Answer to Question #205828 in Statistics and Probability for alicia

Question

The number of calls received by ABC Company follows Poisson distribution with an average of 6 calls per hour.

(a) Find the probability that there will be exactly 3 calls received in an hour. Answer: 

(b) Find the probability that there will be at least 2 calls received in an hour. Answer: 

(c) Find the probability that there will be exactly 2 calls received in 10 minutes. Answer: 

Answer 1

$(a) \\ Solution:\\ P(3)\\ \space Probability\space of\space exactly\space 3\space occurrences\\ mean=λ=6\\ \space and\space \\ x=3\\ \space \space \space apply\space the\space poisson\space probability\space formula:\\ P ( x ) = \frac{e^{ − λ}. ⋅ λ^ x} {x !}\\ where\space \\ x \space is\space the\space number\space of\space occurrences,\space \\ λ \space is\space the\space mean\space number\space of\space occurrences,\space and\space \\ e \space is\space the\space constant\space 2.718.\\\space Substituting\space in\space values\space for\space this\space problem,\space \\ x = 3\\ \space and\space \\ λ = 6\\ ,\space we\space have\\ P ( x ) = \frac{e^{ − 6}. ⋅ 6^ 3} {3 !}\\ Evaluating\space the\space expression,\space we\space have\\ P ( 3 ) = 0.089235078359989\\ -------------------------\\ probability \space \space that \space there \space will \space be \space at \space least \space 2 \space calls \space received \space in \space an \space hour. \space \\ P(x \space \geq \space 2)= \space 1-P(x \space <2) \space \\ Now \space we \space find \space P(x<2)\\ \space P(x<2)=P(x \space =0) \space + \space P(x \space =1) \space \\ = \frac{e^{ − 6}. ⋅ 6^ 0} {0 !}+ \frac{e^{ − 6}. ⋅ 6^ 1} {1 !}\\ Evaluating\space \space the\space \space expression,\space \space we\space \space have\\ P ( 0 )+P(1) = 0.00247875218+0.0148725131\\ =0.01735126528\\ P(x \space \geq \space 2)= \space 1-0.01735126528 \space \\ =0.98264873472 \\ ---------------------\\ (C) \space \\ probability \space that \space there \space will \space be \space exactly \space 2 \space calls \space received \space in \space 10 \space minutes \space \\ Given \space 6 \space phone \space call \space =60 \space min\\ 1 \space phone \space call \space =10 \space min\\ P(x=2)=\frac{e^{(-1) \space } \space 1^2 \space }{2!}\\ \\ 0.183939721$