Respuesta :
Answer:
Step-by-step explanation:
Given, :
x P(x)
0 0.900
1 0.09
2 0.007
3 0.003
The expected variance ; Var(X) = x²*p(x) - E(X)²
The expected mean, E(X) = x*p(x)
E(X) = (0*0.9) + (1*0.09) + (2*0.007) + (2*0.003) = 0.113
Var(X) = [(0^2*0.9) + (1^2*0.09) + (2^2*0.007) + (3^2*0.003)] - 0.113^2
Var(X) = 0.145 - 0.012769
Var(X) = 0.132
Variance is one of the measure of the dispersion in the distribution of a random variable. The variance of the distribution of X is 0.132231
How to find the variance of a random variable?
Supposing that the considered random variable is discrete, we get:
[tex]Mean = E(X) = \sum_{\forall x_i} f(x_i)x_i\\\\Variance = Var(X) = (\sum_{\forall x_i} f(x_i)x^2_i) - (E(X))^2\\[/tex]
As standard deviation is positive root of variance, thus,
[tex]\sigma = \sqrt {Var(X)}[/tex]
where [tex]x_i; \: \: i = 1,2, ... ,n[/tex] is its n data values
and [tex]f(x_i)[/tex] is the probability of [tex]X = x_i[/tex]
For this case, we have the probability distribution of X as:
X P(X=x)
0 0.900
1 0.09
2 0.007
3 0.003
The mean of X is:
[tex]E(X) = 0 \times 0.900 + 1 \times 0.09 + 2 \times 0.007 + 3 \times 0.003\\E(X) = 0.09 + 0.014 + 0.009 = 0.113[/tex]
Thus, we get the variance of X as:
[tex]Var(X) = (\sum_{\forall x_i} f(x_i)x^2_i) - (E(X))^2\\\\Var(X) = (0^2 \times 0.900 + 1^2 \times 0.09 + 2^2 \times 0.007 + 3^2 \times 0.003) - (0.113)^2\\Var(X) = 0.09 + 0.028 + 0.027 - 0.012769 = 0.132231[/tex]
Learn more about variance of a random variable here:
https://brainly.com/question/2496774
