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

Answer 1

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