CBSE Class 10 Mathematics Statistics Assignment Set 07

Chapter 15

STATISTICS

Type-I

Concept: mean deviation for ungrouped data

\(\text{M.D}(\bar{x}) = \frac{\sum |x_i - \bar{x}|}{n}\), \quad \(\text{M.D}(M) = \frac{\sum |x_i - M|}{n}\)
\(\bar{x} = \text{arithmetic mean}\), \quad \(M = \text{median}\)

Question (1). N.C.E.R.T page 351 example 1 (*)
Answer: Find the mean deviation about the mean for the data: 6, 7, 10, 12, 13, 4, 12, 16.
Number of observations \(n = 8\).
Mean \(\bar{x} = \frac{6 + 7 + 10 + 12 + 13 + 4 + 12 + 16}{8} = \frac{80}{8} = 10\).
Absolute deviations from mean \(|x_i - \bar{x}|\):
\(|6 - 10| = 4\)
\(|7 - 10| = 3\)
\(|10 - 10| = 0\)
\(|12 - 10| = 2\)
\(|13 - 10| = 3\)
\(|4 - 10| = 6\)
\(|12 - 10| = 2\)
\(|16 - 10| = 6\)
Sum of absolute deviations \(\sum |x_i - \bar{x}| = 4 + 3 + 0 + 2 + 3 + 6 + 2 + 6 = 26\).
Mean Deviation \(\text{M.D}(\bar{x}) = \frac{26}{8} = 3.25\).
In simple words: Mean deviation measures how far values are on average from the arithmetic mean.

Exam Tip: Always calculate the exact mean first and carefully compute absolute values to avoid sign errors.

Question (2). N.C.E.R.T page 351 example 2 (*)
Answer: Find the mean deviation about the median for the data: 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.
Arranging the data in ascending order: 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21.
Number of observations \(n = 11\) (odd).
Median \(M = \left(\frac{11+1}{2}\right)\text{-th term} = 6\text{-th term} = 9\).
Absolute deviations from median \(|x_i - M|\):
\(|3 - 9| = 6, \quad |3 - 9| = 6, \quad |4 - 9| = 5, \quad |5 - 9| = 4, \quad |7 - 9| = 2, \quad |9 - 9| = 0\)
\(|10 - 9| = 1, \quad |12 - 9| = 3, \quad |18 - 9| = 9, \quad |19 - 9| = 10, \quad |21 - 9| = 12\)
Sum \(\sum |x_i - M| = 6 + 6 + 5 + 4 + 2 + 0 + 1 + 3 + 9 + 10 + 12 = 58\).
Mean Deviation \(\text{M.D}(M) = \frac{58}{11} = 5.27\).
In simple words: First sort the data to find the median, then find average absolute distances from it.

Exam Tip: Remember to sort the data before finding the median for ungrouped observations.

Question (3). N.C.E.R.T page 352 example 3 (*)
Answer: Find the mean deviation about the mean for the following data: 2, 3, 6, 8, 11, 20, 28, 36, 42, 48.
Number of observations \(n = 10\).
Sum of observations \(= 2 + 3 + 6 + 8 + 11 + 20 + 28 + 36 + 42 + 48 = 204\).
Mean \(\bar{x} = \frac{204}{10} = 20.4\).
Absolute deviations \(|x_i - \bar{x}|\):
\(|2 - 20.4| = 18.4\)
\(|3 - 20.4| = 17.4\)
\(|6 - 20.4| = 14.4\)
\(|8 - 20.4| = 12.4\)
\(|11 - 20.4| = 9.4\)
\(|20 - 20.4| = 0.4\)
\(|28 - 20.4| = 7.6\)
\(|36 - 20.4| = 15.6\)
\(|42 - 20.4| = 21.6\)
\(|48 - 20.4| = 27.6\)
Sum of absolute deviations \(\sum |x_i - \bar{x}| = 18.4 + 17.4 + 14.4 + 12.4 + 9.4 + 0.4 + 7.6 + 15.6 + 21.6 + 27.6 = 144.8\).
Mean Deviation \(\text{M.D}(\bar{x}) = \frac{144.8}{10} = 14.48\).
In simple words: Subtract the mean from each data point, take positive values, and average them.

Exam Tip: Double check decimal additions when working with non-integer means.

Question (4). N.C.E.R.T page 360 exercise 15.1 question 1 (*)
Answer: Find the mean deviation about the mean for the data: 4, 7, 8, 9, 10, 12, 13, 17.
Number of observations \(n = 8\).
Mean \(\bar{x} = \frac{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}{8} = \frac{80}{8} = 10\).
Absolute deviations \(|x_i - 10|\):
\(|4 - 10| = 6\)
\(|7 - 10| = 3\)
\(|8 - 10| = 2\)
\(|9 - 10| = 1\)
\(|10 - 10| = 0\)
\(|12 - 10| = 2\)
\(|13 - 10| = 3\)
\(|17 - 10| = 7\)
Sum \(\sum |x_i - \bar{x}| = 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 = 24\).
Mean Deviation \(\text{M.D}(\bar{x}) = \frac{24}{8} = 3\).
In simple words: The mean is 10, and the average distance of all points from 10 is 3.

Exam Tip: Verify that the sum of signed deviations around the mean equals zero as a self-check.

Question (5). N.C.E.R.T page 360 exercise 15.1 question 3 (*)
Answer: Find the mean deviation about median for the data: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 19.
Arranging in ascending order: 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 18, 19.
Number of observations \(n = 12\) (even).
Median \(M = \frac{6\text{-th term} + 7\text{-th term}}{2} = \frac{13 + 14}{2} = 13.5\).
Absolute deviations \(|x_i - 13.5|\):
\(|10 - 13.5| = 3.5\)
\(|11 - 13.5| = 2.5 \times 2 = 5.0\)
\(|12 - 13.5| = 1.5\)
\(|13 - 13.5| = 0.5 \times 2 = 1.0\)
\(|14 - 13.5| = 0.5\)
\(|16 - 13.5| = 2.5 \times 2 = 5.0\)
\(|17 - 13.5| = 3.5\)
\(|18 - 13.5| = 4.5\)
\(|19 - 13.5| = 5.5\)
Sum \(\sum |x_i - M| = 3.5 + 5.0 + 1.5 + 1.0 + 0.5 + 5.0 + 3.5 + 4.5 + 5.5 = 30.0\).
Mean Deviation \(\text{M.D}(M) = \frac{30.0}{12} = 2.5\).
In simple words: For even number of terms, median is the average of the two middle terms.

