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X is a Poisson variate satisfying the following condition 9 P (X = 4) + 90 P (X = 6) = P (X = 2). What is the value of P (X £ 1)?
- a)0.5655
- b)0.6559
- c)0.7358
- d)0.8201
Correct answer is option 'C'. Can you explain this answer?
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X is a Poisson variate satisfying the following condition 9 P (X = 4) ...
Given, 9P(X=4) ≤ 90P(X=6) = P(X=2)
Let λ be the mean of the Poisson distribution.
Then, P(X=4) = e^(-λ)(λ^4)/4! and P(X=6) = e^(-λ)(λ^6)/6!
Also, P(X=2) = e^(-λ)(λ^2)/2!
Dividing the given equations by e^(-λ) we get:
9(λ^4)/4! ≤ 90(λ^6)/6! = (λ^2)/2!
Simplifying, we get:
λ^2/λ^4 ≤ 10/3
λ^2 ≤ 10λ/3
λ ≤ 10/3
Now, we can find P(X≥1) as follows:
P(X≥1) = 1 - P(X=0) = 1 - e^(-λ)
Substituting λ=10/3, we get:
P(X≥1) = 1 - e^(-10/3)
P(X≥1) ≈ 0.7358
Therefore, the value of P(X≥1) is approximately 0.7358, which is option C.
Let λ be the mean of the Poisson distribution.
Then, P(X=4) = e^(-λ)(λ^4)/4! and P(X=6) = e^(-λ)(λ^6)/6!
Also, P(X=2) = e^(-λ)(λ^2)/2!
Dividing the given equations by e^(-λ) we get:
9(λ^4)/4! ≤ 90(λ^6)/6! = (λ^2)/2!
Simplifying, we get:
λ^2/λ^4 ≤ 10/3
λ^2 ≤ 10λ/3
λ ≤ 10/3
Now, we can find P(X≥1) as follows:
P(X≥1) = 1 - P(X=0) = 1 - e^(-λ)
Substituting λ=10/3, we get:
P(X≥1) = 1 - e^(-10/3)
P(X≥1) ≈ 0.7358
Therefore, the value of P(X≥1) is approximately 0.7358, which is option C.
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X is a Poisson variate satisfying the following condition 9 P (X = 4) + 90 P (X = 6) = P (X = 2). What is the value of P (X £ 1)?a)0.5655b)0.6559c)0.7358d)0.8201Correct answer is option 'C'. Can you explain this answer?
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