Question

The random variable x has the following probability distribution: x f(x) 0 .25 1 .20 2 .15 3 .30 4 .10 a. Is this probability distribution valid? Explain and list the requirements for a valid probability distribution. b. Calculate the expected value of x. c. Calculate the variance of x. d. Calculate the standard deviation of x.

Answer 1

Answer and Explanation:

Given : The random variable x has the following probability distribution.

To find :

a. Is this probability distribution valid? Explain and list the requirements for a valid probability distribution.

b. Calculate the expected value of x.

c. Calculate the variance of x.

d. Calculate the standard deviation of x.

Solution :

First we create the table as per requirements,

x      P(x)         xP(x)           x²            x²P(x)

0    0.25           0               0                0

1     0.20        0.20             1              0.20

2    0.15          0.3               4             0.6

3    0.30         0.9               9             2.7

4    0.10          0.4               16             1.6

   ∑P(x)=1     ∑xP(x)=1.8               ∑x²P(x)=5.1

a) To determine that table shows a probability distribution we add up all five probabilities if the sum is 1 then it is a valid distribution.

$\sum P(X)=0.25+0.20+0.15+0.30+0.10$

$\sum P(X)=1$

Yes it is a probability distribution.

b) The expected value of x is defined as

$E(x)=\sum xP(x)=1.8$

c) The variance of x is defined as

$V=\sum x^2P(x)-(\sum xP(x))^2\\V=5.1-(1.8)^2\\V=5.1-3.24\\V=1.86$

d) The standard deviation of x is  defined as

$\sigma=\sqrt{V}$

$\sigma=\sqrt{1.86}$

$\sigma=1.136$