The second moment of a Poisson-distributed random variable is 2. The mean of the random variable is _______.
Explanation:
The Poisson distribution is characterized by the property that its mean (μ) is equal to its variance (σ²). In other words, if X is a Poisson-distributed random variable:
E[X] = μ = λ,
Var[X] = σ² = λ,
where λ is the rate parameter for the Poisson distribution.
Furthermore, the second moment of a random variable X, E[X²], is related to its mean and variance by:
E[X²] = Var[X] + (E[X])².
Substituting the expressions for the mean and variance of a Poisson distribution, we get:
E[X²] = λ + λ².
In your case, you're given that the second moment E[X²] is 2. Equating this to λ + λ² and solving for λ, you get a quadratic equation:
λ² + λ - 2 = 0.
Solving this quadratic equation gives two solutions, λ = 1 and λ = -2. Since the rate parameter λ of a Poisson distribution must be positive (it represents a rate or number of occurrences), we discard the negative solution, so:
λ = 1.
Therefore, the mean of the random variable, which is equal to λ for a Poisson distribution, is 1.
In an examination involving multiple choice questions, a student works out the solution in 50% of the questions. In the remaining questions the student guesses the answer. However, when the answer is guessed the probability that it is correct is 0.30. When the student works out the solutions it may be wrong with probability 0.10.
If the answer to a particular question is correct, what is the probability that the student guessed the answer?
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In a frequency curve, what is plotted on the vertical axis?
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