Answer:
a) 'X' be the random variable in discrete distribution
b) The value of y = 0.2
c) The mean value or Expectation value
E (X) = 1.25
d)
The variance σ² of the discrete distribution function is
Variance ( V(x)= 1.2875
Step-by-step explanation:
Step(i):-
Given data
x : 0 1 2 3 4
P(x) : 0.30 0.35 y 0.10 0.05
a)
Let 'X' be the random variable in discrete distribution
Given data is discrete distribution
i) If the numbers [tex]p(x_{i} )\geq 0[/tex] for all values of 'i'
ii) ∑P(x) = 1
Given data [tex]p(x_{i} )\geq 0[/tex] for all values of 'i'
∑P(x) = 1
0.30+0.35+y+0.10+0.05 =1
y + 0.8 = 1
y = 1 -0.8 = 0.2
b) The value of y = 0.2
Step(ii):-
Expectation:
Given data
x : 0 1 2 3 4
P(x) : 0.30 0.35 0.2 0.10 0.05
The mean value or Expectation value
E (x) = ∑ x p ( X = x)
= 0 × 0.30 + 1 × 0.35 + 2 × 0.2 + 3 × 0.10 + 4 × 0.05
= 0 + 0.35 + 0.4 + 0.30 + 0.2
= 1.25
Variance of X
The variance σ² of the discrete distribution function is defined by
σ² = ∑ x² p(x=x) - μ²
= 0× 0.30 + 1² × 0.35 + 2²× 0.2 +3²× 0.10 + 4²× 0.05 - (1.25)²
= 0 + 0.35 + 0.8 + 0.9 + 0.8 - 1.5625
= 1.2875
conclusion:-
The mean value or Expectation value = 1.25
The variance σ² of the discrete distribution function
V(X) = 1.2875
