Assume that customers arrival has Poisson distribution with "\\lambda = 1.5"
Let "P_k=" the number of arrivals during the given interval.
a)
Let "\\lambda_t" be arrival rate during "t" minutes, "\\lambda_t=1.5\\cdot t";
"P(k\\le4)="
"P_0+P_1+P_2+P_3+P_4="
"\\sum_0^4{\\frac{(1.5t)^k}{k!}e^{-1.5t}}"
b)
"\\lambda=1.5\\cdot 2=3"
"P(k\\ge3)=1-P_0-P_1-P_2="
"1-\\frac{3^0}{0!}\\cdot e^{-3}-\\frac{3^1}{1!}\\cdot e^{-3}-\\frac{3^2}{2!}\\cdot e^{-3}="
"1-e^{-3}(1+3+9\/2)=1-\\frac{17}{2}e^{-3}\\approx 0.577."
c)
"\\lambda=1.5\\cdot6=9"
Use Normal distribution "N(\\lambda,\\lambda)" to approximate Poisson distribution:
"F_{poisson}(x,\\lambda)\\approx F_{normal}(x+1\/2,\\lambda,\\lambda)"
"F_{poisson}(x,9)\\approx F_{normal}(x+1\/2,9,9)"
"F_{normal}(x+1\/2,9,9)=Pr(X_{normal}\\le15.5)="
"Pr(\\frac{X_{normal}-9}{3}\\le\\frac{15.5-9}{3})=Pr(Z\\le2.1667)\\approx0.985"
Another option is to use cumulative Poisson distribution table, this method gives 0.978.
Answer: a) "\\sum_0^4{\\frac{(1.5t)^k}{k!}e^{-1.5t}}", b) 0.423, c) 0.978.
Comments
Dear A, thank you for correcting us.
I'm sure b is ≈0.5768 , ≈0.423 were the answer of the exponential number before 1 deduct by it
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Correct answer. Helped me score perfect marks and made me about about the question.
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