Looking for exercise 13.1 class 10 maths NCERT solutions? This page carries all nine questions of the exercise exactly as printed, each solved step by step with the mean method that fits the data. For every question you get the class marks, the worked frequency table, the formula applied and the final mean with its unit.
These solutions follow the textbook’s “Mean of Grouped Data” section, so the direct method, the assumed mean method and the step-deviation method all appear, with the reason for the choice stated wherever the question asks for it.
Exercise 13.1 Class 10 Maths NCERT Solutions: What This Exercise Teaches
Exercise 13.1 applies one idea, the mean of grouped data, to nine frequency tables. Each class interval is represented by its class mark, the mid-point of the class, because the textbook assumes the frequency of a class is centred at its mid-point (NCERT, p. 173).
The three methods from section 13.2 are all used in this exercise:
- direct method: \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\);
- assumed mean method: \(\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}\), where \(d_i = x_i – a\);
- step-deviation method: \(\bar{x} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right)\), where \(u_i = \frac{x_i – a}{h}\).
Questions 1, 2, 4, 6, 7 and 9 ask for a plain mean, and question 5 also asks you to name your method. Question 3 runs the formula backwards to find a missing frequency, while question 8 mixes unequal class widths.
Two conventions matter throughout. A value on the upper boundary of a class is counted in the next class (NCERT, p. 173), and the grouped mean is approximate, because every observation in a class is replaced by its class mark.
In the textbook’s Example 1, the same 30 marks give 59.3 as ungrouped data but 62 as grouped data; the textbook calls 59.3 the exact mean and 62 the approximate mean (NCERT, p. 174).
Every question and table on this page is reproduced from the official Class 10 Mathematics textbook, so you can cross-check any entry against the source by opening Chapter 13 Statistics in the official NCERT textbook PDF, where Exercise 13.1 is printed on pages 181-183.
Important Concepts: Class Marks and the Three Mean Formulas
A grouped table hides the individual values, so each class interval is replaced by a single representative number, the class mark:
\[ \text{Class mark} = \frac{\text{Upper limit} + \text{Lower limit}}{2} \]
This class mark is the \(x_i\) used in every formula below. The class mark idea and the mid-point assumption come from section 13.2 of the textbook (NCERT, p. 173).
| Method | Formula | Use it when |
|---|---|---|
| Direct method | \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\) | \(x_i\) and \(f_i\) are small; safest when class sizes are unequal |
| Assumed mean method | \(\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}\), where \(d_i = x_i – a\) | \(x_i\) are large; take \(a\) as a central class mark |
| Step-deviation method | \(\bar{x} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right)\), where \(u_i = \frac{x_i – a}{h}\) | all deviations share a common factor \(h\); removes decimals and large numbers |
The textbook’s Remark gives the rule to quote in the exam (NCERT, p. 180): use the direct method when \(x_i\) and \(f_i\) are sufficiently small; use the assumed mean or step-deviation method when the numbers are numerically large; and if class sizes are unequal, step deviation still works by taking \(h\) as a suitable divisor of all the deviations.
All three methods produce the same mean, and the assumed mean and step-deviation methods are simplified forms of the direct method (NCERT, p. 180).
Two habits save marks: always form the class mark before subtracting or multiplying, and always check the final mean lies between the smallest and largest class marks, because the mean of any data lies between the extremes (NCERT, p. 197).
Question 1: A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. Which method did you use for finding the mean, and why?
| Number of plants | 0 – 2 | 2 – 4 | 4 – 6 | 6 – 8 | 8 – 10 | 10 – 12 | 12 – 14 |
|---|---|---|---|---|---|---|---|
| Number of houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |
Concept: In grouped data the individual values are hidden, so each class is represented by its class mark, the mid-point of the interval. The mark for 0 – 2 is \((0+2)/2 = 1\), for 2 – 4 it is 3, and so on up to 13. Each frequency tells us how many houses belong to that mark, which is exactly what the direct formula needs.
Since both the marks and the frequencies are small, the direct method is the natural choice (NCERT, p. 180).
Step 1: Find every class mark and the products \(f_i x_i\).
| Class | Class mark \(x_i\) | Frequency \(f_i\) | \(f_i x_i\) |
|---|---|---|---|
| 0 – 2 | 1 | 1 | 1 |
| 2 – 4 | 3 | 2 | 6 |
| 4 – 6 | 5 | 1 | 5 |
| 6 – 8 | 7 | 5 | 35 |
| 8 – 10 | 9 | 6 | 54 |
| 10 – 12 | 11 | 2 | 22 |
| 12 – 14 | 13 | 3 | 39 |
| Total | \(\sum f_i = 20\) | \(\sum f_i x_i = 162\) |
Step 2: Apply the direct formula.
