In Exercises 1–7, consider a grocery store that can process a ...

Question

In Exercises 1–7, consider a grocery store that can process a total of four customers at its checkout counters each minute.

The mean number of arrivals per minute is four. Find the probability that

b. more than four customers will arrive during the first minute.

Accepted Answer

Welcome back, everyone. In this problem, a ticket counter can assist up to 3 customers in any given minute. The number of arriving customers per minute follows a poison distribution with a mean of 3. What is the chance that over 3 customers come in the first minute? A says it's 0.6472, B 0.3528, C 0.2241, and the D 0.7760. Now what can help us to figure out the chance or probability that over 3 customers come in the first minute? Well, notice that so far we're told the number of arriving customers per minute follows a poison distribution with a mean of 3, OK? So that means lambda in this problem equals 3. And We can let X be the number of customers. That arrived for a minute. Now, since we know it follows a poi and distribution, OK, then. We can say that X follows the poison distribution with a mean of 3, and from it we want to find the probability. OK, so we're solving for the probability. That X is greater than 3, that over 3 customers come in the first minute. No, recall that for the poison distribution formula. It tells us that the probability. That X is equal to some value K is equal toe to the pore of negative lambda multiplied by lambda raised to the pore of K divided by K factorial. OK? So it doesn't tell us, or this formula doesn't tell us what it would be for X greater than K. However, if we use the idea of the complement. Then the probability that X is greater than 3 is going to be equal to 1 minus the probability that x is less than or equal to 3, and the probability X is less than or equal to 3 would be equal to the probability of 0 plus the probability of 1 plus the probability of 2 plus the probability of 3. So if we can figure out these 4 probabilities, then we should be able to substitute the values to solve for the probability that X is greater than 3. Now if we break them up one by one, the probability that X equals 0 then would be equal to e-3 multiplied by 3 to the power of 0 divided by 0 factoria, and that would be equal to 0.0498. The probability X equals 1 is E-3 multiplied by 3 to the power of 1 divided by 1 factorial, which is 0.1494. The probability that X equals 2 would be E to the power of -3 multiplied by 32 divided by 2 factorial, which is 0.2240. And the probability, let me make some space on the right here, the probability that X equals 3 would be equal to the power of -3 multiplied by 3 cubed divided by 3 factorial, which equals 0.2240. So now that we have our 4 probabilities. We can substitute them into our formula for the probability that X is greater than 3. So that means it's going to be equal to 1 minus the sum of all those probabilities. So the sum of 0.0498, 0.1494, 0.2240, and 0.2240. Now, when we add all of these probabilities up, it equals 0.6472 and 1 minus 0.6472 equals 0.3528. Thus, the probability or the chance that over three customers come in the first minute would be 0.3528 or 35.28%. When we look back at our answer choices, that means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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In Exercises 1–7, consider a grocery store that can process a total of four customers at its checkout counters each minute.
The mean number of arrivals per minute is four. Find the probability that
b. more than four customers will arrive during the first minute.

Key Concepts

Poisson Distribution

Mean (λ) in Poisson Distribution

Calculating Probability

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In Exercises 1–7, consider a grocery store that can process a total of four customers at its checkout counters each minute.The mean number of arrivals per minute is four. Find the probability thatb. more than four customers will arrive during the first minute.