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e-lub [12.9K]
2 years ago
8

An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of tw

o 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?
Business
1 answer:
solong [7]2 years ago
7 0

Answer:

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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