For the following discrete random variable X with probability distribution:

X 0 1 2 3

P(X) 0.2 0.4 0.15 0.25

(a) Find the probability distribution for Y = 3X2 − 2X + 1
(i. e, all values of Y and P(Y )).
(b) Find E[Y ] using part (a).
(c) Find E[X] and E[X2].
(d) Find Pr(Y ≤ 2).
(e) Using part (c), verify your result from part (b) using
E[X] and E[X2].

(a) The probability distribution is shown in the attachment.

(b) The value of E (Y) is 7.85.

(c) The value of E (X) and E (X²) are 1.45 and 3.25 respectively.

(d) The value of P (Y ≤ 2) is 0.60.

(e) Verified that the value of E (Y) is 7.85.

Step-by-step explanation:

(a)

The random variable Y is defined as:

For X = {0, 1, 2, 3} the value of Y are:

The probability of Y for different values are as follows:

P (Y = 1) = P (X = 0) = 0.20

P (Y = 2) = P (X = 1) = 0.40

P (Y = 9) = P (X = 2) = 0.15

P (Y = 22) = P (X = 3) = 0.25

The probability distribution of Y is shown below.

(b)

The expected value of a random variable using the probability distribution table is:

Compute the expected value of Y as follows:

Thus, the value of E (Y) is 7.85.

(c)

Compute the expected value of X as follows:

Compute the expected value of X² as follows:

Thus, the value of E (X) and E (X²) are 1.45 and 3.25 respectively.

(d)

Compute the value of P (Y ≤ 2) as follows:

Thus, the value of P (Y ≤ 2) is 0.60.

(e)

The value of E (Y) is 7.85.

Use the values of E (X) and E (X²) computed in part (c) to compute the value of E (Y).

Hence verified.

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step-by-step explanation: