Question

1. Consider the following function: (?) = ?(?³ + 1), ? = 0, 1, 2, 3
2. 1. What is the value of the constant ? so that (?) is a pmf?
   2. Plug-in the value of ? in the expression of (?) and show the pmf in a table. Draw a probability histogram of the pmf.
   3. Find the cdf (?) and write it explicitly defined over the entire real number line. Draw the cdf (?).
   4. Calculate the probabilities: (i). (0 < ? ≤ 2),      (ii) ?(? ≠ 1).
   5. Find the expected value of ?.
   6. Find the expected value of 3?³ − 2? + 1.
   7. Find the variance of ?.

Answer 1

We would be looking at the first 4 parts here as:

a) The value of k is computed by using the property that the sum of all probabilities across the X range is 1. Therefore we have here:

k(0 + 1) + k(1 + 1) + k(2³ + 1) + k(3³ + 1) = 1

k + 2k + 9k + 28k = 1

40k = 1

k = 1/40 = 0.025

Therefore 0.025 is the required value of k here.

b) Using the above value of k, the PMF here is obtained as:

p(x) = 0.025*(x³ + 1)

x	P(x)
0	0.025
1	0.05
2	0.225
3	0.7

This is the required PMF table here.

The histogram for the same is given here as:

c) The CDF here is given as:
P(X < 0) = 0
P(X <= 0) = P(X = 0) = 0.025
P(X <= 1) = 0.025 + 0.05 = 0.075
P(X <= 2) = 0.075 + 0.225 = 0.3
P(X >= 3) = 1

d) The probabilities here are computed as:
(i) P( 0 < X <= 2) = P(X = 1) + P(X = 2) = 0.05 + 0.225 = 0.275
Therefore 0.275 is the required probability here.

(ii) P(X not equal to 1) = 1 - P(X = 1) = 1 - 0.05 = 0.95

Therefore 0.95 is the required probability here.