Customer complain messages arrive at a switching centre at random and at an average rate of 1.2 per minute. Based on the calculation of probability distributions, the exact number of resources to be allocated to serve the customer complaints better and faster. Find the probability of 5 messages arriving in a 2-minute interval.Select one:a.0.01b.0.06c.0.02d.0.07
Question
Customer complain messages arrive at a switching centre at random and at an average rate of 1.2 per minute. Based on the calculation of probability distributions, the exact number of resources to be allocated to serve the customer complaints better and faster. Find the probability of 5 messages arriving in a 2-minute interval.Select one:a.0.01b.0.06c.0.02d.0.07
Solution
To find the probability of 5 messages arriving in a 2-minute interval, we can use the Poisson distribution formula. The Poisson distribution is commonly used to model the number of events occurring in a fixed interval of time or space.
The formula for the Poisson distribution is:
P(x; λ) = (e^(-λ) * λ^x) / x!
Where:
- P(x; λ) is the probability of x events occurring in the given interval
- e is the base of the natural logarithm (approximately 2.71828)
- λ is the average rate of events occurring in the interval
- x is the number of events we are interested in
In this case, the average rate of messages arriving is 1.2 per minute. So, in a 2-minute interval, the average number of messages would be 2.4 (1.2 * 2).
Now, we can plug in the values into the formula:
P(5; 2.4) = (e^(-2.4) * 2.4^5) / 5!
Calculating this expression, we find that P(5; 2.4) is approximately 0.067, which is closest to option d. 0.07.
Therefore, the probability of 5 messages arriving in a 2-minute interval is approximately 0.07.
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