Answer:
To find the probability P(X = 2), we need to use the moment generating function (MGF) and the formula for the nth moment of a random variable:
Mx(t) = E[e^(tx)] = Σ [x^n P(X = x) e^(tx)]
Taking the second derivative of the MGF with respect to t, we get:
Mx''(t) = E[X^2 e^(tx)]
Setting t = 0.5 in the MGF, we get:
Mx(0.5) = 1 - 0.1e
where e is the mathematical constant e = 2.71828...
Taking the second derivative of the MGF with respect to t, we get:
Mx''(t) = 3.6e^(2t) for t < -ln(0.1)
Mx''(t) = ∞ for t ≥ -ln(0.1)
Therefore, we can write:
E[X^2] = Mx''(0) = 3.6e^0 = 3.6
Using the formula for the variance of a random variable:
Var(X) = E[X^2] - E[X]^2
We need to find E[X] first.
Taking the first derivative of the MGF with respect to t, we get:
Mx'(t) = E[X e^(tx)]
Setting t = 0.5 in the MGF, we get:
Mx'(0.5) = 1.8e
Therefore, we can write:
E[X] = Mx'(0) = 1.8
Now we can find the variance:
Var(X) = E[X^2] - E[X]^2 = 3.6 - 1.8^2 = 0.72
Finally, we can find the probability P(X = 2) using the formula for the probability mass function (PMF) of a discrete random variable:
P(X = 2) = e^(-λ) λ^k / k!
where λ is the expected value of the random variable, which is also the parameter of the Poisson distribution.
In this case, λ = E[X] = 1.8, and k = 2.
Therefore, we can write:
P(X = 2) = e^(-1.8) (1.8)^2 / 2! ≈ 0.1638
Rounding to 4 decimal places, we get:
P(X = 2) ≈ 0.1638
hope it helps!!