The random variable [tex]$X$[/tex] follows a Poisson process with the given value of [tex]$\lambda$[/tex] and [tex]$t$[/tex]. Assuming [tex]$\lambda=0.14$[/tex] and [tex]$t=10$[/tex], compute the following.

(a) [tex]$P(6)$[/tex]
(b) [tex]$P(X\ \textless \ 6)$[/tex]
(c) [tex]$P ( X \geq 6)$[/tex]
(d) [tex]$P(5 \leq X \leq 9)$[/tex]
(e) [tex]$\mu_X$[/tex] and [tex]$\sigma_X$[/tex]

(a) [tex]$P (6) \approx$[/tex] $\square$
(Do not round until the final answer. Then round to four decimal places as needed.)
(b) [tex]$P ( X \ \textless \ 6) \approx$[/tex] $\square$
(Do not round until the final answer. Then round to four decimal places as needed.)
(c) [tex]$P ( X \geq 6) \approx$[/tex] $\square$
(Do not round until the final answer. Then round to four decimal places as needed.)
(d) [tex]$P (5 \leq X \leq 9) \approx$[/tex] $\square$
(Do not round until the final answer. Then round to four decimal places as needed.)
(e) [tex]$\mu_X \approx$[/tex] $\square$ (Round to two decimal places as needed.)
[tex]$\sigma_{ X } \approx$[/tex] $\square$ (Round to three decimal places as needed.)

Calculate μ = λ t = 0.14 × 10 = 1.4 .
Use the Poisson probability mass function: P ( X = k ) = k ! e − μ μ k ​ to find the probabilities.
Compute the required probabilities: P ( 6 ) ≈ 0.0026 , P ( X < 6 ) ≈ 0.9968 , P ( X ≥ 6 ) ≈ 0.0032 , P ( 5 ≤ X ≤ 9 ) ≈ 0.0143 .
Calculate the mean and standard deviation: μ X ​ = 1.40 , σ X ​ ≈ 1.183 .
P ( 6 ) ≈ 0.0026 , P ( X < 6 ) ≈ 0.9968 , P ( X ≥ 6 ) ≈ 0.0032 , P ( 5 ≤ X ≤ 9 ) ≈ 0.0143 , μ X ​ ≈ 1.40 , σ X ​ ≈ 1.183 ​

Explanation

Understand the problem First, let's understand the problem. We are given a Poisson process with λ = 0.14 and t = 10 . We need to calculate several probabilities and statistics related to the random variable X , which represents the number of events occurring in the given time interval.

Calculate P(X=6) (a) We need to find P ( X = 6 ) . The formula for the Poisson probability mass function is: P ( X = k ) = k ! e − μ μ k ​ where μ = λ t . In our case, μ = 0.14 × 10 = 1.4 . So, we need to calculate P ( X = 6 ) = 6 ! e − 1.4 ( 1.4 ) 6 ​ .

Calculate P(X<6) (b) We need to find P ( X < 6 ) . This is the sum of probabilities from X = 0 to X = 5 : P ( X < 6 ) = k = 0 ∑ 5 ​ P ( X = k ) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 ) + P ( X = 5 ) . Each term can be calculated using the Poisson probability mass function.

Calculate P(X>=6) (c) We need to find P ( X ≥ 6 ) . This is the complement of P ( X < 6 ) , so P ( X ≥ 6 ) = 1 − P ( X < 6 ) .

Calculate P(5<=X<=9) (d) We need to find P ( 5 ≤ X ≤ 9 ) . This is the sum of probabilities from X = 5 to X = 9 : P ( 5 ≤ X ≤ 9 ) = k = 5 ∑ 9 ​ P ( X = k ) = P ( X = 5 ) + P ( X = 6 ) + P ( X = 7 ) + P ( X = 8 ) + P ( X = 9 ) . Each term can be calculated using the Poisson probability mass function.

Calculate mean and standard deviation (e) We need to find the mean μ X ​ and standard deviation σ X ​ of X . For a Poisson distribution, the mean is equal to μ = λ t , and the standard deviation is σ = μ ​ = λ t ​ .

Final Calculations and Results Now, let's calculate the values using the formulas: (a) P ( X = 6 ) = 6 ! e − 1.4 ( 1.4 ) 6 ​ ≈ 0.0026 (b) P ( X < 6 ) = ∑ k = 0 5 ​ k ! e − 1.4 ( 1.4 ) k ​ ≈ 0.9968 (c) P ( X ≥ 6 ) = 1 − P ( X < 6 ) ≈ 1 − 0.9968 = 0.0032 (d) P ( 5 ≤ X ≤ 9 ) = ∑ k = 5 9 ​ k ! e − 1.4 ( 1.4 ) k ​ ≈ 0.0143 (e) μ X ​ = λ t = 1.4 $\sigma_X = \sqrt{\lambda t} = \sqrt{1.4} \approx 1.183

Final Answer Therefore, the final answers are: (a) P ( 6 ) ≈ 0.0026 (b) P ( X < 6 ) ≈ 0.9968 (c) P ( X ≥ 6 ) ≈ 0.0032 (d) P ( 5 ≤ X ≤ 9 ) ≈ 0.0143 (e) μ X ​ ≈ 1.40 σ X ​ ≈ 1.183


Examples
Imagine you are managing a website and want to model the number of users visiting your site per minute as a Poisson process. Knowing the average rate of visits ( λ ) and the time interval ( t ), you can calculate the probability of observing a specific number of visits, or a range of visits. This helps you prepare your servers for expected traffic and understand the variability in user visits, which is crucial for resource allocation and ensuring a smooth user experience. For example, if λ = 5 visits per minute and t = 1 minute, you can calculate the probability of having exactly 10 visits in that minute, or the probability of having fewer than 3 visits.

Answered by GinnyAnswer | 2025-07-07

To summarize, the calculations for the Poisson process yield the following results: P ( 6 ) ≈ 0.0026 , P ( X < 6 ) ≈ 0.9968 , P ( X ≥ 6 ) ≈ 0.0032 , and P ( 5 ≤ X ≤ 9 ) ≈ 0.0143 . The mean μ X ​ is 1.40 and the standard deviation σ X ​ is approximately 1.183 .
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Answered by Anonymous | 2025-07-17