Respuesta :
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.
[tex]\sum P(X)=0.25+0.20+0.15+0.30+0.10[/tex]
[tex]\sum P(X)=1[/tex]
Yes it is a probability distribution.
b) The expected value of x is defined as
[tex]E(x)=\sum xP(x)=1.8[/tex]
c) The variance of x is defined as
[tex]V=\sum x^2P(x)-(\sum xP(x))^2\\V=5.1-(1.8)^2\\V=5.1-3.24\\V=1.86[/tex]
d) The standard deviation of x is defined as
[tex]\sigma=\sqrt{V}[/tex]
[tex]\sigma=\sqrt{1.86}[/tex]
[tex]\sigma=1.136[/tex]
