Find the mean and variance for the data: 6, 7, 10, 12, 13, 4, 8, 12.
Hint. Find the mean, then average the squared deviations from it.
Mean = (6+7+10+12+13+4+8+12)/8 = 72/8 = 9. Squared deviations: 9,4,1,9,16,25,1,9, summing to 74.
✦ Working through each part gives: Mean = 9, variance = 74/8 = 9.25.
Find the mean and variance for the first n natural numbers.
Hint. Use the standard results for the sum and sum of squares of the first n natural numbers: mean = (n+1)/2, and derive the variance from sum(i^2)=n(n+1)(2n+1)/6.
Mean = (1+2+...+n)/n = [n(n+1)/2]/n = (n+1)/2. Variance = [sum(i^2)]/n - mean^2 = [n(n+1)(2n+1)/6]/n - [(n+1)/2]^2 = (n+1)(2n+1)/6 - (n+1)^2/4. Combining over a common denominator of 12 and simplifying: (n+1)[2(2n+1)-3(n+1)]/12 = (n+1)(n-1)/12 = (n^2-1)/12.
✦ Working through each part gives: Mean = (n+1)/2, variance = (n^2-1)/12.
Find the mean and variance for the first 10 multiples of 3.
Hint. The first 10 multiples of 3 are exactly 3 times the first 10 natural numbers, so use the scaling property with the result from Q2.
The data is 3,6,9,...,30, i.e. 3 times each of the first 10 natural numbers. From Q2 with n=10: mean of first 10 naturals = 11/2 = 5.5, variance = (100-1)/12 = 99/12 = 8.25. By the scaling property (multiplying every observation by 3): new mean = 3(5.5)=16.5, new variance = 9(8.25)=74.25.
✦ Working through each part gives: Mean = 16.5, variance = 74.25.
Find the mean and variance for the data: xi = 6,10,14,18,24,28,30 with fi = 2,4,7,12,8,4,3.
Hint. Compute N and the weighted mean, then the frequency-weighted average of the squared deviations.
N = 2+4+7+12+8+4+3 = 40. Sum(fi.xi) = 12+40+98+216+192+112+90 = 760. Mean = 760/40 = 19. Squared deviations (xi-19)^2: 169,81,25,1,25,81,121. Weighted and summed: 338+324+175+12+200+324+363 = 1736.
✦ Working through each part gives: Mean = 19, variance = 1736/40 = 43.4.
Find the mean and variance for the data: xi = 92,93,97,98,102,104,109 with fi = 3,2,3,2,6,3,3.
Hint. Compute N and the weighted mean, then the frequency-weighted average of the squared deviations.
N = 3+2+3+2+6+3+3 = 22. Sum(fi.xi) = 276+186+291+196+612+312+327 = 2200. Mean = 2200/22 = 100. Squared deviations (xi-100)^2: 64,49,9,4,4,16,81. Weighted and summed: 192+98+27+8+24+48+243 = 640.
✦ Working through each part gives: Mean = 100, variance = 640/22, approximately 29.09.
Find the mean and standard deviation using the short-cut method for: xi = 60,61,62,...,68 with fi = 2,1,12,29,25,12,10,4,5.
Hint. Take an assumed mean A (the middle value 64 is convenient) and h=1, form yi=(xi-A)/h, then apply the shortcut variance formula.
Let A=64, h=1. yi = xi-64 runs -4,-3,-2,-1,0,1,2,3,4. N = 2+1+12+29+25+12+10+4+5 = 100. fi.yi: -8,-3,-24,-29,0,12,20,12,20, summing to 0. fi.yi^2: 32,9,48,29,0,12,40,36,80, summing to 286. Mean = 64 + 1(0/100) = 64. Variance = (1/100^2)[100(286)-0^2] = 28600/10000 = 2.86.
✦ Working through each part gives: Mean = 64, standard deviation = sqrt(2.86), approximately 1.69.
Find the mean and variance for the frequency distribution: classes 0-30,30-60,...,180-210 with frequencies 2,3,5,10,3,5,2.
Hint. Use class midpoints as xi, compute N and the weighted mean, then the frequency-weighted average of the squared deviations.
Midpoints: 15,45,75,105,135,165,195. N = 2+3+5+10+3+5+2 = 30. Sum(fi.xi) = 30+135+375+1050+405+825+390 = 3210. Mean = 3210/30 = 107. Squared deviations (xi-107)^2: 8464,3844,1024,4,784,3364,7744. Weighted and summed: 16928+11532+5120+40+2352+16820+15488 = 68280.
✦ Working through each part gives: Mean = 107, variance = 68280/30 = 2276.
Find the mean and variance for the frequency distribution: classes 0-10,10-20,20-30,30-40,40-50 with frequencies 5,8,15,16,6.
Hint. Use class midpoints as xi, compute N and the weighted mean, then the frequency-weighted average of the squared deviations.
Midpoints: 5,15,25,35,45. N = 5+8+15+16+6 = 50. Sum(fi.xi) = 25+120+375+560+270 = 1350. Mean = 1350/50 = 27. Squared deviations (xi-27)^2: 484,144,4,64,324. Weighted and summed: 2420+1152+60+1024+1944 = 6600.
✦ Working through each part gives: Mean = 27, variance = 6600/50 = 132.
Find the mean, variance and standard deviation using the short-cut method for the height distribution: classes 70-75,75-80,...,110-115 with frequencies 3,4,7,7,15,9,6,6,3.
Hint. Use class midpoints, take the middle class's midpoint as assumed mean A with h equal to the class width, form yi=(xi-A)/h, and apply the shortcut variance formula.
Midpoints: 72.5,77.5,82.5,87.5,92.5,97.5,102.5,107.5,112.5. Let A=92.5, h=5. yi runs -4,-3,-2,-1,0,1,2,3,4. N = 3+4+7+7+15+9+6+6+3 = 60. fi.yi: -12,-12,-14,-7,0,9,12,18,12, summing to 6. fi.yi^2: 48,36,28,7,0,9,24,54,48, summing to 254. Mean = 92.5 + 5(6/60) = 92.5+0.5 = 93. Variance = (25/3600)[60(254)-6^2] = (25/3600)(15240-36) = (25/3600)(15204) = 105.58.
✦ Working through each part gives: Mean = 93, variance is approximately 105.58, standard deviation is approximately 10.28.
The diameters of circles (in mm) drawn in a design are: classes 33-36,37-40,41-44,45-48,49-52 with frequencies 15,17,21,22,25. Calculate the standard deviation and mean diameter of the circles.
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 apply the short-cut method.
Continuous classes: 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5-48.5, 48.5-52.5, with midpoints 34.5,38.5,42.5,46.5,50.5. Let A=42.5, h=4. yi runs -2,-1,0,1,2. N = 15+17+21+22+25 = 100. fi.yi: -30,-17,0,22,50, summing to 25. fi.yi^2: 60,17,0,22,100, summing to 199. Mean = 42.5 + 4(25/100) = 42.5+1 = 43.5. Variance = (16/10000)[100(199)-25^2] = (16/10000)(19900-625) = (16/10000)(19275) = 30.84.
✦ Working through each part gives: Mean diameter = 43.5 mm, standard deviation = sqrt(30.84), approximately 5.55 mm.
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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