An internal study by the Technology Services department at Lahley 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 that Linda Lahey, company president, received exactly 1 email between 4pm and 5pm yesterday?
b. what is the probability she received 5 or more email during the same period?
c. what is the probability she did not receive any email during the period?
To solve these questions, we will use the Poisson distribution formula, which is given by:
P(x; λ) = (e^(-λ) * λ^x) / x!
Where:
- x is the number of events we are interested in (in this case, the number of emails received by Linda Lahey).
- λ is the average rate of events (in this case, 2 emails per hour).
a. To find the probability that Linda Lahey received exactly 1 email between 4pm and 5pm yesterday, we will use the Poisson distribution formula:
P(x = 1; λ = 2) = (e^(-2) * 2^1) / 1!
To calculate this, we can substitute the values into the formula:
P(x = 1) = (e^(-2) * 2^1) / 1!
= (e^(-2) * 2) / 1
= 0.2706706
So, the probability that Linda Lahey received exactly 1 email between 4pm and 5pm yesterday is approximately 0.2707 or 27.07%.
b. To find the probability that Linda Lahey received 5 or more emails during the same period, we need to calculate the sum of probabilities for x = 5, 6, 7, ...
P(x ≥ 5; λ = 2) = 1 - P(x < 5; λ = 2)
= 1 - (P(x = 0; λ = 2) + P(x = 1; λ = 2) + P(x = 2; λ = 2) + P(x = 3; λ = 2) + P(x = 4; λ = 2))
To calculate this, we can substitute the values into the formula:
P(x = 0) = (e^(-2) * 2^0) / 0!
= (1 * 1) / 1
= 1
P(x = 1) = 0.2706706 (as calculated in part a)
P(x = 2) = (e^(-2) * 2^2) / 2!
= (0.1353353 * 4) / 2
= 0.2706706
P(x = 3) = (e^(-2) * 2^3) / 3!
= (0.1353353 * 8) / 6
≈ 0.180447
P(x = 4) = (e^(-2) * 2^4) / 4!
= (0.1353353 * 16) / 24
≈ 0.090223
P(x ≥ 5) = 1 - (1 + 0.2706706 + 0.2706706 + 0.180447 + 0.090223)
= 1 - 1.812011
= -0.812011 (negative value means it does not make sense)
So, there is no valid probability for Linda Lahey receiving 5 or more emails during the same period, as it cannot be negative.
c. To find the probability that Linda Lahey did not receive any email during the period, we need to calculate P(x = 0; λ = 2) using the Poisson distribution formula:
P(x = 0; λ = 2) = (e^(-2) * 2^0) / 0!
= 1 * 1 / 1
= 1
So, the probability that Linda Lahey did not receive any email during the period is 1 or 100%.