Question

Let X and Y be two Independent random variables such that V(X) =1 and V(Y) =2. Then V(3X-2Y+5) is equal: a. 25 b. 20 17 d. 15 C. O a d
please be clear and solved all

Answer 1

Solution :

1) If X and Y are two independent random variables and a, b and c are constants then,

V(aX - bY + C) = a² V(X) + b² V(Y)

We have, V(X) = 1, V(Y) = 2, a = 3, b = 2 and C = 5

V(3X - 2Y + 5) = (3² × 1) + (2² × 2)

V(3X - 2Y + 5) = 9 + 8

V(3X - 2Y + 5) = 17

Option (c) is correct.

2) The Cove(X, Y) is given as follows :

Cov(X, Y) = E(XY) - E(X).E(Y)

We have, E(XY) = 7, E(X) = 2 and E(Y) = 5

Cov(X, Y) = 7 - (2 × 5)

Cov(X, Y) = -3

Option (d) is correct.

3) Let X represents the number of rejected bulbs in a sample of 15 bulbs.

Probability of rejection = 0.04

Let's consider "rejecting a bulb" as success.

Probability of success (p) = 0.04

Number of trials (n) = 15

Since, probability of success remains constant in each trial, number of trials are finite, outcomes are independent and we have only two mutually exclusive outcomes (success and failure) for each trials, therefore we can consider that X follows binomial distribution.

According to binomial distribution, the probability of occurrence of exactly x success in n trials is given by,

P(X = z) = (2)(•)*(1 – p)*-*

Where, p is probability of success.

We have to find P(X = 2).

We have, n = 15 and p = 0.04

15 P(X = 2) = (0.04)?(1 – 0.04)15-2 2

15! P(X = 2) (2!)(15 – 2)?(0.04)? (0.96)13

P(X = 2) 15 x 14 x 13! (2 x 1) (139) Х (0.04)? (0.96)13-

P(X = 2) 15 x 14 (2 x 1) x (0.04)2(0.96)13-

P(X = 2) = 0.0988

Hence, the required probability is 0.0988.

Second option is correct.