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

Here \(, \frac{1+3 p}{4}, \frac{1-p}{4}, \frac{1+2 p}{4}\) and \(\frac{1-4 p}{4}\) 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 .
\(\text { i.e., } 0 \leq \frac{1+3 p}{4} \leq 1,0 \leq \frac{1-p}{4} \leq 1 \)
\( 0 \leq \frac{1+2 p}{4} \leq 1 \text { and } 0 \leq \frac{1-4 p}{4} \leq 1 \)
\( \Rightarrow-\frac{1}{3} \leq p \leq \frac{1}{4}\)
\(\operatorname{Mean}(X)=-1 \times \frac{1+3 p}{4}+0 \times \frac{1-p}{4} +1 \times \frac{1+2 p}{4} \) \( +2 \times \frac{1-4 p}{4}\)
\(=\frac{2-9 p}{4}\)
Now, \(-\frac{1}{3} \leq \mathrm{p} \leq \frac{1}{4}\)
\(\Rightarrow 3 \geq-9 p \geq-\frac{9}{4} \)
\( \Rightarrow-\frac{1}{4} \leq 2-9 p \leq 5 \)
\( \Rightarrow-\frac{1}{16} \leq \frac{2-9 p}{4} \leq \frac{5}{4}\)