An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of two emails per hour. Assume the arrival of these emails is approximated by the Poisson distribution. a. What is the probability Linda Lahey, company president, received exactly 1 email between 4 P. M. and 5 P. M. yesterday?
b. What is the probability she received 5 or more emails during the same period?
c. What is the probability she did not receive any emails during the period?

0.2706 ; 0.05265 ; 0.1353

Explanation:

Given that :

λ = 2

According to the poisson distribution formula :

P(x = x) = (λ^x * e^-λ) / x!

P(x = 1) = (2^1 *e^-2) / 1!

P(x = 1) = (2 * 0.1353352) = 0.2706

P(x ≥ 5) = 1 - P(x < 5)

1 - P(x < 5) = 1 - [p(x = 0) + p(x = 1) + p(x = 2) + p(x = 3) + p(x = 4)]

We obtain and add the individual probabilities. To save computation time, we can use a poisson distribution calculator :

1 - P(x < 5) = 1 - (0.13534+0.27067+0.27067+0.18045+0.09022)

1 - P(x < 5) = 1 - 0.94735 = 0.05265

P(x ≥ 5) = 1 - P(x < 5) = 0.05265

Probability that no emails was received :

x = 0

P(x = 0) = (2^0 *e^-2) / 0!

P(x = 0) = (1 * 0.1353352) / 1 = 0.1353

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