1. **Stating the problem:** We are given probability distributions and asked to find missing probabilities and means (expected values) for two distributions, and then to find the variance for a third distribution. 2. **Important formulas:** - Mean (Expected value) of a discrete random variable $X$ is given by: $$E(X) = \sum X_i P(X_i)$$ - Variance is given by: $$\text{Var}(X) = E\left[(X - \mu)^2\right] = \sum (X_i - \mu)^2 P(X_i)$$ where $\mu = E(X)$. --- ### Problem 1: Find missing probabilities and mean Given partial data: - For $X=5$, $P(X)=0.14$, $X.P(X)=?$ - For $X=7$, $P(X)=0.2$, $X.P(X)=?$ **Step 1:** Calculate missing $X.P(X)$ values: - For $X=5$: $X.P(X) = 5 \times 0.14 = 0.7$ - For $X=7$: $X.P(X) = 7 \times 0.2 = 1.4$ **Step 2:** Calculate mean by summing all $X.P(X)$ values: Given values: - $4 \times 0.25 = 1.0$ - $5 \times 0.14 = 0.7$ - $6 \times 0.31 = 1.86$ - $7 \times 0.2 = 1.4$ - $8 \times 0.1 = 0.8$ Sum: $$1.0 + 0.7 + 1.86 + 1.4 + 0.8 = 5.76$$ So, mean $= 5.76$. --- ### Problem 2: Number of Books Borrowed Given: $$X = \{0,1,2,3,4,5\}$$ $$P(X) = \left\{\frac{1}{10}, \frac{3}{10}, \frac{1}{5}, \frac{1}{10}, \frac{1}{5}, \frac{1}{10}\right\}$$ **Step 1:** Calculate $X.P(X)$ for each: - $0 \times \frac{1}{10} = 0$ - $1 \times \frac{3}{10} = 0.3$ - $2 \times \frac{1}{5} = 0.4$ - $3 \times \frac{1}{10} = 0.3$ - $4 \times \frac{1}{5} = 0.8$ - $5 \times \frac{1}{10} = 0.5$ **Step 2:** Sum to find mean: $$0 + 0.3 + 0.4 + 0.3 + 0.8 + 0.5 = 2.3$$ Mean $= 2.3$. --- ### Problem 3: Variance of Number of Text Messages Received Given partial table: - $X = 15, 18$ - $P(X) = \frac{1}{8}$ each - $X.P(X)$ given as 1.875 and 2.25 respectively **Step 1:** Calculate mean $\mu$ (assuming full data not given, we use partial mean): Sum $X.P(X)$ so far: $$1.875 + 2.25 = 4.125$$ Since only two points given, mean $\mu = 4.125 / (1/8 + 1/8) = 4.125 / 0.25 = 16.5$ **Step 2:** Calculate $X - \mu$: - For $X=15$: $15 - 16.5 = -1.5$ - For $X=18$: $18 - 16.5 = 1.5$ **Step 3:** Calculate $(X - \mu)^2$: - $(-1.5)^2 = 2.25$ - $(1.5)^2 = 2.25$ **Step 4:** Calculate $(X - \mu)^2 P(X)$: - $2.25 \times \frac{1}{8} = 0.28125$ - $2.25 \times \frac{1}{8} = 0.28125$ **Step 5:** Sum for variance: $$0.28125 + 0.28125 = 0.5625$$ Variance $= 0.5625$ (based on partial data). --- **Final answers:** - (a) $X.P(5) = 0.7$ - (b) $X.P(7) = 1.4$ - Mean of first distribution $= 5.76$ - Mean of books borrowed $= 2.3$ - Variance of text messages (partial) $= 0.5625$