1. **Problem Statement:** We have a frequency distribution of lifetimes of 500 light bulbs. We need to calculate the mean, standard deviation, construct a percentage ogive to find median, quartile deviation, percentage in a range, and confirm these values by computation. Also, analyze risks and probabilities related to guarantees. 2. **Calculate the Mean:** The midpoint $x_i$ of each class interval is calculated as the average of the lower and upper bounds. | Class Interval | Midpoint $x_i$ | Frequency $f_i$ | |---------------|----------------|----------------| | 400-499 | 450 | 10 | | 500-599 | 550 | 16 | | 600-699 | 650 | 38 | | 700-799 | 750 | 56 | | 800-899 | 850 | 63 | | 900-999 | 950 | 67 | | 1000-1099 | 1050 | 92 | | 1100-1199 | 1150 | 68 | | 1200-1299 | 1250 | 57 | | 1300-1399 | 1350 | 33 | Mean formula: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ Calculate $\sum f_i x_i$: $$10\times450 + 16\times550 + 38\times650 + 56\times750 + 63\times850 + 67\times950 + 92\times1050 + 68\times1150 + 57\times1250 + 33\times1350$$ $$= 4500 + 8800 + 24700 + 42000 + 53550 + 63650 + 96600 + 78200 + 71250 + 44550 = 487800$$ Total frequency $\sum f_i = 500$ Mean: $$\bar{x} = \frac{487800}{500} = 975.6$$ 3. **Calculate the Standard Deviation:** Use formula: $$\sigma = \sqrt{\frac{\sum f_i x_i^2}{\sum f_i} - \bar{x}^2}$$ Calculate $\sum f_i x_i^2$: $$10\times450^2 + 16\times550^2 + 38\times650^2 + 56\times750^2 + 63\times850^2 + 67\times950^2 + 92\times1050^2 + 68\times1150^2 + 57\times1250^2 + 33\times1350^2$$ $$= 10\times202500 + 16\times302500 + 38\times422500 + 56\times562500 + 63\times722500 + 67\times902500 + 92\times1102500 + 68\times1322500 + 57\times1562500 + 33\times1822500$$ $$= 2025000 + 4840000 + 16055000 + 31500000 + 45517500 + 60467500 + 101430000 + 89930000 + 89062500 + 60142500 = 484500000$$ Calculate variance: $$\frac{484500000}{500} - (975.6)^2 = 969000 - 951792.36 = 17207.64$$ Standard deviation: $$\sigma = \sqrt{17207.64} \approx 131.2$$ 4. **Construct Percentage Ogive and Find Median, Quartile Deviation, Percentage in Range:** Cumulative frequencies: | Upper Class Boundary | Cumulative Frequency | |---------------------|----------------------| | 500 | 10 | | 600 | 26 | | 700 | 64 | | 800 | 120 | | 900 | 183 | | 1000 | 250 | | 1100 | 342 | | 1200 | 410 | | 1300 | 467 | | 1400 | 500 | - Median is the value at cumulative frequency $\frac{500}{2} = 250$. Using linear interpolation in 900-1000 class: $$L = 900, F = 183, f = 67, n = 250$$ $$\text{Median} = L + \frac{n - F}{f} \times \text{class width} = 900 + \frac{250 - 183}{67} \times 100 = 900 + 0.985 \times 100 = 998.5$$ - Quartiles: $Q_1$ at $\frac{500}{4} = 125$ lies in 800-900 class: $$L=800, F=120, f=63$$ $$Q_1 = 800 + \frac{125 - 120}{63} \times 100 = 800 + 7.94 = 807.94$$ $Q_3$ at $\frac{3\times500}{4} = 375$ lies in 1100-1200 class: $$L=1100, F=342, f=68$$ $$Q_3 = 1100 + \frac{375 - 342}{68} \times 100 = 1100 + 48.53 = 1148.53$$ Quartile deviation: $$QD = \frac{Q_3 - Q_1}{2} = \frac{1148.53 - 807.94}{2} = 170.3$$ - Percentage of bulbs with lifetime between 700 and 1200 hours: Cumulative frequency at 1200 = 410, at 700 = 64 Number in range = 410 - 64 = 346 Percentage: $$\frac{346}{500} \times 100 = 69.2\%$$ 5. **Confirm Values by Computation:** Already computed median, quartiles, interquartile range, quartile deviation, and percentage in range using formulas above. 6. **Risk if Guarantee Replacement for Bulbs Lasting Less Than 1000 Hours:** Number of bulbs lasting less than 1000 hours = cumulative frequency at 1000 = 250 Risk = fraction of bulbs replaced = $\frac{250}{500} = 0.5$ or 50% 7. **Probability of Refunds with 100-day Guarantee (7 hours/day):** Total hours in 100 days: $$100 \times 7 = 700$$ Number of bulbs lasting less than 700 hours = cumulative frequency at 700 = 64 Probability of refund: $$\frac{64}{500} = 0.128 = 12.8\%$$ **Final answers:** - Mean = 975.6 hours - Standard deviation = 131.2 hours - Median = 998.5 hours - Quartile deviation = 170.3 hours - Percentage between 700 and 1200 hours = 69.2% - Risk of replacement under 1000 hours = 50% - Probability of refund under 100-day guarantee = 12.8%