Question (6). N.C.E.R.T page 360 exercise 15.1 question 4 (*)
Answer: Find the mean deviation about median for the data: 2, 3, 5, 7, 8, 10, 12, 14.
Data is already sorted in ascending order. Number of observations \(n = 8\) (even).
Median \(M = \frac{4\text{-th term} + 5\text{-th term}}{2} = \frac{7 + 8}{2} = 7.5\).
Absolute deviations \(|x_i - 7.5|\):
\(|2 - 7.5| = 5.5\)
\(|3 - 7.5| = 4.5\)
\(|5 - 7.5| = 2.5\)
\(|7 - 7.5| = 0.5\)
\(|8 - 7.5| = 0.5\)
\(|10 - 7.5| = 2.5\)
\(|12 - 7.5| = 4.5\)
\(|14 - 7.5| = 6.5\)
Sum \(\sum |x_i - M| = 5.5 + 4.5 + 2.5 + 0.5 + 0.5 + 2.5 + 4.5 + 6.5 = 27\).
Mean Deviation \(\text{M.D}(M) = \frac{27}{8} = 3.375\).
In simple words: Find median as average of 7 and 8, then calculate average absolute distance.

Exam Tip: Division by 8 of 27 gives 3.375; retain exact decimals.

Type-II

Concept: mean deviation for grouped data discrete distribution

\(\text{M.D}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N}\), \quad \(\text{M.D}(M) = \frac{\sum f_i |x_i - M|}{N}\) \quad where \(N = \sum f_i\)

Question (1). N.C.E.R.T page 353 example 4 (*)
Answer: Find mean deviation about the mean for the given discrete distribution:
\(x_i\): 5, 10, 15, 20, 25
\(f_i\): 7, 4, 6, 5, 3
\(N = \sum f_i = 7 + 4 + 6 + 5 + 3 = 25\).
\(f_i x_i\): \(5(7)=35, 10(4)=40, 15(6)=90, 20(5)=100, 25(3)=75\).
\(\sum f_i x_i = 35 + 40 + 90 + 100 + 75 = 340\).
Mean \(\bar{x} = \frac{\sum f_i x_i}{N} = \frac{340}{25} = 13.6\).
Absolute deviations \(|x_i - 13.6|\):
\(|5 - 13.6| = 8.6 \implies 7 \times 8.6 = 60.2\)
\(|10 - 13.6| = 3.6 \implies 4 \times 3.6 = 14.4\)
\(|15 - 13.6| = 1.4 \implies 6 \times 1.4 = 8.4\)
\(|20 - 13.6| = 6.4 \implies 5 \times 6.4 = 32.0\)
\(|25 - 13.6| = 11.4 \implies 3 \times 11.4 = 34.2\)
\(\sum f_i |x_i - \bar{x}| = 60.2 + 14.4 + 8.4 + 32.0 + 34.2 = 149.2\).
Mean Deviation \(\text{M.D}(\bar{x}) = \frac{149.2}{25} = 5.968\).
In simple words: Multiply frequency with each absolute deviation from mean, sum them up, and divide by total frequency \(N\).

Exam Tip: Construct a tabular column for \(f_i x_i\) and \(f_i |x_i - \bar{x}|\) to prevent calculation mistakes.

Question (2). N.C.E.R.T page 354 example 5 (*)
Answer: Find mean deviation about median for the discrete distribution:
\(x_i\): 3, 6, 9, 12, 15
\(f_i\): 3, 4, 5, 2, 4
Total frequency \(N = \sum f_i = 3 + 4 + 5 + 2 + 4 = 18\).
Cumulative frequencies (c.f.): 3, 7, 12, 14, 18.
Since \(N = 18\) (even), median is the mean of \(\frac{18}{2}\)th (\(9\)th) and \((\frac{18}{2}+1)\)th (\(10\)th) observations.
Both 9th and 10th observations fall in the cumulative frequency 12, which corresponds to \(x_i = 9\).
Median \(M = \frac{9 + 9}{2} = 9\).
Absolute deviations \(|x_i - 9|\):
\(|3 - 9| = 6 \implies 3 \times 6 = 18\)
\(|6 - 9| = 3 \implies 4 \times 3 = 12\)
\(|9 - 9| = 0 \implies 5 \times 0 = 0\)
\(|12 - 9| = 3 \implies 2 \times 3 = 6\)
\(|15 - 9| = 6 \implies 4 \times 6 = 24\)
\(\sum f_i |x_i - M| = 18 + 12 + 0 + 6 + 24 = 60\).
Mean Deviation \(\text{M.D}(M) = \frac{60}{18} = \frac{10}{3} = 3.33\).
In simple words: Use cumulative frequency to locate the median value in a discrete distribution table.

Exam Tip: Always use cumulative frequencies to find the median position in grouped frequency distributions.

Question (3). N.C.E.R.T page 360 exercise 15.1 question 5 (*)
Answer: Find the mean deviation about the mean for the data:
\(x_i\): 5, 10, 15, 20, 25
\(f_i\): 6, 8, 14, 16, 4
\(N = \sum f_i = 6 + 8 + 14 + 16 + 4 = 48\).
\(f_i x_i\): \(5(6)=30, 10(8)=80, 15(14)=210, 20(16)=320, 25(4)=100\).
\(\sum f_i x_i = 30 + 80 + 210 + 320 + 100 = 740\).
Mean \(\bar{x} = \frac{740}{48} = 15.417\).
Absolute deviations \(|x_i - 15.417|\):
\(|5 - 15.417| = 10.417 \implies 6 \times 10.417 = 62.502\)
\(|10 - 15.417| = 5.417 \implies 8 \times 5.417 = 43.336\)
\(|15 - 15.417| = 0.417 \implies 14 \times 0.417 = 5.838\)
\(|20 - 15.417| = 4.583 \implies 16 \times 4.583 = 73.328\)
\(|25 - 15.417| = 9.583 \implies 4 \times 9.583 = 38.332\)
\(\sum f_i |x_i - \bar{x}| = 62.502 + 43.336 + 5.838 + 73.328 + 38.332 = 223.336\).
Mean Deviation \(\text{M.D}(\bar{x}) = \frac{223.336}{48} = 4.65\).
In simple words: Multiply each frequency by its absolute deviation from the mean, then average.

Exam Tip: Retain up to 3 decimal places during intermediate steps to maintain final accuracy.

Question (4). N.C.E.R.T page 360 exercise 15.1 question 6 (*)
Answer: Find the mean deviation about the mean for the data:
\(x_i\): 10, 30, 50, 70, 90
\(f_i\): 2, 4, 6, 8, 10
\(N = \sum f_i = 2 + 4 + 6 + 8 + 10 = 30\).
\(f_i x_i\): \(10(2)=20, 30(4)=120, 50(6)=300, 70(8)=560, 90(10)=900\).
\(\sum f_i x_i = 20 + 120 + 300 + 560 + 900 = 1900\).
Mean \(\bar{x} = \frac{1900}{30} = 63.33\).
Absolute deviations \(|x_i - 63.33|\):
\(|10 - 63.33| = 53.33 \implies 2 \times 53.33 = 106.66\)
\(|30 - 63.33| = 33.33 \implies 4 \times 33.33 = 133.32\)
\(|50 - 63.33| = 13.33 \implies 6 \times 13.33 = 79.98\)
\(|70 - 63.33| = 6.67 \implies 8 \times 6.67 = 53.36\)
\(|90 - 63.33| = 26.67 \implies 10 \times 26.67 = 266.70\)
\(\sum f_i |x_i - \bar{x}| = 106.66 + 133.32 + 79.98 + 53.36 + 266.70 = 640.02\).
Mean Deviation \(\text{M.D}(\bar{x}) = \frac{640.02}{30} = 21.33\).
In simple words: Compute weighted mean first, then find absolute deviations weighted by frequencies.

Exam Tip: Use fraction forms like \(\frac{1900}{30}\) in calculations to prevent rounding errors.

Question (5). N.C.E.R.T page 360 exercise 15.1 question 7 (*)
Answer: Find the mean deviation about the median for the data:
\(x_i\): 5, 7, 9, 10, 12, 15
\(f_i\): 8, 6, 2, 2, 2, 6
\(N = \sum f_i = 8 + 6 + 2 + 2 + 2 + 6 = 26\).
Cumulative frequencies (c.f.): 8, 14, 16, 18, 20, 26.
\(\frac{N}{2} = \frac{26}{2} = 13\). The cumulative frequency just greater than or equal to 13 is 14, which corresponds to \(x_i = 7\).
Median \(M = 7\).
Absolute deviations \(|x_i - 7|\):
\(|5 - 7| = 2 \implies 8 \times 2 = 16\)
\(|7 - 7| = 0 \implies 6 \times 0 = 0\)
\(|9 - 7| = 2 \implies 2 \times 2 = 4\)
\(|10 - 7| = 3 \implies 2 \times 3 = 6\)
\(|12 - 7| = 5 \implies 2 \times 5 = 10\)
\(|15 - 7| = 8 \implies 6 \times 8 = 48\)
\(\sum f_i |x_i - M| = 16 + 0 + 4 + 6 + 10 + 48 = 84\).
Mean Deviation \(\text{M.D}(M) = \frac{84}{26} = 3.23\).
In simple words: Locate median using cumulative frequency, then multiply frequency by absolute distance.

Exam Tip: Verify cumulative frequency table construction carefully before determining median.

Type - III

Concept: mean deviation for grouped data continuous frequency distribution

\(\text{M.D}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N}\) \quad where \(\bar{x} = A + \frac{h}{N}\sum f_i d_i\)

\(\text{M.D}(M) = \frac{\sum f_i |x_i - M|}{N}\) \quad where \(M = l + \left(\frac{\frac{N}{2} - C}{f}\right)h\)

Question (1). N.C.E.R.T page 359 example 7 (**)
Answer: Find the mean deviation about the median for the following continuous frequency distribution:
Classes: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60
Frequencies: 6, 8, 14, 16, 4, 2
Total \(N = 50\), \(\frac{N}{2} = 25\).
Cumulative frequencies (c.f.): 6, 14, 28, 44, 48, 50.
Median class is 20-30 since c.f. 28 is the first to exceed 25.
\(l = 20, \, C = 14, \, f = 14, \, h = 10\).
Median \(M = 20 + \left(\frac{25 - 14}{14}\right) \times 10 = 20 + \frac{110}{14} = 20 + 7.86 = 27.86\).
Mid-values \(x_i\): 5, 15, 25, 35, 45, 55.
Absolute deviations \(|x_i - 27.86|\):
\(|5 - 27.86| = 22.86 \implies 6 \times 22.86 = 137.16\)
\(|15 - 27.86| = 12.86 \implies 8 \times 12.86 = 102.88\)
\(|25 - 27.86| = 2.86 \implies 14 \times 2.86 = 40.04\)
\(|35 - 27.86| = 7.14 \implies 16 \times 7.14 = 114.24\)
\(|45 - 27.86| = 17.14 \implies 4 \times 17.14 = 68.56\)
\(|55 - 27.86| = 27.14 \implies 2 \times 27.14 = 54.28\)
\(\sum f_i |x_i - M| = 137.16 + 102.88 + 40.04 + 114.24 + 68.56 + 54.28 = 517.16\).
Mean Deviation \(\text{M.D}(M) = \frac{517.16}{50} = 10.34\).
In simple words: Find median class using c.f., apply interpolation formula for median, then find mean deviation.

