According to email logs, one employee at your company receives an average of 2.15 emails per week. Suppose the count of emails received can be adequately modeled as a Poisson random variable.

a. What is the distribution of the number of emails in a two-week period?
b. What is the probability of receiving 4 or fewer emails in a two-week period?

a) P(x=r) = e^-4.3 × 4.3^r/(r!), b) 0.7637

Step-by-step explanation: The employee receives an average of 2.15 emails in one week.

This is the constant rate at which the event is occurring and it does not affect any other.

Hence λ = 2.15.

If she receives 2.15 emails in 1 week, then in 2 week, she will receive an average of 2.15×2 = 4.3.

Hence for the 2 week period, λ = 4.3 ( that's she receives and average of 4.3 emails in 2 weeks).

a) the probability distribution that defines a possion probability distribution is given below as

P(x=r) = e^-λ × λ^r/(r!)

Our question is to get the distribution, hence our answer will be in terms of r and λ= 4.3

P(x=r) = e^-4.3 × 4.3^r/(r!)

b) P(4 or fewer emails) = P(x≤4)

P(x≤4) = P(x=4) + P(x=3) + P(x=2) + P(x=1) +P(x=0)

Recall that

P(x=r) = e^-4.3 × 4.3^r/(r!)

At x = 4

P(x=4) = e^-4.3 × 4.3^4/(4!)

P(x=4) =e^-4.3 × 4.3^4/24

P(x=4) = 0.3866

At x = 3

P(x=3) = e^-4.3 × 4.3^3/(3!)

P(x=3) = e^-4.3 × 4.3^3/(6)

P(x=3) = 0.1798.

At x = 2

P(x=2) = e^-4.3 × 4.3^2/(2!)

P(x=2) = e^-4.3 × 4.3^2/(2)

P(x=2) = 0.1254

At x = 1

P(x=1) = e^-4.3 × 4.3^1/(1!)

P(x=1) = e^-4.3 × 4.3^1/(1)

P(x=1) = 0.0583

At x = 0

P(x=0) = e^-4.3 × 4.3^0/(0!)

P(x=0) = e^-4.3 × 1/(1)

P(x=0) = 0.0136

P(x≤4) = 0.3866+0.1798+0.1254+0.0583+0.0136 = 0.7637

3201

step-by-step explanation:

coins of silver or gold

step-by-step explanation: