1. **State the problem:** We are comparing the stability of weekly sales for Store A and Store B using their sales data: Store A: 10000, 12000, 11000, 13000, 12000; Store B: 5000, 15000, 8000, 20000, 9000. 2. **Calculate the mean sales for each store:** For Store A: $$\bar{x}_A = \frac{10000 + 12000 + 11000 + 13000 + 12000}{5} = \frac{58000}{5} = 11600$$ For Store B: $$\bar{x}_B = \frac{5000 + 15000 + 8000 + 20000 + 9000}{5} = \frac{57000}{5} = 11400$$ 3. **Calculate the standard deviation (SD) for each store to measure variability:** For Store A: Calculate squared differences from mean: $(10000 - 11600)^2 = 160000$ $(12000 - 11600)^2 = 160000$ $(11000 - 11600)^2 = 360000$ $(13000 - 11600)^2 = 1960000$ $(12000 - 11600)^2 = 160000$ Sum: $160000 + 160000 + 360000 + 1960000 + 160000 = 2800000$ Variance: $$s_A^2 = \frac{2800000}{5 - 1} = \frac{2800000}{4} = 700000$$ Standard deviation: $$s_A = \sqrt{700000} \approx 836.66$$ For Store B: Squared differences: $(5000 - 11400)^2 = 40960000$ $(15000 - 11400)^2 = 12960000$ $(8000 - 11400)^2 = 11560000$ $(20000 - 11400)^2 = 73960000$ $(9000 - 11400)^2 = 5760000$ Sum: $40960000 + 12960000 + 11560000 + 73960000 + 5760000 = 145600000$ Variance: $$s_B^2 = \frac{145600000}{4} = 36400000$$ Standard deviation: $$s_B = \sqrt{36400000} \approx 6033.15$$ 4. **Answer the questions:** - **Which store is more stable?** Store A has a smaller standard deviation ($836.66$) indicating more stable weekly sales. - **Which is riskier to invest in?** Store B has a larger standard deviation ($6033.15$), so it is riskier. 5. **How can the owner stabilize sales?** - Diversify product offerings and promotions to smooth out demand. - Analyze causes of sales fluctuations (seasonality, marketing) and adjust operations. - Implement inventory and staffing adjustments to better match demand. **Final conclusion:** Store A is more stable due to lower variability, while Store B is riskier because of higher fluctuations in sales.