1. **State the problem:** We have a two-way table with orders placed (Online or In Store) and customer satisfaction (Satisfied or Not Satisfied). The totals for each order type are 50, and the grand total is 100. 2. **Given:** - Total orders online = 50 - Total orders in store = 50 - Total customers = 100 - Probability(order online or satisfied) > 0.7 - More customers satisfied with online orders than in-store orders 3. **Define variables:** Let $x$ = number of satisfied customers who ordered online. Let $y$ = number of satisfied customers who ordered in store. 4. **Fill in the table with variables:** | | Satisfied | Not Satisfied | Total | |----------------|-----------|---------------|-------| | Order Online | $x$ | $50 - x$ | 50 | | Order in Store | $y$ | $50 - y$ | 50 | | Total | $x + y$ | $100 - (x + y)$ | 100 | 5. **Use the probability condition:** Probability(order online or satisfied) = $$P(\text{online}) + P(\text{satisfied}) - P(\text{online and satisfied})$$ $$= \frac{50}{100} + \frac{x + y}{100} - \frac{x}{100} = 0.5 + \frac{y}{100}$$ We want this to be greater than 0.7: $$0.5 + \frac{y}{100} > 0.7$$ $$\frac{y}{100} > 0.2$$ $$y > 20$$ 6. **Use the satisfaction comparison:** More customers satisfied with online orders than in store: $$x > y$$ 7. **Find integer values satisfying both conditions:** Since $y > 20$ and $x > y$, possible values are: - $y = 21$, $x = 22$ (minimum satisfying values) 8. **Complete the table:** | | Satisfied | Not Satisfied | Total | |----------------|-----------|---------------|-------| | Order Online | 22 | 28 | 50 | | Order in Store | 21 | 29 | 50 | | Total | 43 | 57 | 100 | 9. **Verify the probability:** $$P(\text{online or satisfied}) = 0.5 + \frac{21}{100} = 0.71 > 0.7$$ **Final answer:** The completed table is: | | Satisfied | Not Satisfied | Total | |----------------|-----------|---------------|-------| | Order Online | 22 | 28 | 50 | | Order in Store | 21 | 29 | 50 | | Total | 43 | 57 | 100 |