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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

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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 respectively A. - 7 4 and 1 2 B. - 1 16 and 5 16 C. - 7 4 and 5 16 D. - 1 16 and 5 4 Select the correct answer from above options

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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≤ 1+3p 4 ≤1,0≤ 1-p 4 ≤1,0≤ 1+2p 4 ≤1and0≤ 1-4p 4 ≤1⇒- 1 3 ≤p≤ 1 4 0 ≤ 1 + 3 p 4 ≤ 1 , 0 ≤ 1 − p 4 ≤ 1 , 0 ≤ 1 + 2 p 4 ≤ 1 and 0 ≤ 1 − 4 p 4 ≤ 1 ⇒ − 1 3 ≤ p ≤ 1 4 Let ˉ X be the mean of X. Then, ˉ X =-1× 1+3p 4 +0× 1p 4 +1× 1+2p 4 +2× 1-4p 4 ⇒ ˉ X = 2-9p 4 Now, - 1 3 ≤p≤ 1 4 ⇒3≤-9p≤- 9 4 ⇒- 1 4 ≤2-9p≤5⇒- 1 16 ≤ 2-9p 4 ≤5⇒- 1 16 ≤X≤ 5 4 .

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