An internal study by the Technology Services department at L

An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of 3.3 emails per hour. Assume the arrival of these emails is approximated by the Poisson distribution.

What is the probability Linda Lahey, company president, received exactly 1 email between 4 P.M. and 5 P.M. yesterday? (Round your answer to 4 decimal places.)

What is the probability she received 7 or more emails during the same period? (Round your answer to 4 decimal places.)

What is the probability she received one or less emails during the period? (Round your answer to 4 decimal places.)

Solution

Given X~Poisson(mean=3.3 per hour)

P(X=x)=(3.3^x)*exp(-3.3)/x!

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(a) P(X=1) = (3.3^1)*exp(-3.3)/1 =0.1217

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(b) P(X>=7)=1-P(X=0)-P(X=1)-...-P(X=6)

=1-(3.3^0)*exp(-3.3)/1-....-(3.3^6)*exp(-3.3)/6!

=0.0510

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(c) P(X<=1)= P(X=0)+P(X=1)

=(3.3^0)*exp(-3.3)/1 + (3.3^1)*exp(-3.3)/1

=0.1586