A random variable X takes values $-1,0,1,2$ with probabilities $\frac{1+3 p}{4}, \frac{1-p}{4}, \frac{1+2…
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.
$-\frac{7}{4}$ and $\frac{1}{2}$
$-\frac{1}{16}$ and $\frac{5}{16}$
$-\frac{7}{4}$ and $\frac{5}{16}$
$-\frac{1}{16}$ and $\frac{5}{4}$
Solution
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 .
$\begin{aligned}
& \text { i.e., } 0 \leq \frac{1+3 p}{4} \leq 1,0 \leq \frac{1-p}{4} \leq 1 \\
& 0 \leq \frac{1+2 p}{4} \leq 1 \text { and } 0 \leq \frac{1-4 p}{4} \leq 1 \\
& \Rightarrow-\frac{1}{3} \leq p \leq \frac{1}{4}
\end{aligned}$
$\begin{aligned} \operatorname{Mean}(X)=-1 \times \frac{1+3 p}{4}+0 \times \frac{1-p}{4} & +1 \times \frac{1+2 p}{4} \\ & +2 \times \frac{1-4 p}{4}\end{aligned}$
$=\frac{2-9 p}{4}$ Now, $-\frac{1}{3} \leq \mathrm{p} \leq \frac{1}{4}$
$\begin{aligned}
& \Rightarrow 3 \geq-9 p \geq-\frac{9}{4} \\
& \Rightarrow-\frac{1}{4} \leq 2-9 p \leq 5 \\
& \Rightarrow-\frac{1}{16} \leq \frac{2-9 p}{4} \leq \frac{5}{4}
\end{aligned}$