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A discrete random variable X has the following probability distribution the value of C and the mean of the distribution are A. `1/(10) and 3.66`B. `1/(20) and 2.66`C. `1/(15) and 1.33`D. None of these |
Answer» Correct Answer - A Since `Segmap=1` we have `C+2C+2C+3C+C^2+2C^2+7C^2+C=1` `rArr" "10C^2+9C-1=0` `rArr" "(10C-1)(C+1)=0` `rArr" "C=1/(10),C=-1` Therefore, the permissible value of `C=1/(10)` Mean `sum_(i=1)^nx_ip_1=sum_(i=1)^7x_ip_i` `=1xx1/(10)+2xx2/(10)+3xx2/(10)+4xx3/(10)+5(1/(10))^2+6xx2(1/(10))^2+7(7(1/(10))^2+1/(10))` `=1/(10)+4/(10)+6/(10)+(12)/(10)+5/(100)+(12)/(100)+(49)/(100)+7/(10)` =3.66 |
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