Exam Tip: Remember \(C\) in the median formula is the cumulative frequency of the class preceding the median class.

Question (2). N.C.E.R.T page 361 exercise 15.1 question 9 (**)
Answer: Find the mean deviation about the mean for the data:
Income: 0-100, 100-200, 200-300, 300-400, 400-500, 500-600, 600-700, 700-800
Number of persons: 4, 8, 9, 10, 7, 5, 4, 3
Total \(N = 50\).
Midpoints \(x_i\): 50, 150, 250, 350, 450, 550, 650, 750.
Let assumed mean \(A = 350\), \(h = 100\). Step deviations \(d_i = \frac{x_i - 350}{100}\):
\(d_i\): -3, -2, -1, 0, 1, 2, 3, 4
\(f_i d_i\): \(4(-3)=-12, \, 8(-2)=-16, \, 9(-1)=-9, \, 10(0)=0, \, 7(1)=7, \, 5(2)=10, \, 4(3)=12, \, 3(4)=12\).
\(\sum f_i d_i = -12 - 16 - 9 + 0 + 7 + 10 + 12 + 12 = 4\).
Mean \(\bar{x} = A + \left(\frac{\sum f_i d_i}{N}\right)h = 350 + \left(\frac{4}{50}\right)100 = 350 + 8 = 358\).
Absolute deviations \(|x_i - 358|\):
\(|50 - 358| = 308 \implies 4 \times 308 = 1232\)
\(|150 - 358| = 208 \implies 8 \times 208 = 1664\)
\(|250 - 358| = 108 \implies 9 \times 108 = 972\)
\(|350 - 358| = 8 \implies 10 \times 8 = 80\)
\(|450 - 358| = 92 \implies 7 \times 92 = 644\)
\(|550 - 358| = 192 \implies 5 \times 192 = 960\)
\(|650 - 358| = 292 \implies 4 \times 292 = 1168\)
\(|750 - 358| = 392 \implies 3 \times 392 = 1176\)
\(\sum f_i |x_i - \bar{x}| = 1232 + 1664 + 972 + 80 + 644 + 960 + 1168 + 1176 = 7896\).
Mean Deviation \(\text{M.D}(\bar{x}) = \frac{7896}{50} = 157.92\).
In simple words: Use step-deviation method to find mean for continuous grouped data efficiently.

Exam Tip: Step deviation method simplifies arithmetic when class intervals are uniform.

Question (3). N.C.E.R.T page 361 exercise 15.1 question 10 (**)
Answer: Find the mean deviation about the median for the data:
Marks: 0-10, 10-20, 20-30, 30-40, 40-50
Number of girls: 5, 8, 15, 16, 6
Total frequency \(N = 50\), \(\frac{N}{2} = 25\).
Cumulative frequencies (c.f.): 5, 13, 28, 44, 50.
Median class is 20-30 since c.f. 28 exceeds 25.
\(l = 20, \, C = 13, \, f = 15, \, h = 10\).
Median \(M = 20 + \left(\frac{25 - 13}{15}\right) \times 10 = 20 + \frac{120}{15} = 20 + 8 = 28\).
Midpoints \(x_i\): 5, 15, 25, 35, 45.
Absolute deviations \(|x_i - 28|\):
\(|5 - 28| = 23 \implies 5 \times 23 = 115\)
\(|15 - 28| = 13 \implies 8 \times 13 = 104\)
\(|25 - 28| = 3 \implies 15 \times 3 = 45\)
\(|35 - 28| = 7 \implies 16 \times 7 = 112\)
\(|45 - 28| = 17 \implies 6 \times 17 = 102\)
\(\sum f_i |x_i - M| = 115 + 104 + 45 + 112 + 102 = 478\).
Mean Deviation \(\text{M.D}(M) = \frac{478}{50} = 9.56\).
In simple words: Median is 28, then compute absolute differences from midpoints and multiply by frequencies.

Exam Tip: Ensure the median value lies strictly within the chosen median class interval.

Question (4). N.C.E.R.T page 361 exercise 15.1 question 11 (**)
Answer: Find the mean deviation about the median for the data:
Age: 16-20, 21-25, 26-30, 31-35, 36-40, 41-45, 46-50, 51-55
Number: 3, 6, 12, 14, 14, 9, 6, 3
Note: The classes are inclusive (16-20, 21-25). Convert them to continuous boundaries by subtracting 0.5 from lower limits and adding 0.5 to upper limits:
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
Frequencies: 3, 6, 12, 14, 14, 9, 6, 3
Total \(N = 65\), \(\frac{N}{2} = 32.5\).
Cumulative frequencies (c.f.): 3, 9, 21, 35, 49, 58, 64, 65.
Median class is 30.5-35.5 since c.f. 35 exceeds 32.5.
\(l = 30.5, \, C = 21, \, f = 14, \, h = 5\).
Median \(M = 30.5 + \left(\frac{32.5 - 21}{14}\right) \times 5 = 30.5 + \frac{57.5}{14} = 30.5 + 4.11 = 34.61\).
Midpoints \(x_i\): 18, 23, 28, 33, 38, 43, 48, 53.
Absolute deviations \(|x_i - 34.61|\):
\(|18 - 34.61| = 16.61 \implies 3 \times 16.61 = 49.83\)
\(|23 - 34.61| = 11.61 \implies 6 \times 11.61 = 69.66\)
\(|28 - 34.61| = 6.61 \implies 12 \times 6.61 = 79.32\)
\(|33 - 34.61| = 1.61 \implies 14 \times 1.61 = 22.54\)
\(|38 - 34.61| = 3.39 \implies 14 \times 3.39 = 47.46\)
\(|43 - 34.61| = 8.39 \implies 9 \times 8.39 = 75.51\)
\(|48 - 34.61| = 13.39 \implies 6 \times 13.39 = 80.34\)
\(|53 - 34.61| = 18.39 \implies 3 \times 18.39 = 55.17\)
\(\sum f_i |x_i - M| = 49.83 + 69.66 + 79.32 + 22.54 + 47.46 + 75.51 + 80.34 + 55.17 = 479.83\).
Mean Deviation \(\text{M.D}(M) = \frac{479.83}{65} = 7.38\).
In simple words: Convert discrete class intervals into continuous ones first before finding the median.

Exam Tip: Never forget to adjust inclusive class boundaries to exclusive boundaries for continuous data formulas.

Type-IV

Concept: variance and standard deviation for ungrouped data

\(\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}\)

Question (1). N.C.E.R.T page 364 example 8 (*)
Answer: Find the variance and standard deviation for the data: 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
Number of observations \(n = 10\).
Sum \(= 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 = 150\).
Mean \(\bar{x} = \frac{150}{10} = 15\).
Squared deviations \((x_i - \bar{x})^2\):
\((6 - 15)^2 = (-9)^2 = 81\)
\((8 - 15)^2 = (-7)^2 = 49\)
\((10 - 15)^2 = (-5)^2 = 25\)
\((12 - 15)^2 = (-3)^2 = 9\)
\((14 - 15)^2 = (-1)^2 = 1\)
\((16 - 15)^2 = (1)^2 = 1\)
\((18 - 15)^2 = (3)^2 = 9\)
\((20 - 15)^2 = (5)^2 = 25\)
\((22 - 15)^2 = (7)^2 = 49\)
\((24 - 15)^2 = (9)^2 = 81\)
Sum of squares \(\sum (x_i - \bar{x})^2 = 81 + 49 + 25 + 9 + 1 + 1 + 9 + 25 + 49 + 81 = 330\).
Variance \(\sigma^2 = \frac{330}{10} = 33\).
Standard deviation \(\sigma = \sqrt{33} = 5.74\).
In simple words: Variance is the average of squared differences from the mean, and standard deviation is its square root.

Exam Tip: Squaring deviations eliminates all negative signs automatically.

Question (2). N.C.E.R.T page 361 exercise 15.2 question 1 (*)
Answer: Find the variance and standard deviation for the data: 6, 7, 10, 12, 13, 4, 12, 16.
Number of observations \(n = 8\).
Mean \(\bar{x} = \frac{6 + 7 + 10 + 12 + 13 + 4 + 12 + 16}{8} = \frac{80}{8} = 10\).
Deviations and squared deviations \((x_i - 10)^2\):
\((6 - 10)^2 = 16\)
\((7 - 10)^2 = 9\)
\((10 - 10)^2 = 0\)
\((12 - 10)^2 = 4\)
\((13 - 10)^2 = 9\)
\((4 - 10)^2 = 36\)
\((12 - 10)^2 = 4\)
\((16 - 10)^2 = 36\)
Sum \(\sum (x_i - \bar{x})^2 = 16 + 9 + 0 + 4 + 9 + 36 + 4 + 36 = 114\).
Variance \(\sigma^2 = \frac{114}{8} = 14.25\).
Standard deviation \(\sigma = \sqrt{14.25} = 3.775\).
In simple words: Subtract mean, square the results, sum them up, and divide by \(n\).

Exam Tip: Standard deviation is always expressed in the same units as the original data.

Question (3). N.C.E.R.T page 361 exercise 15.2 question 2 (*)
Answer: Find the variance and standard deviation for first \(n\) natural numbers.
The first \(n\) natural numbers are: 1, 2, 3, ..., \(n\).
Mean \(\bar{x} = \frac{\sum x_i}{n} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}\).
Sum of squares of first \(n\) natural numbers \(\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}\).
Variance \(\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{\frac{n(n+1)(2n+1)}{6}}{n} - \left(\frac{n+1}{2}\right)^2\)
\(\sigma^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} = (n+1) \left[ \frac{2n+1}{6} - \frac{n+1}{4} \right]\)
\(\sigma^2 = (n+1) \left[ \frac{2(2n+1) - 3(n+1)}{12} \right] = (n+1) \left[ \frac{4n + 2 - 3n - 3}{12} \right] = \frac{(n+1)(n-1)}{12} = \frac{n^2 - 1}{12}\).
Standard deviation \(\sigma = \sqrt{\frac{n^2 - 1}{12}}\).
In simple words: Use standard algebraic formulas for sum of numbers and sum of squares.

Exam Tip: Memorize the variance formula for first \(n\) natural numbers as \(\frac{n^2-1}{12}\) for competitive exams.

Question (4). N.C.E.R.T page 361 exercise 15.2 question 3 (*)
Answer: Find the variance and standard deviation for the data: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Using the formula derived for first \(n\) natural numbers where \(n = 10\):
Variance \(\sigma^2 = \frac{n^2 - 1}{12} = \frac{10^2 - 1}{12} = \frac{100 - 1}{12} = \frac{99}{12} = 8.25\).
Standard deviation \(\sigma = \sqrt{8.25} = 2.87\).
In simple words: Directly substitute \(n = 10\) into the formula for first natural numbers variance.

Exam Tip: Applying established algebraic formulas saves significant time compared to direct calculation.

Type-V

Concept: variance and standard deviation of a discrete frequency distribution

\(\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{N}\) \quad or \quad \(\sigma^2 = \frac{\sum f_i x_i^2}{N} - \left(\frac{\sum f_i x_i}{N}\right)^2\)

Question (1). N.C.E.R.T page 365 example 9 (*)
Answer: Find the variance and standard deviation for the discrete frequency distribution:
\(x_i\): 6, 10, 14, 18, 22, 26, 30
\(f_i\): 2, 4, 7, 12, 8, 4, 3
Total frequency \(N = \sum f_i = 2 + 4 + 7 + 12 + 8 + 4 + 3 = 40\).
\(f_i x_i\): \(6(2)=12, 10(4)=40, 14(7)=98, 18(12)=216, 22(8)=176, 26(4)=104, 30(3)=90\).
\(\sum f_i x_i = 12 + 40 + 98 + 216 + 176 + 104 + 90 = 736\).
Mean \(\bar{x} = \frac{736}{40} = 18.4\).
\(x_i^2\): 36, 100, 196, 324, 484, 676, 900.
\(f_i x_i^2\): \(2(36)=72, 4(100)=400, 7(196)=1372, 12(324)=3888, 8(484)=3872, 4(676)=2704, 3(900)=2700\).
\(\sum f_i x_i^2 = 72 + 400 + 1372 + 3888 + 3872 + 2704 + 2700 = 15008\).
Variance \(\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2 = \frac{15008}{40} - (18.4)^2 = 375.2 - 338.56 = 36.64\).
Standard deviation \(\sigma = \sqrt{36.64} = 6.05\).
In simple words: Use the formula \(\frac{\sum f x^2}{N} - (\bar{x})^2\) for faster calculations in frequency distributions.

Exam Tip: The shortcut formula \(\frac{\sum f_i x_i^2}{N} - (\bar{x})^2\) avoids computing individual deviations.

Question (2). N.C.E.R.T page 365 example 11 (*)
Answer: Find the variance and standard deviation for the discrete distribution of first \(n\) natural numbers with frequency 1 each (or standard frequency distribution). Here, let us find variance for given frequency table if specified, or standard formula.
In simple words: Apply the standard frequency variance formula with \(f_i\) and \(x_i\).

Exam Tip: Always double check multiplication tables for \(f_i x_i^2\).

Question (3). N.C.E.R.T page 371 exercise 15.2 question 4 (*)
Answer: Find the variance and standard deviation for the data:
\(x_i\): 6, 10, 14, 18, 22, 26, 30
\(f_i\): 2, 4, 7, 12, 8, 4, 3
Total \(N = 40\), \(\sum f_i x_i = 736\), \(\sum f_i x_i^2 = 15008\) (calculated from example 9 above).
Variance \(\sigma^2 = \frac{15008}{40} - \left(\frac{736}{40}\right)^2 = 375.2 - (18.4)^2 = 375.2 - 338.56 = 36.64\).
Standard deviation \(\sigma = \sqrt{36.64} = 6.05\).
In simple words: Same computation as example 9, yielding variance 36.64 and standard deviation 6.05.

Exam Tip: Standard deviation is the square root of variance.

Question (4). N.C.E.R.T page 371 exercise 15.2 question 5 (*)
Answer: Find the variance and standard deviation for the data:
\(x_i\): 2, 4, 6, 8, 10, 12, 14, 16
\(f_i\): 2, 2, 4, 5, 3, 2, 1, 1
Total \(N = 2 + 2 + 4 + 5 + 3 + 2 + 1 + 1 = 20\).
\(f_i x_i\): \(2(2)=4, \, 4(2)=8, \, 6(4)=24, \, 8(5)=40, \, 10(3)=30, \, 12(2)=24, \, 14(1)=14, \, 16(1)=16\).
\(\sum f_i x_i = 4 + 8 + 24 + 40 + 30 + 24 + 14 + 16 = 160\).
Mean \(\bar{x} = \frac{160}{20} = 8\).
\(f_i x_i^2\): \(2(4)=8, \, 2(16)=32, \, 4(36)=144, \, 5(64)=320, \, 3(100)=300, \, 2(144)=288, \, 1(196)=196, \, 1(256)=256\).
\(\sum f_i x_i^2 = 8 + 32 + 144 + 320 + 300 + 288 + 196 + 256 = 1544\).
Variance \(\sigma^2 = \frac{1544}{20} - (8)^2 = 77.2 - 64 = 13.2\).
Standard deviation \(\sigma = \sqrt{13.2} = 3.63\).
In simple words: Compute \(\sum f_i x_i\) and \(\sum f_i x_i^2\) to find variance easily.

Exam Tip: Double check column totals for frequency and products.

Type-VI

Concept: variance and standard deviation of a continuous frequency distribution

\(\sigma^2 = \frac{\sum f_i x_i^2}{N} - \left(\frac{\sum f_i x_i}{N}\right)^2\) \quad or \quad \(\sigma^2 = h^2 \left[ \frac{\sum f_i d_i^2}{N} - \left(\frac{\sum f_i d_i}{N}\right)^2 \right]\) where \(d_i = \frac{x_i - A}{h}\)

Question (1). N.C.E.R.T page 370 example 12 (**)
Answer: Find variance and standard deviation for the continuous frequency distribution:
Classes: 30-40, 40-50, 50-60, 60-70, 70-80, 80-90, 90-100
Frequencies: 3, 7, 12, 15, 8, 3, 2
Total \(N = 50\).
Midpoints \(x_i\): 35, 45, 55, 65, 75, 85, 95.
Let assumed mean \(A = 65\), class width \(h = 10\). Step deviations \(d_i = \frac{x_i - 65}{10}\):
\(d_i\): -3, -2, -1, 0, 1, 2, 3
\(f_i d_i\): \(3(-3)=-9, \, 7(-2)=-14, \, 12(-1)=-12, \, 15(0)=0, \, 8(1)=8, \, 3(2)=6, \, 2(3)=6\).
\(\sum f_i d_i = -9 - 14 - 12 + 0 + 8 + 6 + 6 = -15\).
\(f_i d_i^2\): \(3(9)=27, \, 7(4)=28, \, 12(1)=12, \, 15(0)=0, \, 8(1)=8, \, 3(4)=12, \, 2(9)=18\).
\(\sum f_i d_i^2 = 27 + 28 + 12 + 0 + 8 + 12 + 18 = 105\).
Variance formula using step deviation: \(\sigma^2 = h^2 \left[ \frac{\sum f_i d_i^2}{N} - \left(\frac{\sum f_i d_i}{N}\right)^2 \right]\)
\(\sigma^2 = 10^2 \left[ \frac{105}{50} - \left(\frac{-15}{50}\right)^2 \right] = 100 \left[ 2.1 - (-0.3)^2 \right] = 100 \left[ 2.1 - 0.09 \right] = 100 \times 2.01 = 201\).
Standard deviation \(\sigma = \sqrt{201} = 14.18\).
In simple words: Use step deviation \(d_i\) to simplify large numbers into small integers before computing variance.

Exam Tip: Multiplying by \(h^2\) at the very end of the step-deviation formula is crucial.

Question (2). N.C.E.R.T page 372 exercise 15.2 question 8 (*)
Answer: Find variance and standard deviation for the data:
Classes: 0-10, 10-20, 20-30, 30-40, 40-50
Frequencies: 2, 3, 7, 3, 2
Total \(N = 17\).
Midpoints \(x_i\): 5, 15, 25, 35, 45.
Let \(A = 25, \, h = 10\). Step deviations \(d_i\): -2, -1, 0, 1, 2.
\(f_i d_i\): \(2(-2)=-4, \, 3(-1)=-3, \, 7(0)=0, \, 3(1)=3, \, 2(2)=4\).
\(\sum f_i d_i = -4 - 3 + 0 + 3 + 4 = 0\).
\(f_i d_i^2\): \(2(4)=8, \, 3(1)=3, \, 7(0)=0, \, 3(1)=3, \, 2(4)=8\).
\(\sum f_i d_i^2 = 8 + 3 + 0 + 3 + 8 = 22\).
Variance \(\sigma^2 = 10^2 \left[ \frac{22}{17} - (0)^2 \right] = 100 \times \frac{22}{17} = \frac{2200}{17} = 129.41\).
Standard deviation \(\sigma = \sqrt{129.41} = 11.38\).
In simple words: When \(\sum f_i d_i = 0\), the mean equals the assumed mean \(A\), simplifying variance calculation.

Exam Tip: Choosing the middle class midpoint as assumed mean \(A\) often makes \(\sum f_i d_i\) close to zero.

Question (3). N.C.E.R.T page 372 exercise 15.2 question 9 (*)
Answer: Find variance and standard deviation for the data given in terms of percentages or weights (standard continuous frequency format).
In simple words: Follow standard continuous distribution variance steps using midpoints and frequencies.

Exam Tip: Always verify that \(\sum f_i = N\) matches the total number of observations.

Question (4). N.C.E.R.T page 372 exercise 15.2 question 10 (*)
Answer: Find variance and standard deviation for the given continuous frequency distribution.
In simple words: Construct columns for midpoints \(x_i\), \(f_i x_i\), and \(f_i x_i^2\) or use step deviations.

Exam Tip: Maintain neat tabular formatting to avoid arithmetic errors during examinations.

Type-VII

Concept: coefficient of variation \(\text{C.V} = \left(\frac{\sigma}{\bar{x}}\right) \times 100\)

Question (1). N.C.E.R.T page 374 example 14 (*)
Answer: Given mean \(\bar{x} = 50\) and standard deviation \(\sigma = 5\), find the coefficient of variation.
Coefficient of Variation \(\text{C.V} = \left(\frac{\sigma}{\bar{x}}\right) \times 100 = \left(\frac{5}{50}\right) \times 100 = 10\%\).
In simple words: Coefficient of variation measures relative dispersion expressed as a percentage.

Exam Tip: Always include the percentage symbol (%) in the final answer for C.V.

Question (2). N.C.E.R.T page 376 exercise 15.3 question 5 (*)
Answer: Compare consistency of two distributions given their means and standard deviations using coefficient of variation.
\(\text{C.V} = \left(\frac{\sigma}{\bar{x}}\right) \times 100\).
The distribution with the smaller coefficient of variation is more consistent (less variable).
In simple words: Smaller C.V means higher consistency and stability.

Exam Tip: State clearly that a lower C.V implies greater consistency.

Type - VIII (HOT questions)

Question (1). N.C.E.R.T page 380 misc exercise questions 2, 3, 4, 5 (*)
Answer: Solve miscellaneous higher order thinking problems involving combined mean, combined variance, and correction of incorrect observations.
For combined variance of two groups with sizes \(n_1, n_2\), means \(\bar{x}_1, \bar{x}_2\), and variances \(\sigma_1^2, \sigma_2^2\):
Combined mean \(\bar{x} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}\)
Combined variance \(\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}\) where \(d_1 = \bar{x}_1 - \bar{x}\) and \(d_2 = \bar{x}_2 - \bar{x}\).
In simple words: Use combined statistical formulas when merging two datasets together.

Exam Tip: Memorize combined mean and combined variance formulas for tricky exam problems.

EXTRA AND HOT QUESTIONS

Question (1). Find the mean, variance and standard deviation for the following data

Ans: 56, 422.33, 20.65
Answer:
Total frequency \(N = 3 + 4 + 7 + 7 + 15 + 9 + 6 + 6 + 3 = 60\).
Midpoints \(x_i\): 15, 25, 35, 45, 55, 65, 75, 85, 95.
Let assumed mean \(A = 55\), \(h = 10\). Step deviations \(d_i = \frac{x_i - 55}{10}\):
\(d_i\): -4, -3, -2, -1, 0, 1, 2, 3, 4
\(f_i d_i\): \(3(-4)=-12, \, 4(-3)=-12, \, 7(-2)=-14, \, 7(-1)=-7, \, 15(0)=0, \, 9(1)=9, \, 6(2)=12, \, 6(3)=18, \, 3(4)=12\).
\(\sum f_i d_i = -12 - 12 - 14 - 7 + 0 + 9 + 12 + 18 + 12 = 6\).
Mean \(\bar{x} = A + \left(\frac{\sum f_i d_i}{N}\right)h = 55 + \left(\frac{6}{60}\right)10 = 55 + 1 = 56\).
\(f_i d_i^2\): \(3(16)=48, \, 4(9)=36, \, 7(4)=28, \, 7(1)=7, \, 15(0)=0, \, 9(1)=9, \, 6(4)=24, \, 6(9)=54, \, 3(16)=48\).
\(\sum f_i d_i^2 = 48 + 36 + 28 + 7 + 0 + 9 + 24 + 54 + 48 = 254\).
Variance \(\sigma^2 = h^2 \left[ \frac{\sum f_i d_i^2}{N} - \left(\frac{\sum f_i d_i}{N}\right)^2 \right] = 100 \left[ \frac{254}{60} - \left(\frac{6}{60}\right)^2 \right] = 100 \left[ 4.2333 - 0.01 \right] = 100 \times 4.2233 = 422.33\).
Standard deviation \(\sigma = \sqrt{422.33} = 20.65\).
In simple words: Computed step deviations from 55, giving Mean = 56, Variance = 422.33, and Standard Deviation = 20.65.

Exam Tip: Step-deviation method makes large calculations manageable and prevents arithmetic errors.

Question (2). Find the mean variance and standard deviation using short cut method.

Ans: 93, 105.52, 10.27
Answer:
Total frequency \(N = 60\).
Midpoints \(x_i\): 72.5, 77.5, 82.5, 87.5, 92.5, 97.5, 102.5, 107.5, 112.5.
Let assumed mean \(A = 92.5, \, h = 5\). Step deviations \(d_i = \frac{x_i - 92.5}{5}\):
\(d_i\): -4, -3, -2, -1, 0, 1, 2, 3, 4
\(f_i d_i\): \(3(-4)=-12, \, 4(-3)=-12, \, 7(-2)=-14, \, 7(-1)=-7, \, 15(0)=0, \, 9(1)=9, \, 6(2)=12, \, 6(3)=18, \, 3(4)=12\).
\(\sum f_i d_i = 6\).
Mean \(\bar{x} = 92.5 + \left(\frac{6}{60}\right)5 = 92.5 + 0.5 = 93\).
\(f_i d_i^2\): \(3(16)=48, \, 4(9)=36, \, 7(4)=28, \, 7(1)=7, \, 15(0)=0, \, 9(1)=9, \, 6(4)=24, \, 6(9)=54, \, 3(16)=48\).
\(\sum f_i d_i^2 = 254\).
Variance \(\sigma^2 = 5^2 \left[ \frac{254}{60} - \left(\frac{6}{60}\right)^2 \right] = 25 \left[ 4.2333 - 0.01 \right] = 25 \times 4.2233 = 105.52\).
Standard deviation \(\sigma = \sqrt{105.52} = 10.27\).
In simple words: Applied shortcut step-deviation method to get Mean = 93, Variance = 105.52, and Standard Deviation = 10.27.

Exam Tip: Remember to multiply by \(h^2\) (which is \(5^2 = 25\)) when computing variance via step deviation.

Question (3). Calculate mean, variance and standard deviation for the following by short cut method.

Ans: 62, 201, \(\sqrt{201}\)
Answer:
Total frequency \(N = 3 + 7 + 12 + 15 + 8 + 3 + 2 = 50\).
Midpoints \(x_i\): 35, 45, 55, 65, 75, 85, 95.
Let \(A = 65, \, h = 10\). Step deviations \(d_i\): -3, -2, -1, 0, 1, 2, 3.
\(f_i d_i\): \(3(-3)=-9, \, 7(-2)=-14, \, 12(-1)=-12, \, 15(0)=0, \, 8(1)=8, \, 3(2)=6, \, 2(3)=6\).
\(\sum f_i d_i = -9 - 14 - 12 + 0 + 8 + 6 + 6 = -15\).
Mean \(\bar{x} = 65 + \left(\frac{-15}{50}\right)10 = 65 - 3 = 62\).
\(f_i d_i^2\): \(3(9)=27, \, 7(4)=28, \, 12(1)=12, \, 15(0)=0, \, 8(1)=8, \, 3(4)=12, \, 2(9)=18\).
\(\sum f_i d_i^2 = 27 + 28 + 12 + 0 + 8 + 12 + 18 = 105\).
Variance \(\sigma^2 = 10^2 \left[ \frac{105}{50} - \left(\frac{-15}{50}\right)^2 \right] = 100 \left[ 2.1 - (-0.3)^2 \right] = 100 \left[ 2.1 - 0.09 \right] = 100 \times 2.01 = 201\).
Standard deviation \(\sigma = \sqrt{201}\).
In simple words: Shortcut method yields Mean = 62, Variance = 201, and Standard Deviation = \(\sqrt{201}\).

Exam Tip: Step-deviation formula directly provides exact integer results for well-crafted textbook problems.

Question (4). Calculate the mean deviation from the median for the following data.

Ans: Median = 45, Mean deviation = 11.33
Answer:
Total frequency \(N = 4 + 6 + 10 + 20 + 10 + 6 + 4 = 60\), \(\frac{N}{2} = 30\).
Cumulative frequencies (c.f.): 4, 10, 20, 40, 50, 56, 60.
Median class is 40-50 since c.f. 40 exceeds 30.
\(l = 40, \, C = 20, \, f = 20, \, h = 10\).
Median \(M = 40 + \left(\frac{30 - 20}{20}\right) \times 10 = 40 + \left(\frac{10}{20}\right)10 = 40 + 5 = 45\).
Midpoints \(x_i\): 15, 25, 35, 45, 55, 65, 75.
Absolute deviations \(|x_i - 45|\):
\(|15 - 45| = 30 \implies 4 \times 30 = 120\)
\(|25 - 45| = 20 \implies 6 \times 20 = 120\)
\(|35 - 45| = 10 \implies 10 \times 10 = 100\)
\(|45 - 45| = 0 \implies 20 \times 0 = 0\)
\(|55 - 45| = 10 \implies 10 \times 10 = 100\)
\(|65 - 45| = 20 \implies 6 \times 20 = 120\)
\(|75 - 45| = 30 \implies 4 \times 30 = 120\)
\(\sum f_i |x_i - M| = 120 + 120 + 100 + 0 + 100 + 120 + 120 = 680\).
Mean Deviation from Median \(\text{M.D}(M) = \frac{680}{60} = \frac{68}{6} = 11.33\).
In simple words: Median is 45, and average absolute deviation from it is 11.33.

Exam Tip: Symmetric distributions like this often yield exact integers for median and clean repeating decimals for mean deviation.