Let the random variable X be from a binomial distribution with n=12 trials and p=0.35. A second random variable Y is defined as Y=-2X+7. Find the moment-generating function for Y.
a) Omyt = e(-2⋅0.35t+7)
b) my = 0.35e(0.65t)+e⁽¹²ᵗ⁾
c) Omyt = e(t-0.7t-2)
d) Omyt =e(0.65+0.35e⁽⁻²ᵗ⁾)

Respuesta :

Answer:

Step-by-step explanation: To find the moment-generating function (MGF) for the random variable Y, which is defined as Y = -2X + 7, we can use the properties of MGFs.

1. The MGF of a linear transformation of a random variable X is given by:

M_Y(t) = E[e^(tY)] = E[e^(t(-2X+7))] = E[e^(-2tX+7t)]

2. Since X follows a binomial distribution with n=12 trials and p=0.35, the MGF of X is:

M_X(t) = E[e^(tX)] = (0.35e^t + 0.65)^12

3. Now, substitute the MGF of X into the expression for M_Y(t):

M_Y(t) = E[e^(-2tX+7t)] = E[e^(-2tX)e^(7t)]

4. Using the property that the MGF of the sum of independent random variables is the product of their MGFs, we get:

M_Y(t) = M_X(-2t) e^(7t) = (0.35e^(-2t) + 0.65)^12 e^(7t)

5. Therefore, the moment-generating function for Y, denoted as M_Y(t), is:

M_Y(t) = (0.35e^(-2t) + 0.65)^12 * e^(7t)

The correct option related to the MGF for Y among the given choices is not explicitly mentioned in the question. However, the accurate expression derived above matches closest to the provided options.

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