The number of emails follows a Poisson distribution with an average rate of 0.2 emails per minute. The…

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The number of emails follows a Poisson distribution with an average rate of 0.2 emails per minute. The probability mass function for a Poisson distribution is given by $P(X=k) = lambda^k e^-

ScanSolve AI Tutor · Accepted Answer

Here's how to solve the problem using the Poisson distribution: The number of emails follows a Poisson distribution with an average rate of 0.2 emails per minute. The probability mass function for a Poisson distribution is given by P(X=k) = (λ^k e^(−λ))/(k!), where λ is the average number of events in the given interval. a) What is the probability that I get no emails in an interval of length 5 minutes? Step 1: Calculate the average rate λ for a 5-minute interval. The rate is 0.2 emails per minute. For a 5-minute interval, the average number of emails is: λ_a = 0.2 emails/minute × 5 minutes = 1 email; Step 2: Calculate the probability of getting no emails (k=0) using λ_a = 1. P(X=0) = (λ_a⁰ e^(−λ_a))/(0!) = (1⁰ e⁻¹)/(0!) = (1 × e⁻¹)/(1) = e⁻¹; P(X=0) ≈ 0.367879 The probability of getting no emails in a 5-minute interval is approximately 0.3679. b) What is the probability that I get more than 3 emails in an interval of length 10 minutes? Step 1: Calculate the average rate λ for a 10-minute interval. The rate is 0.2 emails per minute. For a 10-minute interval, the average number of emails is: λ_b = 0.2 emails/minute × 10 minutes = 2 emails Step 2: Calculate the probability of getting more than 3 emails. P(X > 3) = 1 − P(X ≤ 3). This means P(X > 3) = 1 − [P(X=0) + P(X=1) + P(X=2) + P(X=3)]. We use λ_b = 2 for these calculations. Step 3: Calculate P(X=0), P(X=1), P(X=2), and P(X=3). P(X=0) = (2⁰ e⁻²)/(0!) = (1 × e⁻²)/(1) = e⁻² ≈ 0.135335; P(X=1) = (2¹ e⁻²)/(1!) = (2 × e⁻²)/(1) = 2e⁻² ≈ 0.270671; P(X=2) = (2² e⁻²)/(2!) = (4 × e⁻²)/(2) = 2e⁻² ≈ 0.270671; P(X=3) = (2³ e⁻²)/(3!) = (8 × e⁻²)/(6) = 4/3e⁻² ≈ 0.180447 Step 4: Sum the probabilities and subtract from 1. P(X ≤ 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3); P(X ≤ 3) ≈ 0.135335 + 0.270671 + 0.270671 + 0.180447 = 0.857124; P(X > 3) = 1 − P(X ≤ 3) = 1 − 0.857124 = 0.142876 The probability of getting more than 3 emails in a 10-minute interval is approximately 0.1429.

The number of emails follows a Poisson distribution with an average rate of 0.2 emails per minute. The probability mass function for a Poisson distribution is given by $P(X=k) = {lambda^k e^{-

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Step-by-step solution

Part a· 2 steps

Calculate the average rate $\lambda$ for a 5-minute interval.

Calculate the probability of getting no emails ( $k=0$ ) using $\lambda_a = 1$ .

Part b· 4 steps
$\text{0.1429}$

Calculate the average rate $\lambda$ for a 10-minute interval.

Calculate the probability of getting more than 3 emails.

Calculate $P(X=0)$ , $P(X=1)$ , $P(X=2)$ , and $P(X=3)$ .

Sum the probabilities and subtract from 1.

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