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Customers arrive at the checkout counter of a supermarket at an average rate of 10 per hour, and these arrivals follow a Poisson distribution. Using each of the following two methods, find the probability that exactly 4 customers will arrive at this checkout counter during a 2 -hour period. a. Use the arrivals in each of the two nonoverlapping 1 -hour periods and then add these. (Note that the numbers of arrivals in two nonoverlapping periods are independent of each other.) b. Use the arrivals in a single 2 -hour period.

Short Answer

Expert verified
The probability of exactly 4 customers arriving at the checkout counter during a 2-hour period is given by \( \frac{e^{-20} * 20^{4}}{4!} \), which you calculate in method b. Adding up the probabilities of 4 customers arriving in each of two 1-hour periods according to method a gives incorrect results.

Step by step solution

01

Calculate probability for each 1 hour period method a

First, calculate the probability of exactly 4 customers arriving in each 1-hour period. The average rate \( \lambda \) is 10 customers per hour. As instructed, first calculate the probability for two non-overlapping 1 hour periods and then add these. The formula to use is P(X=k) = \( \frac{e^{-\lambda} * \lambda^{k}}{k!} \) , where k=4. Therefore, the calculation for each hour will be \( \frac{e^{-10} * 10^{4}}{4!} \). Find this value. However, it must be realized that since the Poisson distribution only considers 'non-overlapping periods' as independent, adding the probabilities for the first hour and the second hour does not give the correct result.
02

Calculate the probability for the total 2 hour period method b

Then, calculate the probability of exactly 4 customers arriving in a full 2-hour period. In this case, the average rate \( \lambda \) is 20 customers per 2-hour period (since the rate is 10 customers per hour). The formula to use remains the same, but now with k=4 and \( \lambda = 20 \). The calculation will thus be \( \frac{e^{-20} * 20^{4} }{4!} \). Find this value.
03

Compare and analyze the results.

Once the above calculations are completed, you compare the results. You will observe that the results of method a is incorrect as adding non-overlapping periods does not provide the correct result for a Poisson Distribution. The correct probability is given by the calculation in method b.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculation
Probability calculation can feel complex, but breaking it down is a great starting point. In our problem of determining customer arrivals using a Poisson distribution, we use a specific formula. The Poisson probability formula is \[ P(X=k) = \frac{e^{-\lambda} \cdot \lambda^{k}}{k!} \]where:
  • \( \lambda \) represents the average number of events (customer arrivals in our case) in a given time period.
  • \( k \) is the exact number of events you want to find the probability for, which is 4 in this scenario.
The exponential base \( e \) (approximately equal to 2.718) is common to all Poisson calculations. For method b, when events are considered for a single complete interval (a 2-hour time span here), the calculations reflect the more accurate application of the Poisson distribution.
Customer Arrival Rates
Understanding customer arrival rates is essential. In this problem, an average of 10 customers arrive per hour. This average arrival rate \( \lambda \) directly feeds into our probability calculations. It defines the frequency of arrival occurrences over the measured time span.
For multi-hour calculations, it’s important to adjust \( \lambda \) accordingly. For example, over a 2-hour period, the average rate changes from \( \lambda = 10 \) for one hour to \( \lambda = 20 \) for two hours. This adjustment is crucial for accurate probability results.
When working with probability distributions, always first determine the average rate to set your calculations up for success. By accurately defining \( \lambda \), you ensure that you're correctly applying statistical methods.
Independent Events
One of the key characteristics of the Poisson distribution is its handling of independent events. Events are independent when the occurrence of one does not affect the probability of another occurring.
In our exercise, this principle affects how calculations are made over different time periods. When we looked at non-overlapping 1-hour periods, we assumed independence—what happens in one hour shouldn't impact the next.
However, simply adding up the probabilities from two separate 1-hour calculations doesn't yield the correct result for a 2-hour period. This is because Poisson treats the entire period as one event, and independence should only be applied as part of that broader analysis. Following just this principle, without proper calculation over the extended period, can lead to incorrect results.
Statistical Methods
Statistical methods like the Poisson distribution are fundamental in analyzing and predicting events over time. They allow us to model situations where events occur independently and sporadically within fixed intervals.
Poisson is particularly useful for events happening at a constant average rate, like customer arrivals. But careful application matters.
  • When using statistical methods, check assumptions—such as event independence and uniformity of rates.
  • Ensure appropriate calculations for your timeframe. Here, a single 2-hour calculation was more effective than summing shorter periods.
  • Such methods help predict real-world situations and offer insights based on data, which can guide decision-making in business, science, and engineering.
Learning to think statistically allows better interpretation of daily life patterns, aiding in decision-making and resource allocation.

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