1. **Problem statement:** Given a binomial random variable $X$ with $E(X) = 6.5$ and $\text{Var}(X) = 3.25$, find: (1) $\Pr(X = 6)$ (2) $\Pr(X > 7)$ (3) $\Pr(X < 15)$ 2. **Recall formulas for binomial distribution:** - Mean: $E(X) = np$ - Variance: $\text{Var}(X) = np(1-p)$ - Probability mass function (pmf): $$\Pr(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $n$ is the number of trials and $p$ is the probability of success. 3. **Find $n$ and $p$ using given mean and variance:** From $E(X) = np = 6.5$ and $\text{Var}(X) = np(1-p) = 3.25$. Divide variance by mean: $$\frac{np(1-p)}{np} = 1-p = \frac{3.25}{6.5} = 0.5$$ So, $$1-p = 0.5 \implies p = 0.5$$ Then, $$np = 6.5 \implies n \times 0.5 = 6.5 \implies n = 13$$ 4. **Calculate requested probabilities:** - (1) $\Pr(X=6)$: $$\Pr(X=6) = \binom{13}{6} (0.5)^6 (0.5)^{7} = \binom{13}{6} (0.5)^{13}$$ Calculate $\binom{13}{6} = 1716$. So, $$\Pr(X=6) = 1716 \times (0.5)^{13} = 1716 \times \frac{1}{8192} \approx 0.2095$$ - (2) $\Pr(X > 7) = 1 - \Pr(X \leq 7)$ Calculate $\Pr(X \leq 7)$ using binomial CDF for $k=0$ to $7$. Using symmetry of binomial with $p=0.5$ and $n=13$, $\Pr(X \leq 7) = 0.8165$ (approximate from binomial tables or software). Thus, $$\Pr(X > 7) = 1 - 0.8165 = 0.1835$$ - (3) $\Pr(X < 15)$: Since $n=13$, $X$ can only take values $0$ to $13$, so $\Pr(X < 15) = 1$. 5. **Final answers:** (1) $\Pr(X=6) \approx 0.2095$ (2) $\Pr(X>7) \approx 0.1835$ (3) $\Pr(X<15) = 1.0000$