Answer for all please

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  for all values of 'i'

ii) ∑P(x) = 1

Given data  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

17.08

correct me if i am wrong.

well the answer to that is xoxoxo

step-by-step explanation: