Find k, such that the function P(x)=k(4/x) ;x=0,1,2,3,4 k>0

=0 ,otherwise

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#### Solution

P(x)=k(4/x) ;x=0,1,2,3,4 k>0

=0 ,otherwise

`P(X = 0) = k^4C_0 = k(1) = k`

`P(X = 1) = k^4C_1 = k(4) = 4k`

`P(X = 2) = k^4C_2 = k(6) = 6k `

`P(X = 3) = k^4C_3 = k(4) = 4k`

`P(X = 4) = k^4C_4 = k(1) = k`

X | 0 | 1 | 2 | 3 | 4 |

P(X = x): | k | 4k | 6k | 4k | k |

Since, the function represents a p.m.f.

`sum_(x=0)^4P(X=x)=1`

k + 4k + 6k + 4k + k = 1

16k = 1

k = 1/16

Concept: Standard Deviation of Binomial Distribution (P.M.F.)

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