Miscellaneous Exercise — Statistics

1. M.15 marksNCERT Class 11 Mathematics, Statistics, Reprint 2026-27

   The mean and variance of eight observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

   Hint. Use the mean to find the total sum and the variance to find the total sum of squares, then subtract off the six known values to get a system in the two unknowns.

   Total sum = 8(9) = 72. Known sum = 6+7+10+12+12+13 = 60, so the two unknowns sum to 12. Total sum of squares = 8(9.25+81) = 8(90.25) = 722. Known sum of squares = 36+49+100+144+144+169 = 642, so the unknowns' squares sum to 80. Let the unknowns be a,b: a+b=12, a^2+b^2=80. Then 2ab=(a+b)^2-(a^2+b^2)=144-80=64, so ab=32. a,b are roots of t^2-12t+32=0, which factors as (t-8)(t-4)=0.

   ✦ Working through each part gives: The remaining two observations are 4 and 8.

2. M.25 marksNCERT Class 11 Mathematics, Statistics, Reprint 2026-27

   The mean and variance of 7 observations are 8 and 16 respectively. If five of the observations are 2, 4, 10, 12, 14, find the remaining two observations.

   Hint. Use the mean to find the total sum and the variance to find the total sum of squares, then subtract off the five known values to get a system in the two unknowns.

   Total sum = 7(8) = 56. Known sum = 2+4+10+12+14 = 42, so the two unknowns sum to 14. Total sum of squares = 7(16+64) = 7(80) = 560. Known sum of squares = 4+16+100+144+196 = 460, so the unknowns' squares sum to 100. Let the unknowns be a,b: a+b=14, a^2+b^2=100. Then 2ab=(a+b)^2-(a^2+b^2)=196-100=96, so ab=48. a,b are roots of t^2-14t+48=0, which factors as (t-8)(t-6)=0.

   ✦ Working through each part gives: The remaining two observations are 6 and 8.

3. M.33 marksNCERT Class 11 Mathematics, Statistics, Reprint 2026-27

   The mean and standard deviation of six observations are 8 and 4 respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

   Hint. Apply the scaling property: multiplying every observation by a constant a multiplies the mean by a and the standard deviation by |a|.

   New mean = 3(8) = 24. New variance = 3^2(4^2) = 9(16) = 144.

   ✦ Working through each part gives: New mean = 24, new standard deviation = sqrt(144) = 12.

4. M.46 marksNCERT Class 11 Mathematics, Statistics, Reprint 2026-27

   Given that x_bar is the mean and sigma^2 is the variance of n observations x1, x2, ..., xn. Prove that the mean and variance of the observations ax1, ax2, ax3, ..., axn are a.x_bar and a^2.sigma^2 respectively (a is not 0).

   Hint. Write the new mean directly from the definition, then express each new deviation in terms of the old deviation before squaring and averaging.

   Let yi = a.xi for i=1 to n. The new mean is y_bar = (1/n)sum(yi) = (1/n)sum(a.xi) = a.(1/n)sum(xi) = a.x_bar. For the variance, each new deviation is yi - y_bar = a.xi - a.x_bar = a(xi - x_bar). Squaring: (yi-y_bar)^2 = a^2(xi-x_bar)^2. Averaging over all n observations: variance(y) = (1/n)sum[(yi-y_bar)^2] = (1/n)sum[a^2(xi-x_bar)^2] = a^2.(1/n)sum[(xi-x_bar)^2] = a^2.sigma^2.

   ✦ Working through each part gives: The mean of ax1,...,axn is a.x_bar and the variance is a^2.sigma^2, as required.

5. M.56 marksNCERT Class 11 Mathematics, Statistics, Reprint 2026-27

   The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) if the wrong item is omitted, (ii) if it is replaced by 12.

   Hint. Recover the original total sum and sum of squares from the given mean and standard deviation, then adjust each by removing or replacing the value 8, and recompute.

   Original sum = 20(10) = 200. Original sum of squares = 20(2^2+10^2) = 20(104) = 2080. (i) Omitting 8: new n=19, new sum = 200-8=192, new mean = 192/19, approximately 10.11. New sum of squares = 2080-64=2016. New variance = 2016/19 - (192/19)^2, approximately 106.11-102.11 = 3.99, so new SD is approximately 2.00. (ii) Replacing 8 with 12: n stays 20, new sum = 200-8+12=204, new mean = 204/20=10.2. New sum of squares = 2080-64+144=2160. New variance = 2160/20 - 10.2^2 = 108-104.04=3.96, so new SD is approximately 1.99.

   ✦ Working through each part gives: (i) Omitted: mean is approximately 10.11, SD is approximately 2.00. (ii) Replaced by 12: mean = 10.2, SD is approximately 1.99.

6. M.66 marksNCERT Class 11 Mathematics, Statistics, Reprint 2026-27

   The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

   Hint. Recover the original total sum and sum of squares from the given mean and standard deviation, then subtract off the three incorrect values and recompute over the remaining 97 observations.

   Original sum = 100(20) = 2000. Original sum of squares = 100(3^2+20^2) = 100(409) = 40900. Removing the three incorrect values (21,21,18): new n=97, new sum = 2000-(21+21+18)=2000-60=1940, new mean = 1940/97 = 20 exactly. New sum of squares = 40900-(441+441+324)=40900-1206=39694. New variance = 39694/97 - 20^2 = 409.22-400 = 9.22, so new SD is approximately 3.04.

   ✦ Working through each part gives: Corrected mean = 20, corrected standard deviation is approximately 3.04.

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.