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Question from Poisson probability distribution

7. A sales form receives on the average of 3 calls per hour on its toll-free number. For any given hour, find the probability that it will receive a. At most 3 calls; b. At least 3 calls; and c. Five or more calls.

 

To solve this problem, imagine you're running this bustling sales firm, and your trusty toll-free number is ringing off the hook, with an average of 3 calls per hour. Now, we want to figure out the probability of different scenarios happening during any given hour.

First up, let's find out the probability of getting at most 3 calls. This includes 0 calls, 1 call, 2 calls, and 3 calls. To do this, we're going to use a nifty little formula called the Poisson distribution. We'll plug in our average number of calls (λ = 3) and calculate the probability for each possible number of calls. Then, we'll add them all up to get our answer.

Next, we'll tackle the probability of getting at least 3 calls. This includes scenarios with 3, 4, 5, and so on up to infinity calls. Again, we'll use the Poisson distribution to calculate the probabilities for each case and add them up.

Lastly, let's find the probability of getting five or more calls. This one's a bit more straightforward. We'll just calculate the probabilities for 5, 6, 7, and so on calls using the Poisson distribution, and then add them up.

Remember, probability is like predicting the future based on past patterns, so don't worry if it feels a bit like fortune-telling. Just follow the steps, plug in the numbers, and you'll be a probability pro in no time!

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It's true that this is a question from Poisson's probability distribution. 

Are you looking for someone to do your excel homework? Our probability tutors provided the solution below to this question. If your parameters have changed, then you can still contact us for probability homework help. We can provide help with any excel assignment you might have. 

This follows a poisson Distribution     
Average  =  3  
Atmost 3 Calls   =  p(x<=3)
p(x<=3)   0.6472
b) At least 3 calls:  =  1-P(x<=2)
p(x>=3) = 0.57681
c. Probability of 5 or more calls   = 1-P(x<=4)
p(x>=5) = 0.1847
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