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In a Poisson distribution if 2P(X = 1) = P(X = 2), then what is the mean?
- a)0
- b)2
- c)4
- d)± 2
Correct answer is option 'C'. Can you explain this answer?
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In a Poisson distribution if 2P(X = 1) = P(X = 2), then what is the me...
If X is random variable, then Poisson’s distribution is given by:
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In a Poisson distribution if 2P(X = 1) = P(X = 2), then what is the me...
To solve this problem, we can use the fact that the mean of a Poisson distribution is equal to λ, where λ is the rate parameter.
Let's assume that the rate parameter is λ.
The probability mass function of a Poisson distribution is given by P(X = k) = (e^(-λ) * λ^k) / k!, where k is a non-negative integer.
Given that 2P(X = 1) = P(X = 2), we can write the equation as:
2 * (e^-λ * λ^1) / 1! = (e^-λ * λ^2) / 2!
Simplifying this equation, we get:
2 * λ = λ^2 / 2
Multiplying both sides by 2, we get:
4 * λ = λ^2
Rearranging the equation, we get:
λ^2 - 4 * λ = 0
Factoring out λ, we get:
λ * (λ - 4) = 0
This equation has two solutions: λ = 0 and λ = 4.
However, the mean of a Poisson distribution cannot be zero, so the only valid solution is λ = 4.
Therefore, the mean of the Poisson distribution is 4.
Answer: c) 4
Let's assume that the rate parameter is λ.
The probability mass function of a Poisson distribution is given by P(X = k) = (e^(-λ) * λ^k) / k!, where k is a non-negative integer.
Given that 2P(X = 1) = P(X = 2), we can write the equation as:
2 * (e^-λ * λ^1) / 1! = (e^-λ * λ^2) / 2!
Simplifying this equation, we get:
2 * λ = λ^2 / 2
Multiplying both sides by 2, we get:
4 * λ = λ^2
Rearranging the equation, we get:
λ^2 - 4 * λ = 0
Factoring out λ, we get:
λ * (λ - 4) = 0
This equation has two solutions: λ = 0 and λ = 4.
However, the mean of a Poisson distribution cannot be zero, so the only valid solution is λ = 4.
Therefore, the mean of the Poisson distribution is 4.
Answer: c) 4
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In a Poisson distribution if 2P(X = 1) = P(X = 2), then what is the mean?a)0b)2c)4d)± 2Correct answer is option 'C'. Can you explain this answer?
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