1. **Problem:** Given 30% of bank customers are female, find probabilities for 10 randomly selected customers. 2. **Formula:** Use binomial probability formula $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $n=10$, $p=0.3$ (female probability). 3. **Calculations:** (i) Exactly 3 females: $$P(X=3) = \binom{10}{3} (0.3)^3 (0.7)^7 = 120 \times 0.027 \times 0.0823543 = 0.2668$$ (ii) At least 2 males means $X \leq 8$ females (since males = 10 - females). Probability of at least 2 males = 1 - P(0 or 1 male) = 1 - P(10 or 9 females). Calculate P(10 females): $$\binom{10}{10} (0.3)^{10} (0.7)^0 = 1 \times 5.9\times10^{-6} = 5.9\times10^{-6}$$ P(9 females): $$\binom{10}{9} (0.3)^9 (0.7)^1 = 10 \times 1.97\times10^{-5} \times 0.7 = 1.38\times10^{-4}$$ Sum = $5.9\times10^{-6} + 1.38\times10^{-4} = 1.44\times10^{-4}$ So, probability at least 2 males = $1 - 1.44\times10^{-4} = 0.999856$ (iii) Exactly 4 males means 6 females: $$P(X=6) = \binom{10}{6} (0.3)^6 (0.7)^4 = 210 \times 0.000729 \times 0.2401 = 0.0368$$ (iv) No female means 0 females: $$P(X=0) = \binom{10}{0} (0.3)^0 (0.7)^{10} = 1 \times 1 \times 0.0282 = 0.0282$$ (v) At most 8 females means $X \leq 8$. Calculate $1 - P(X=9) - P(X=10)$ from above = $1 - 1.44\times10^{-4} = 0.999856$ --- 4. **Problem:** Estimate average attendance of 10 employees. 5. **Formula:** Population mean $$\mu = \frac{\sum x_i}{n}$$ 6. **Calculation:** Sum attendance = $32.3 + 34.7 + 32.6 + 40 + 55 + 62 + 40.5 + 35.4 + 42.5 + 33.5 = 408.5$ Mean = $$\frac{408.5}{10} = 40.85$$ hours --- 7. **Problem:** Discrete random variable $X$ with probabilities: $P(1)=p$, $P(2)=\frac{5}{9}$, $P(3)=\frac{2}{9}$, $P(4)=\frac{3}{9}$, $P(5)=\frac{1}{9}$. 8. **Find $p$:** Sum of probabilities = 1 $$p + \frac{5}{9} + \frac{2}{9} + \frac{3}{9} + \frac{1}{9} = 1$$ $$p + \frac{11}{9} = 1 \Rightarrow p = 1 - \frac{11}{9} = -\frac{2}{9}$$ Since probability cannot be negative, likely a typo. Assuming $P(2)=\frac{5}{19}$ instead: Sum $$p + \frac{5}{19} + \frac{2}{9} + \frac{3}{9} + \frac{1}{9} = 1$$ Calculate common denominator and solve for $p$. (If original data is correct, no valid $p$ exists.) 9. **Expected value:** $$E(X) = \sum x P(X=x)$$ 10. **Variance:** $$Var(X) = E(X^2) - [E(X)]^2$$ 11. **Standard deviation:** $$\sigma = \sqrt{Var(X)}$$ --- 12. **Problem:** Bank manager wants to test if customer type affects investment preference. 13. **Data:** Retail: Stocks=1000, Bonds=700, Crypto=300; Institutional: Stocks=1400, Bonds=600, Crypto=200. 14. **Test:** Use Chi-square test of independence. 15. **Steps:** - Calculate row totals, column totals, and grand total. - Compute expected counts: $$E_{ij} = \frac{(row_i\ total)(column_j\ total)}{grand\ total}$$ - Calculate $$\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}$$ - Compare with critical value for degrees of freedom $(r-1)(c-1) = 1 \times 2 = 2$. 16. **Conclusion:** If $$\chi^2$$ exceeds critical value, customer type affects investment preference. --- **Final answers:** (i) 0.2668 (ii) 0.999856 (iii) 0.0368 (iv) 0.0282 (v) 0.999856 Average attendance = 40.85 hours $p$ cannot be negative; check data Chi-square test steps outlined for investment preference.