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  • Please Solve R D Sharma class 12 Chapter 31 Mean and Variance of a random Variable Exercise Very Short Answer Type Question 1 Maths Textbook Solution

Please Solve R.D. Sharma class 12 Chapter 31 Mean and Variance of a random Variable  Exercise Very Short Answer Type Question 1 Maths Textbook Solution.

Answers (1)

Answer:--\frac{1}{2}\leq a\leq \frac{1}{2}

Hint: -You must know the rules of solving probability distributions problems.

Given:

            \begin{array}{|c|c|c|c|c|} \hline X=x_{i} & -2 & -1 & 0 & 1 \\ \hline P\left(X=x_{i}\right) & \frac{1-a}{4} & \frac{1+2 a}{4} & \frac{1-2 a}{4} & \frac{1+a}{4} \\ \hline \end{array}

Solution: - We know that the sum of probabilities in a probability distribution is always 1.

\begin{aligned} &\therefore P(x=-2)+P(x=-1)+P(x=0)+P(x=1)=1 \\\\ &\frac{1-a}{4}+\frac{1+2 a}{4}+\frac{1-2 a}{4}+\frac{1+a}{4}=1 \\\\ &\frac{1-a+1+2 a+1-2 a+1+a}{4}=1 \\\\ &\frac{4}{4}=1 \\\\ &1=1 \end{aligned}

\begin{aligned} &0 \leq \frac{1-a}{4} \leq 1 \\ &0 \leq 1-a \leq 4 \\ &-1 \leq-a \leq 3 \\ &+1 \geq a \geq-3 \\ &-3 \leq a \leq 1--(1) \end{aligned}

\begin{aligned} &0 \leq \frac{1+a}{4} \leq 1 \\ &0 \leq 1+a \leq 4 \\ &-1 \leq a \leq 3--(2) \end{aligned}

\begin{aligned} &0 \leq \frac{1-2 a}{4} \leq 1 \\ &0 \leq 1-2 a \leq 4 \\ &-1 \leq-2 a \leq 3 \\ &-\frac{3}{2} \leq a \leq \frac{1}{2}--(3) \end{aligned}

\begin{aligned} &0 \leq \frac{1+2 a}{4} \leq 1 \\ &0 \leq 1+2 a \leq 4 \\ &-1 \leq 2 a \leq 3 \\ &-\frac{1}{2} \leq a \leq \frac{3}{2}--(4) \end{aligned}

From (1), (2), (3) and (4)

-\frac{1}{2}\leq a\leq \frac{1}{2}

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