Solution to i
∑p(x)=1
Thus;
0.1+k+0.2+2k+0.3+k=10.6+4k=1∴4k=0.4k=0.1 Mean, xˉ
xˉ=∑xp(x)
=−2(0.1)−1(0.1)+0(0.2)+1(0.2)+2(0.3)+3(0.1)=0.8 variance
var=∑x2p(x)−xˉ2
∑x2p(x) =4(0.1)+1(0.1)+0(0.2)+1(0.2)+4(0.3)+9(0.1)
=2.8
var=2.8−(0.82)=2.16 Solution to ii
Cumulative Distribution Function
Cumulative Distribution Graph
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