Question:
For the probability distribution given by following
the value of $\text{Var}(X) =$
For the probability distribution given by following
the value of $\text{Var}(X) =$
the value of $\text{Var}(X) =$
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Always double check your value of $E(X)$. Since the distribution is heavily weighted around the values 7 and 8 (each having a probability of $0.3$), the mean must fall very close to $7.5$. Finding $E(X) = 7.3$ tells you that your preliminary arithmetic is well on track!
Updated On: Jun 12, 2026
- 2.65
- 2.85
- 1.65
- 3.85
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
We are given a discrete random variable $X$ with its values ranging from 5 to 10 and their corresponding probabilities. One probability value is an unknown constant $k$. We need to find the variance of $X$, denoted as $\text{Var}(X)$.
Step 2: Key Formula or Approach:
1. The sum of all individual probabilities in a valid probability distribution must equal 1: $$\sum P(X = x_i) = 1$$ 2. The formula for the Mean (Expectation) of $X$ is: $$\mu = E(X) = \sum x_i p_i$$ 3. The formula for the Variance of $X$ is: $$\text{Var}(X) = E(X^2) - [E(X)]^2 = \sum x_i^2 p_i - \left(\sum x_i p_i\right)^2$$
Step 3: Detailed Explanation:
The given distribution table provides: $$\begin{array}{c|c|c|c|c|c|c|c} x_i & 5 & 6 & 7 & 8 & 9 & 10 & 11 \ \\ \hline p_i & 0.07 & 0.2 & 0.3 & k & 0.07 & 0.04 & 0.02 \end{array}$$ 4. Find the value of $k$: $$0.07 + 0.2 + 0.3 + k + 0.07 + 0.04 + 0.02 = 1$$ $$0.7 + k = 1 \implies k = 0.3$$ 5. Construct a calculation table to compute $\sum p_i x_i$ and $\sum p_i x_i^2$: $$\begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2 \ \\ \hline 5 & 0.07 & 0.35 & 1.75 \ \\ 6 & 0.20 & 1.20 & 7.20 \ \\ 7 & 0.30 & 2.10 & 14.70 \ \\ 8 & 0.30 & 2.40 & 19.20 \ \\ 9 & 0.07 & 0.63 & 5.67 \ \\ 10 & 0.04 & 0.40 & 4.00 \ \\ 11 & 0.02 & 0.22 & 2.42 \ \\ \hline \textbf{Total} & \mathbf{1.00} & \mathbf{7.30} & \mathbf{54.94} \end{array}$$ 6. From the tabulated outputs, we have: $$E(X) = \sum p_i x_i = 7.3$$ $$E(X^2) = \sum p_i x_i^2 = 54.94$$ 7. Calculate the variance: $$\text{Var}(X) = E(X^2) - [E(X)]^2$$ $$\text{Var}(X) = 54.94 - (7.3)^2$$ $$\text{Var}(X) = 54.94 - 53.29 = 1.65$$
Step 4: Final Answer:
The variance of the distribution is 1.65, which matches option (C).
We are given a discrete random variable $X$ with its values ranging from 5 to 10 and their corresponding probabilities. One probability value is an unknown constant $k$. We need to find the variance of $X$, denoted as $\text{Var}(X)$.
Step 2: Key Formula or Approach:
1. The sum of all individual probabilities in a valid probability distribution must equal 1: $$\sum P(X = x_i) = 1$$ 2. The formula for the Mean (Expectation) of $X$ is: $$\mu = E(X) = \sum x_i p_i$$ 3. The formula for the Variance of $X$ is: $$\text{Var}(X) = E(X^2) - [E(X)]^2 = \sum x_i^2 p_i - \left(\sum x_i p_i\right)^2$$
Step 3: Detailed Explanation:
The given distribution table provides: $$\begin{array}{c|c|c|c|c|c|c|c} x_i & 5 & 6 & 7 & 8 & 9 & 10 & 11 \ \\ \hline p_i & 0.07 & 0.2 & 0.3 & k & 0.07 & 0.04 & 0.02 \end{array}$$ 4. Find the value of $k$: $$0.07 + 0.2 + 0.3 + k + 0.07 + 0.04 + 0.02 = 1$$ $$0.7 + k = 1 \implies k = 0.3$$ 5. Construct a calculation table to compute $\sum p_i x_i$ and $\sum p_i x_i^2$: $$\begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2 \ \\ \hline 5 & 0.07 & 0.35 & 1.75 \ \\ 6 & 0.20 & 1.20 & 7.20 \ \\ 7 & 0.30 & 2.10 & 14.70 \ \\ 8 & 0.30 & 2.40 & 19.20 \ \\ 9 & 0.07 & 0.63 & 5.67 \ \\ 10 & 0.04 & 0.40 & 4.00 \ \\ 11 & 0.02 & 0.22 & 2.42 \ \\ \hline \textbf{Total} & \mathbf{1.00} & \mathbf{7.30} & \mathbf{54.94} \end{array}$$ 6. From the tabulated outputs, we have: $$E(X) = \sum p_i x_i = 7.3$$ $$E(X^2) = \sum p_i x_i^2 = 54.94$$ 7. Calculate the variance: $$\text{Var}(X) = E(X^2) - [E(X)]^2$$ $$\text{Var}(X) = 54.94 - (7.3)^2$$ $$\text{Var}(X) = 54.94 - 53.29 = 1.65$$
Step 4: Final Answer:
The variance of the distribution is 1.65, which matches option (C).
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