InterviewSolution
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InterviewSolution
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A random variable X takes values `-1,0,1,2` with probabilities `(1+3p)/4,(1-p)/4,(1+2p)/4,(-14p)/4` respectively, where `p` varies over `R.` Then the minimum and maximum values of the mean of `X` are respectivelyA. `-(7)/(4)and (1)/(2)`B. `-(1)/(16)and (5)/(16)`C. `-(7)/(4) and (5)/(16)`D. `-(1)/(16)and (5)/(4)` |
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Answer» Correct Answer - D Since `(1+3p)/(4),(1-p)/(4),(1+2p)/(4)and(1-4p)/(4)` are probabilites when X takes values `-1,0,1` and 2 respectively. Therefore, each is greater tahn or equal to 0 and less than or equal to 1. i.e., `0 le (1+3p)/(4) le 1,0le (1-p)/(4)le 1,0 le (1+2p)/(4)le 1 and 0 le (1-4p)/(4)le 1 rArr -(1)/(3)le p le (1)/(4)` Let `overlineX` be the mean of X. Then, `overline X=-1xx(1+3p)/(4)+0xx(1p)/(4)+1xx(1+2p)/(4)+2xx(1-4p)/(4)` `rArr overline(X) =(2-9p)/(4)` Now, `-(1)/(3)le p le (1)/(4)` `rArr 3 le -9p le -(9)/(4)` `rArr-(1)/(4) le 2-9p le 5rArr -(1)/(16)le (2-9p)/(4)le 5rArr-(1)/(16) le X le (5)/(4)`. |
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