Question

The discrete random variable B has a probability distribution function given by P(B=b)=k(4−b) for b=0,1,2,3. Find k and E(B).

a. k=1/8 , E(B)= 5/2
b. k= 1/12 , E(B)= 5/2
c. k= 1/10 , E(B)= 3/2
d. k= 1/15 , E(B)= 4/3

Answer 1

To solve the problem, the value of k is first determined by ensuring the sum of the probabilities equals 1. After finding k, the expected value E(B) of the distribution is calculated by summing the product of each value of b and its probability. The correct values are k = 1/10 and E(B) = 1.

Step-by-step explanation:

The discrete random variable B has a probability distribution function (PDF) given by P(B=b)=k(4−b) for b=0,1,2,3. To find k, we must ensure that the sum of all probabilities equals 1 according to the characteristics of a PDF for a discrete random variable:

• P(B=0) = k(4-0) = 4k
• P(B=1) = k(4-1) = 3k
• P(B=2) = k(4-2) = 2k
• P(B=3) = k(4-3) = k

Sum of probabilities: 4k + 3k + 2k + k = 10k = 1

Hence, k = 1/10.

Now to find the expected value E(B), we multiply each value of b by its corresponding probability and sum them up:

• E(B) = 0*(4k) + 1*(3k) + 2*(2k) + 3*(k)

This simplifies to:

E(B) = 0 + 3k + 4k + 3k = 10k

Since we know k = 1/10, E(B) = 10*(1/10) = 1.