/3 , compute the following: a. If no one comes to the drive-up window in the next 15 minutes (starting now), what is the chance that no one will show up during the next 20 minutes (starting now)? b. Find the probability that two people arrive in the next minute. c. How many people...
Let X be the time between successive arrivals to an intersection in a rural area. Suppose...
Let X be the time between successive arrivals to an intersection in a rural area. Suppose cars arrive to the intersection via a Poisson process at a rate of 1 every 5 minutes. (a) What distribution does X have? (b) What is the mean time between arrivals? (c) What is the probability that more than 7 minutes will pass between arrivals to the intersection?
Suppose that the time between successive occurrences of an event follows an exponential distribution with mean...
Suppose that the time between successive occurrences of an event follows an exponential distribution with mean number of occurrences per minute given by λ = 5. Assume that an event occurs. (A) Derive the probability that more than 2 minutes elapses before the occurrence of the next event. Derive the probability that more than 4 minutes elapses before the occurrence of the next event. (B) Use to previous results to show: Given that 2 minutes have already elapsed, what is...
Let X = the time between two successive arrivals at the drive-up window of a local...
Let X = the time between two successive arrivals at the drive-up window of a local bank. Suppose that X has an exponential distribution with an average time between arrivals of 4 minutes. a. A car has just left the window. What is probability that it will take more than 4 minutes before the next drive-up to the window? b. A car has just left the window. What is the probability that it will take more than 5 minutes before...
Let X = the time between two successive arrivals at the drive-up window of a local...
Let X = the time between two successive arrivals at the drive-up window of a local bank. If X has an exponential distribution with λ=1, compute the following: a) The expected time between two successive arrivals b) The standard deviation of the time between success arrivals c) Calculate the probability P (X≤3) d) Calculate the probability P(2≤X≤5)
Customers arrive at random times, with an exponential distribution for the time between arrivals. Currently the...
Customers arrive at random times, with an exponential distribution for the time between arrivals. Currently the mean time between customers is 6.34 minutes. a. Since the last customer arrived, 3 minutes have gone by. Find the mean time until the next customer arrives. b. Since the last customer arrived, 10 minutes have gone by. Find the mean time until the next customer arrives.
The time between customer arrivals at a furniture store has an approximate exponential distribution with mean...
The time between customer arrivals at a furniture store has an approximate exponential distribution with mean θ = 9.9 minutes. [Round to 4 decimal places where necessary.] If a customer just arrived, find the probability that the next customer will arrive in the next 9.4 minutes. Tries 0/5 If a customer just arrived, find the probability that the next customer will arrive within next 13 to 17 minutes? Tries 0/5 If after the previous customer, no customer arrived in next...
The exponential distribution is frequently applied to the waiting times between successes in a Poisson process....
The exponential distribution is frequently applied to the waiting times between successes in a Poisson process. If the number of calls received per hour by a telephone answering service is a Poisson random variable with parameter λ = 6, we know that the time, in hours, between successive calls has an exponential distribution with parameter β =1/6. What is the probability of waiting more than 15 minutes between any two successive calls?