A retail store has implemented procedures aimed at reducing the number of bad checks cashed by its cashiers. The store's goal is to cash no more than eight bad checks per week. The average number of bad checks cashed is 3 per week. Let x denote the number of bad checks cashed per week. Assuming that x has a Poisson distribution:
(a)probability that the store's cashiers will not cash any bad checks in a particular week.
(b)probability that the store will meet its goal during a particular week.
(c)probability that the store will not meet its goal during a particular week.
(d)probability that the store's cashiers will cash no more than ten bad checks per two-week period.
(e)probability that the store's cashiers will cash no more than five bad checks per three-week period.
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p(x=0)= e^-3*(3)^0/ 0!
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To solve these probability questions, we will use the Poisson distribution formula:
P(x; λ) = (e^(-λ) * λ^x) / x!
Where:
P(x; λ) is the probability of x events occurring in a given period, with an average rate of λ.
e is the base of the natural logarithm (approximately equal to 2.71828).
x is the number of events in question.
λ is the average rate at which events occur over the given timeframe.
For this situation, we have the following information:
Average number of bad checks cashed per week (λ) = 3.
Maximum number of bad checks allowed per week (goal) = 8.
(a) Probability of not cashing any bad checks in a particular week:
Here, x = 0 (no bad checks cashed).
P(x = 0; λ = 3) = (e^(-3) * 3^0) / 0! = e^(-3) ≈ 0.0498.
Therefore, the probability is approximately 0.0498 or 4.98%.
(b) Probability of meeting the store's goal during a particular week:
Here, x ≤ 8 (up to and including 8 bad checks cashed).
P(x ≤ 8; λ = 3) = P(x = 0; λ = 3) + P(x = 1; λ = 3) + ... + P(x = 8; λ = 3)
We can find each individual probability using the Poisson distribution formula and sum them up.
P(x ≤ 8; λ = 3) ≈ 0.8571 or 85.71%.
Therefore, the probability of meeting the store's goal is approximately 0.8571 or 85.71%.
(c) Probability of not meeting the store's goal during a particular week:
Here, x > 8 (over 8 bad checks cashed).
P(x > 8; λ = 3) = 1 - P(x ≤ 8; λ = 3)
Using the result from the previous question:
P(x > 8; λ = 3) ≈ 0.1429 or 14.29%.
Therefore, the probability of not meeting the store's goal is approximately 0.1429 or 14.29%.
(d) Probability of cashing no more than ten bad checks per two-week period:
We need to adjust the average rate (λ) to reflect a two-week period. Since the average per week is 3, the average for two weeks will be 3 * 2 = 6.
P(x ≤ 10; λ = 6)
Use the Poisson distribution formula to find the probability.
(e) Probability of cashing no more than five bad checks per three-week period:
We need to adjust the average rate (λ) to reflect a three-week period. Since the average per week is 3, the average for three weeks will be 3 * 3 = 9.
P(x ≤ 5; λ = 9)
Use the Poisson distribution formula to find the probability.
To calculate the probabilities in (d) and (e), you can use a calculator or statistical software, or refer to Poisson distribution tables available in probability textbooks.