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Question
The average number of phone calls per minute into the switchboard of a company between 10.00 am and 2.30 pm is 2.5. Find the probability that during one particular minute there will be atleast 5 calls
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Solution
The average number of phone cells per minute into the switchboard of a company is λ = 2.5
x follows poisson distribution with
P(x) = `("e"^(-lambda) lambda^x)/(x!)`
= `((2^(-2.5) (2.5)^x)/(x!))`
P(atleast 5 calls) = P(X ≥ 5)
= P(X = 5) + P(X = 6) + …………..
= 1 – P(X < 5)
= `1 - {{:("P"("X" = 0) + "P"("X" = 1)"P"("X" = 2)),(+"p"("X" = 3) + "P"("X" = 4)):}}`
= `1 - {{:(0.08208 + ("e"^(-2.5) (2.5)^1)/(1!) +),(("e"^(-2.5) (2.5)^2)/(2!) + 0.2138 + ("e"^(-2.5) (2.5)^4)/(4!)):}}`
= `1 - {{:(0.08208 + (0.08208)(2.5) +),(((0.08208)(6.25))/2 + 0.2138 +),(((0.08208)(39.0625))/24):}}`
= `1 - {{:(0.08208 + 0.2052 + 0.2565),(+ 0.2138 + 0.1336):}}`
= `1 - 89094`
= 0.1091
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