A random variable X takes values -1, 0, 1, 2 with probabilities $\frac{1+3p}{4},\frac{1-p}{4},\frac{1+2p}{4},\frac{1-4p}{4}$ respectively, where p varies over R. Then the minimum and maximum values of the mean of X are respectively

$-\frac{1}{16}$ and $\frac{5}{4}$

Since $\frac{1+3p}{4},\frac{1-p}{4},\frac{1+2p}{4}$ and $\frac{1-4p}{4}$ are probabilities when X takes values -1, 0, 1 and 2 respectively.

Therefore, each is greater than or equal to 0 and less than or equal to 1.

$i.e. 0≤\frac{1+3p}{4}≤1, 0 ≤\frac{1-p}{4}≤1,0≤\frac{1+2p}{4}≤1$ and $0 ≤ \frac{1-4p}{4}≤1$

$⇒ -\frac{1}{3}≤p≤\frac{1}{4}$

Let $\overline{X}$ be the mean of X. Then,

$\overline{X}= -1×\frac{1+3p}{4}+0×\frac{1-p}{4}+1 ×\frac{1+2p}{4}+2×\frac{1-4p}{4}$

$⇒\overline{X} =\frac{2-9p}{4}$

Now, $-\frac{1}{3}≤p≤\frac{1}{4}$

$⇒ 3 ≤ -9p ≥ -\frac{9}{4}$

$⇒ -\frac{1}{4}≤2-9p ≤ 5 ⇒ -\frac{1}{16}≤\frac{2-9p}{4}≤5 ⇒-\frac{1}{16}≤X≤\frac{5}{4}$