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The probability distribution of a random variable X is given as:

P(X=x)={kx2forx=1,2,32kxforx=4,5,60otherwiseP(X=x) = \begin{cases} kx^2 & for & x=1,2,3 \\ 2kx & for & x=4,5,6 \\ 0 & otherwise \end{cases}

Where kk is an arbitrary constant, then E(X)E(X) is equal to:

Solution

✅ Correct Option: 4

Probabilities sum to 11: k(1+4+9)+2k(4+5+6)=14k+30k=44k=1k(1+4+9)+2k(4+5+6)=14k+30k=44k=1, so k=144k=\frac{1}{44}. Then E(X)=∑xP(x)=k(1+8+27)+2k(16+25+36)=36k+154k=190k=19044=9522E(X)=\sum xP(x)=k(1+8+27)+2k(16+25+36)=36k+154k=190k=\frac{190}{44}=\frac{95}{22}.

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