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Question
The number of complaints which a bank manager receives per day is a Poisson random variable with parameter m = 4. Find the probability that the manager will receive -
(a) only two complaints on any given day.
(b) at most two complaints on any given day
[Use e-4 =0.0183]
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Solution
`P(X=k)=e^(-m) xx m^k/(k!)`
(a) only two complaints on any given day.
k=2, m=4
`P(X=2)=e^(-4) xx 4^2/2`
`=0.0183 xx 8 `
`=0.1464`
Probability that the manager will recieve only two complaints on any given day = 0.1464
(b) at most two complaints on any given day
`=P(x<=2)`
`=P(x=0)+P(x=1)+P(x=2)`
`= e^(-4) + 4e^(-4) + 4^2 (e^(-4))/(2!)`
`= e^(-4) ( 1 + 4 + 8)`
`= 13 x e^(-4)`
= 13 x 0.0183
= 0.2379
Probab ility that the manager will receive · at most two complaints on any given day = 0.2379.
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The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives only two complaints on a given day
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