Question

2. (a) the P(X ≤ 4), P(X ≥ 5) and P(X=3). and also find moment generating function M_X=3(t = 3) for same given problem.

(b)X: -2, -1, 0, 1, 2, 3       Here : The base of M is 3, M_X=3                                                 

p(x): 0.1, k, 0.2, 2k, 0.3, k

1) Find the value of k, mean and variance. (2). Construct CDF and drar the graph.

3.(a)X: -0, -1, 2, 3 p(x): 0.1, 0.3, 0.5, 0.1 Find probability function of Y=X^2+2X . Also find P(X ≤ 3), P(1 < X ≤ 3) and P(X = 3). (b) f(x)=cx^2(1 − x), 0 < x < 1 then find the constant c, P(−1 < X < 0.5), P(0.2 < X < 0.8).

Answer 1

2) In the absence of other information I'm assuming these probabilities are associated with rolling a fair die. So dataset=1, 2, 3, 4, 5, 6.

Therefore P(X≤4)=4/6=⅔; P(X≥5)=2/6=⅓; P(X=3)=⅙.

MGF is M_X(t)=∑e^tXP(X)=⅙∑e^tX, because P(X)=⅙ for all X between 1 and 6 (rectangular or uniform distribution).

M_X(t) = ⅙(e^t+e^2t+e^3t+e^4t+e^5t+e^6t). From this if we put X=3 and t=3 we get: M_X=3(t=3)=⅙e⁹ (not sure if this what is meant in your question).

2b) p(-2)=0.1, p(-1)=k, p(0)=0.2, p(1)=2k, p(2)=0.3, p(3)=k.

If the given values of X and p(X=x) represent the entire population of discrete probabilities, then ∑p(x)=1.

Therefore, 0.1+k+0.2+2k+0.3+k=1, 0.6+4k=1, 4k=0.4, k=0.1.

2b1) If M_X(t)=∑e^tXp(X) for X=-2, -1, 0, 1, 2, 3.

So M_X(t)=∑e^tXp(X)=0.1e^-2t+ke^-t+0.2+2ke^t+0.3e^2t+ke^3t, and, substituting for k:

M_X(t)=0.1e^-2t+0.1e^-t+0.2+0.2e^t+0.3e^2t+0.1e^3t.

To find the mean μ=E(X), we need to differentiate M_X(t) wrt t to get the first moment:

-0.2e^-2t-0.1e^-t+0.2e^t+0.6e^2t+0.3e^3t, then set t=0: -0.2-0.1+0.2+0.6+0.3=0.8, so mean, μ=0.8.

Variance, σ², is calculated from the second moment. Differentiate again wrt t:

0.4e^-2t+0.1e^-t+0.2e^t+1.2e^2t+0.9e^3t then set t=0: 0.4+0.1+0.2+1.2+0.9=2.8.

Variance is 2.8.

2b2) To construct the CDF we need to accumulate probabilities:

X:	-2	-1	0	1	2	3
p(X):	0.1	0.1	0.2	0.2	0.3	0.1
CDF:	0.1	0.2	0.4	0.6	0.9	1.0

The CDF graph is represented by the points A to E.

More to follow in due course...