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Jet001 [13]
3 years ago
5

Answer for all please

Mathematics
1 answer:
malfutka [58]3 years ago
3 0

Answer:

a) 'X' be the random variable in discrete distribution

b) <em> The value of y = 0.2</em>

<em>c) The mean value or  Expectation value</em>

<em>E (X) = 1.25</em>

<em>d) </em>

<em>The variance σ² of the discrete distribution function is</em>

<em>Variance ( V(x)= 1.2875</em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

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

<em>i) If the numbers </em>p(x_{i} )\geq 0  for all values of 'i'

ii) ∑P(x) = 1

Given data <em> </em>p(x_{i} )\geq 0  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

<em>b) The value of y = 0.2</em>

<u><em>Step(ii):-</em></u>

<u><em>Expectation:</em></u>

Given data

 x    :      0              1          2        3        4

P(x)  :     0.30      0.35      0.2      0.10   0.05

<em>The mean value or  Expectation value</em>

<em>E (x) = ∑ x p ( X = x)</em>

<em>         =  0 × 0.30 + 1 × 0.35 + 2 × 0.2 + 3 × 0.10 + 4 × 0.05</em>

<em>        =     0 + 0.35 + 0.4 + 0.30 + 0.2</em>

<em>       =       1.25</em>

<u><em>Variance of X</em></u>

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

<u><em>conclusion:-</em></u>

<em>The mean value or  Expectation value = 1.25</em>

<em>The variance σ² of the discrete distribution function</em>

<em>  V(X) = 1.2875</em>

   

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