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8. Your friend sends you text messages at an average rate of 3 per mimite; the Poisson distribution is a in the next two minu
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Answer #1

here time between two text messages follow exponential distribution

a)P(no text message in two minutes) =P(X>2)=e-\lambdax =e-3*2 =0.0025

b)P(2nd text more than 2 minutes after first)=no text message in two minutes) =P(X>2)=e-\lambdax =e-3*2 =0.0025

c)P(4th arrive within 30 second(0.5 min) of 3rd)=P(T<0.5)=1-e-3*0.5 =0.7769

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