Answer to Question #298742 in Statistics and Probability for Diksha

Question

Mr. Vijay, a retired government servant, is considering investing his money in two proposals. He wants to choose the one that has higher average net present value and lower standard deviations. The relevant data are given below. Can you help him choosing the proposal?

Proposal A

Net present value (NPV) Chance of the possible outcome of NPV

1559 0.30

5662 0.40

9175 0.30

Proposal B

Net present value (NPV) Chance of the possible outcome of NPV

-10050 0.30

5812 0.40

20584 0.30

Answer 1

Proposal A

Given

X 1559 5662 9175

P(X) 0.30 0.40 0.30

Therefore, the mean;

μ_X  = Σ X_iP_i = 1559*0.30 + 5662*0.40 + 9175= 5485

The variance σ² = Σ(x – μ)2 ⋅ P(x)

= 0.30*(1559 - 5485)² + 0.4*(5662-5485)² + 0.3*(9175-5485)²

=8721404.4

The standard deviation is the square root of the variance

standard deviation = (8721404.4)^1/2 = 2953.20

Therefore, the average net present value for proposal A = 5485, with a standard deviation = 2953.20

Proposal B

Also, given

X -10050 5812 20584

P(X) 0.30 0.40 0.30

Therefore, the mean;

μ_X  = Σ X_iP_i = -10050*0.30 + 5812*0.40 + 20584= 5485

The variance σ² = Σ(x – μ)²⋅ P(x)

= 0.30*(-10050- 5485)² + 0.4*(5812-5485)² + 0.3*(20584-5485)²

= 140837579.4

The standard deviation is the square root of the variance

standard deviation = (140837579.4)^1/2 = 11867.50

The average net present value for proposal B = 5485, with a standard deviation = 11867.50.

Although both proposal A and B have the same average net present value (5485), Mr. Vijay should consider investing in proposal A since it has a lower standard deviation (2953.20) as compared to proposal B which has a higher standard deviation (11867.50)