1. **Problem statement:** We have three parts involving different probability distributions: binomial, Poisson, and normal. --- ### a) Binomial distribution Given: Probability a student stays in hostel $p=0.25$, sample size $n=15$. **Formula:** $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ **(i) Exactly 8 students stay:** $$P(X=8) = \binom{15}{8} (0.25)^8 (0.75)^7$$ Calculate $\binom{15}{8} = 6435$. $$P(X=8) = 6435 \times (0.25)^8 \times (0.75)^7 \approx 0.0003$$ **(ii) At most 2 students do not stay:** "At most 2 do not stay" means "at least 13 stay" because $15 - 2 = 13$. Calculate: $$P(X \geq 13) = P(X=13) + P(X=14) + P(X=15)$$ Use binomial formula for each term: $$P(X=13) = \binom{15}{13} (0.25)^{13} (0.75)^2$$ $$P(X=14) = \binom{15}{14} (0.25)^{14} (0.75)^1$$ $$P(X=15) = \binom{15}{15} (0.25)^{15} (0.75)^0$$ Calculate each and sum: These probabilities are very small, sum approximately $0.00001$. --- ### a) (ii) Approximation for $n=60$, $p=0.25$ to find $P(X>20)$ Use normal approximation to binomial: Mean: $$\mu = np = 60 \times 0.25 = 15$$ Variance: $$\sigma^2 = np(1-p) = 60 \times 0.25 \times 0.75 = 11.25$$ Standard deviation: $$\sigma = \sqrt{11.25} \approx 3.354$$ Apply continuity correction for $P(X>20)$: $$P(X>20) \approx P(Y > 20.5)$$ Standardize: $$Z = \frac{20.5 - 15}{3.354} \approx 1.64$$ From standard normal table: $$P(Z > 1.64) = 1 - P(Z \leq 1.64) \approx 1 - 0.9495 = 0.0505$$ --- ### b) Poisson distribution Given average outages per 6-hour shift $\lambda=0.8$. **Formula:** $$P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$$ **(i) Exactly 3 outages in 6 hours:** $$P(X=3) = \frac{e^{-0.8} 0.8^3}{3!} = \frac{e^{-0.8} 0.512}{6} \approx 0.057$$ **(ii) Less than 2 outages in 12 hours:** For 12 hours, $\lambda = 0.8 \times 2 = 1.6$. Calculate: $$P(X<2) = P(X=0) + P(X=1)$$ $$P(X=0) = e^{-1.6} \frac{1.6^0}{0!} = e^{-1.6} \approx 0.2019$$ $$P(X=1) = e^{-1.6} \frac{1.6^1}{1!} = 1.6 e^{-1.6} \approx 0.3230$$ Sum: $$P(X<2) = 0.2019 + 0.3230 = 0.5249$$ --- ### c) Normal distribution Given mean $\mu=8$ years, standard deviation $\sigma=1.2$ years. **(i) Probability less than 11 years:** $$Z = \frac{11 - 8}{1.2} = 2.5$$ $$P(X<11) = P(Z<2.5) \approx 0.9938$$ **(ii) Probability more than 6 years:** $$Z = \frac{6 - 8}{1.2} = -1.67$$ $$P(X>6) = 1 - P(Z < -1.67) = 1 - 0.0475 = 0.9525$$ **(iii) Probability between 7 and 8.5 years:** Calculate Z-scores: $$Z_1 = \frac{7 - 8}{1.2} = -0.83$$ $$Z_2 = \frac{8.5 - 8}{1.2} = 0.42$$ $$P(7 < X < 8.5) = P(-0.83 < Z < 0.42) = P(Z<0.42) - P(Z<-0.83)$$ From tables: $$P(Z<0.42) = 0.6628$$ $$P(Z<-0.83) = 0.2033$$ Difference: $$0.6628 - 0.2033 = 0.4595$$ --- **Final answers:** - a(i) $\approx 0.0003$ - a(ii) $\approx 0.00001$ - a(ii) approx $P(X>20) \approx 0.0505$ - b(i) $\approx 0.057$ - b(ii) $\approx 0.5249$ - c(i) $\approx 0.9938$ - c(ii) $\approx 0.9525$ - c(iii) $\approx 0.4595$