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Zielflug [23.3K]
2 years ago
12

Customers arrivals at a checkout counter in a department store per hour have a Poisson distribution with parameter λ = 7. Calcul

ate the probabilities for the following events.(1) (2 points) Exactly seven customers arrive in a random 1-hour period.(2) (4 points) No more than two customers arrive in a random 1-hour period.(3) (4 points) At least three customers arrive in a random 1-hour period
Mathematics
1 answer:
IgorLugansk [536]2 years ago
4 0

Answer:

(1)14.9% (2) 2.96% (3) 97.04%

Step-by-step explanation:

Formula for Poisson distribution: P(k) = \frac{\lambda^ke^{-k}}{k!} where k is a number of guests coming in at a particular hour period.

(1) We can substitute k = 7 and \lambda = 7 into the formula:

P(k=7) = \frac{7^7e^{-7}}{7!}

P(k=7) = \frac{823543*0.000911882}{5040} = 0.149 = 14.9\%

(2)To calculate the probability of maximum 2 customers, we can add up the probability of 0, 1, and 2 customers coming in at a random hours

P(k\leq2) = P(k=0)+P(k=1)+P(k=2)

P(k\leq2) = \frac{7^0e^{-7}}{0!} + \frac{7^1e^{-7}}{1!} + \frac{7^2e^{-7}}{2!}

P(k \leq 2) = \frac{0.000911882}{1} + \frac{7*0.000911882}{1} + \frac{49*0.000911882}{2}

P(k\leq2) = 0.000911882+0.006383174+0.022341108 \approx 0.0296=2.96\%

(3) The probability of having at least 3 customers arriving at a random hour would be the probability of having more than 2 customers, which is the invert of probability of having no more than 2 customers. Therefore:

P(k\geq 3) = P(k>2) = 1 - P(k\leq2) = 1 - 0.0296 = 0.9704 = 97.04\%

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