Find the mean deviation about the mean for the data: 4, 7, 8, 9, 10, 12, 13, 17.
Hint. Find the mean first, then average the absolute deviations from it.
Mean = (4+7+8+9+10+12+13+17)/8 = 80/8 = 10. Absolute deviations: 6,3,2,1,0,2,3,7, summing to 24.
✦ Working through each part gives: Mean deviation about the mean = 24/8 = 3.
Find the mean deviation about the mean for the data: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
Hint. Find the mean first, then average the absolute deviations from it.
Mean = (38+70+48+40+42+55+63+46+54+44)/10 = 500/10 = 50. Absolute deviations: 12,20,2,10,8,5,13,4,4,6, summing to 84.
✦ Working through each part gives: Mean deviation about the mean = 84/10 = 8.4.
Find the mean deviation about the median for the data: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.
Hint. Sort the data, find the median (12 values, so average the 6th and 7th), then average the absolute deviations from it.
Sorted: 10,11,11,12,13,13,14,16,16,17,17,18. With 12 (even) values, median = average of 6th and 7th = (13+14)/2 = 13.5. Absolute deviations from 13.5: 3.5,2.5,2.5,1.5,0.5,0.5,0.5,2.5,2.5,3.5,3.5,4.5, summing to 28.
✦ Working through each part gives: Mean deviation about the median = 28/12 = 7/3, approximately 2.33.
Find the mean deviation about the median for the data: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.
Hint. Sort the data, find the median (10 values, so average the 5th and 6th), then average the absolute deviations from it.
Sorted: 36,42,45,46,46,49,51,53,60,72. With 10 (even) values, median = average of 5th and 6th = (46+49)/2 = 47.5. Absolute deviations from 47.5: 11.5,5.5,2.5,1.5,1.5,1.5,3.5,5.5,12.5,24.5, summing to 70.
✦ Working through each part gives: Mean deviation about the median = 70/10 = 7.
Find the mean deviation about the mean for the data: xi = 5,10,15,20,25 with fi = 7,4,6,3,5.
Hint. Compute N, the weighted mean, then the frequency-weighted average of the absolute deviations.
N = 7+4+6+3+5 = 25. Sum(fi.xi) = 35+40+90+60+125 = 350. Mean = 350/25 = 14. Absolute deviations |xi-14|: 9,4,1,6,11. Weighted: 63,16,6,18,55, summing to 158.
✦ Working through each part gives: Mean deviation about the mean = 158/25 = 6.32.
Find the mean deviation about the mean for the data: xi = 10,30,50,70,90 with fi = 4,24,28,16,8.
Hint. Compute N, the weighted mean, then the frequency-weighted average of the absolute deviations.
N = 4+24+28+16+8 = 80. Sum(fi.xi) = 40+720+1400+1120+720 = 4000. Mean = 4000/80 = 50. Absolute deviations |xi-50|: 40,20,0,20,40. Weighted: 160,480,0,320,320, summing to 1280.
✦ Working through each part gives: Mean deviation about the mean = 1280/80 = 16.
Find the mean deviation about the median for the data: xi = 5,7,9,10,12,15 with fi = 8,6,2,2,2,6.
Hint. Find N and the cumulative frequencies to locate the median observation, then compute the frequency-weighted average of the absolute deviations from it.
N = 8+6+2+2+2+6 = 26, so N/2 = 13. Cumulative frequencies: 8,14,16,18,20,26. The value where cumulative frequency first reaches or exceeds 13 is xi=7 (cumulative 14). So median = 7. Absolute deviations |xi-7|: 2,0,2,3,5,8. Weighted: 16,0,4,6,10,48, summing to 84.
✦ Working through each part gives: Mean deviation about the median = 84/26, approximately 3.23.
Find the mean deviation about the median for the data: xi = 15,21,27,30,35 with fi = 3,5,6,7,8.
Hint. Find N and the cumulative frequencies to locate the median observation, then compute the frequency-weighted average of the absolute deviations from it.
N = 3+5+6+7+8 = 29, so N/2 = 14.5. Cumulative frequencies: 3,8,14,21,29. The value where cumulative frequency first reaches or exceeds 14.5 is xi=30 (cumulative 21). So median = 30. Absolute deviations |xi-30|: 15,9,3,0,5. Weighted: 45,45,18,0,40, summing to 148.
✦ Working through each part gives: Mean deviation about the median = 148/29, approximately 5.10.
Find the mean deviation about the mean for the income distribution: classes 0-100,100-200,...,700-800 (Rs./day) with frequencies 4,8,9,10,7,5,4,3.
Hint. Use class midpoints as xi, find N and the weighted mean, then the frequency-weighted average of the absolute deviations.
Midpoints: 50,150,250,350,450,550,650,750. N = 4+8+9+10+7+5+4+3 = 50. Sum(fi.xi) = 200+1200+2250+3500+3150+2750+2600+2250 = 17900. Mean = 17900/50 = 358. Absolute deviations |xi-358|: 308,208,108,8,92,192,292,392. Weighted and summed: 1232+1664+972+80+644+960+1168+1176 = 7896.
✦ Working through each part gives: Mean = Rs. 358, mean deviation about the mean = 7896/50 = Rs. 157.92.
Find the mean deviation about the mean for the height distribution: classes 95-105,105-115,...,145-155 (cm) with frequencies 9,13,26,30,12,10.
Hint. Use class midpoints as xi, find N and the weighted mean, then the frequency-weighted average of the absolute deviations.
Midpoints: 100,110,120,130,140,150. N = 9+13+26+30+12+10 = 100. Sum(fi.xi) = 900+1430+3120+3900+1680+1500 = 12530. Mean = 12530/100 = 125.3. Absolute deviations |xi-125.3|: 25.3,15.3,5.3,4.7,14.7,24.7. Weighted and summed: 227.7+198.9+137.8+141+176.4+247 = 1128.8.
✦ Working through each part gives: Mean = 125.3 cm, mean deviation about the mean = 1128.8/100 = 11.288 cm.
Find the mean deviation about the median for the marks distribution: classes 0-10,10-20,...,50-60 with frequencies 6,8,14,16,4,2.
Hint. Find the median class using cumulative frequency, compute the median by interpolation, then find the frequency-weighted average of the absolute deviations of the midpoints from the median.
N = 6+8+14+16+4+2 = 50, so N/2 = 25. Cumulative frequencies: 6,14,28,44,48,50. The median class (where cumulative first reaches 25) is 20-30, with l=20, C=14, f=14, h=10. Median = 20 + (25-14)/14 = 20+7.857 = 27.857. Using midpoints 5,15,25,35,45,55: absolute deviations from 27.857 are 22.857,12.857,2.857,7.143,17.143,27.143. Weighted and summed: 137.14+102.86+40+114.29+68.57+54.29 = 517.14.
✦ Working through each part gives: Median is approximately 27.86, mean deviation about the median = 517.14/50, approximately 10.34.
Calculate the mean deviation about the median age for the age distribution of 100 persons: classes 16-20,21-25,...,51-55 with frequencies 5,6,12,14,26,12,16,9.
Hint. First convert to continuous classes by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit, as the book's own hint states, then find the median class and compute the mean deviation from the midpoints.
Continuous classes: 15.5-20.5, 20.5-25.5, 25.5-30.5, 30.5-35.5, 35.5-40.5, 40.5-45.5, 45.5-50.5, 50.5-55.5, with midpoints 18,23,28,33,38,43,48,53. N=100, so N/2=50. Cumulative frequencies: 5,11,23,37,63,75,91,100. The median class is 35.5-40.5, with l=35.5, C=37, f=26, h=5. Median = 35.5 + (50-37)/26 = 35.5+2.5 = 38. Absolute deviations of midpoints from 38: 20,15,10,5,0,5,10,15. Weighted and summed: 100+90+120+70+0+60+160+135 = 735.
✦ Working through each part gives: Median age = 38, mean deviation about the median = 735/100 = 7.35.
Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh113.pdf) — Exercise 13.1 (12 questions), Exercise 13.2 (10 questions), plus the chapter's Miscellaneous Exercise (6 questions), 28 questions total. The old stub had fabricated two entire extra exercises (13.3 with 5 questions, 13.4 with 7 questions on Coefficient of Variation) that do not exist in the book, and never mentioned the real Miscellaneous Exercise — corrected. Confirmed via full-text search that Coefficient of Variation appears nowhere in the book and is not named in the syllabus line for this chapter at all (not even as a formative-only topic), unlike Quartile Deviation, which the book names explicitly before stating it will not be taught. Every grouped-data mean-deviation, variance, and shortcut-method answer in this file was independently computed with a Python script (numpy) before being written up as a step-by-step solution, including the two continuity-correction problems (Exercise 13.1 Q12, Exercise 13.2 Q10) that require converting gapped class intervals to continuous ones before the median or midpoints can be found, exactly as the book's own hints state. Miscellaneous Exercise Q1's answer (4 and 8) was cross-verified as an internal consistency check against Exercise 13.2 Q1's own dataset, which turns out to be the same eight observations.. Questions are referenced from the NCERT textbook for identification.
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