1. **Problem statement:** We have a binomial random variable $X$ representing the number of Canadians (out of 20) who will spend money on Halloween. The probability of success (spending money) is $p=0.63$. 2. **Part (a):** Compute $P(X=14)$. The binomial probability formula is: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $n=20$, $k=14$, $p=0.63$. Calculate: $$P(X=14) = \binom{20}{14} (0.63)^{14} (0.37)^6$$ Using a calculator or software: $$\binom{20}{14} = 38760$$ $$0.63^{14} \approx 0.0067$$ $$0.37^{6} \approx 0.0022$$ So, $$P(X=14) \approx 38760 \times 0.0067 \times 0.0022 = 0.5721$$ 3. **Part (b):** Compute $P(8 \leq X \leq 14)$. This is the sum of probabilities from $k=8$ to $k=14$: $$P(8 \leq X \leq 14) = \sum_{k=8}^{14} \binom{20}{k} (0.63)^k (0.37)^{20-k}$$ Using binomial cumulative distribution function (CDF) or software, we find: $$P(8 \leq X \leq 14) \approx 0.8283$$ 4. **Part (c):** Compute expected value $E(X)$ and standard deviation $SD(X)$. For a binomial distribution: $$E(X) = np = 20 \times 0.63 = 12.60$$ $$SD(X) = \sqrt{np(1-p)} = \sqrt{20 \times 0.63 \times 0.37} = \sqrt{4.662} \approx 2.16$$ 5. **Part (d):** Compute the probability that the 10-th Canadian chosen is the 6-th to say they will spend money. This is a negative binomial probability: $$P(Y=10) = \binom{10-1}{6-1} p^6 (1-p)^{10-6} = \binom{9}{5} (0.63)^6 (0.37)^4$$ Calculate: $$\binom{9}{5} = 126$$ $$0.63^6 \approx 0.0635$$ $$0.37^4 \approx 0.0187$$ So, $$P(Y=10) = 126 \times 0.0635 \times 0.0187 \approx 0.1497$$ **Final answers:** - $P(X=14) = 0.5721$ - $P(8 \leq X \leq 14) = 0.8283$ - $E(X) = 12.60$ - $SD(X) = 2.16$ - $P(\text{10-th is 6-th success}) = 0.1497$