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20} = 8.1 \]
Final answer: The mean number of plants per house is 8.1. On average, a house in this locality has 8.1 plants.
Common error: using the class limits (0, 2, 4, …) as the \(x_i\) values instead of the mid-points. Also remember the boundary convention: a house with exactly 2 plants is counted in the class 2 – 4, not 0 – 2 (NCERT, p. 173).
Check: the mean of any data lies between the extremes, and 8.1 does lie between the smallest mark (1) and the largest mark (13).
Question 2: Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.
| Daily wages (in ₹) | 500 – 520 | 520 – 540 | 540 – 560 | 560 – 580 | 580 – 600 |
|---|---|---|---|---|---|
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Concept: The wages lie in the hundreds (510 to 590), so direct products get tedious. The assumed mean method shrinks the numbers by subtracting a fixed value \(a\) from every class mark, and step deviation then divides those differences by the common class size \(h = 20\). The mean is unchanged because \(a\) is added back at the end.
Step 1: Class marks: 510, 530, 550, 570, 590.
Take the central mark \(a = 550\) and form \(u_i = (x_i – 550)/20\).
Step 2: Multiply each \(u_i\) by its frequency and add the column.
| Class | Class mark \(x_i\) | \(f_i\) | \(u_i\) | \(f_i u_i\) |
|---|---|---|---|---|
| 500 – 520 | 510 | 12 | -2 | -24 |
| 520 – 540 | 530 | 14 | -1 | -14 |
| 540 – 560 | 550 | 8 | 0 | 0 |
| 560 – 580 | 570 | 6 | 1 | 6 |
| 580 – 600 | 590 | 10 | 2 | 20 |
| Total | 50 | -12 |
Step 3: Apply the step-deviation formula.
\[ \bar{x} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right) = 550 + 20\left(\frac{-12}{50}\right) = 550 – 4.8 = 545.2 \]
Final answer: The mean daily wage of the 50 workers is ₹545.20.
Common error: subtracting the assumed mean from the class limits (500, 520, …) instead of from the class marks. The deviation is always \(d_i = x_i – a\), where \(x_i\) is the class mark.
Check: 545.2 lies between the extremes 510 and 590. The direct method confirms it: \(\sum f_i x_i = 27260\), so \(27260/50 = 545.2\).
Question 3: The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.
| Daily pocket allowance (in ₹) | 11 – 13 | 13 – 15 | 15 – 17 | 17 – 19 | 19 – 21 | 21 – 23 | 23 – 25 |
|---|---|---|---|---|---|---|---|
| Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |
Concept: This problem runs the mean formula backwards. The unknown frequency \(f\) appears twice: it adds \(f\) children to the total count \(\sum f_i\), and it adds \(20f\) to \(\sum f_i x_i\), because the mark of the class 19 – 21 is 20. So one equation in \(f\) must be formed and solved.
Step 1: Class marks: 12, 14, 16, 18, 20, 22, 24.
Complete the product column.
| Class | Class mark \(x_i\) | \(f_i\) | \(f_i x_i\) |
|---|---|---|---|
| 11 – 13 | 12 | 7 | 84 |
| 13 – 15 | 14 | 6 | 84 |
| 15 – 17 | 16 | 9 | 144 |
| 17 – 19 | 18 | 13 | 234 |
| 19 – 21 | 20 | f | 20f |
| 21 – 23 | 22 | 5 | 110 |
| 23 – 25 | 24 | 4 | 96 |
| Total | 44 + f | 752 + 20f |
- Step 1: The mean is given as 18, so \[ 18 = \frac{752 + 20f}{44 + f} \]
- Step 2: Cross-multiply and solve.
\[ 18(44 + f) = 752 + 20f \;\Rightarrow\; 792 + 18f = 752 + 20f \;\Rightarrow\; 2f = 40 \;\Rightarrow\; f = 20 \]
Final answer: The missing frequency is \(f = 20\).
Common error: forgetting that \(f\) also increases \(\sum f_i\). Using 44 instead of \(44 + f\) in the denominator gives \(f = 2\), which fails the check.
Check: put \(f = 20\) back in: \(\sum f_i = 64\) and \(\sum f_i x_i = 752 + 400 = 1152\), and \(1152/64 = 18\). The mean comes out exactly as stated.
Question 4: Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.
| Number of heartbeats per minute | 65 – 68 | 68 – 71 | 71 – 74 | 74 – 77 | 77 – 80 | 80 – 83 | 83 – 86 |
|---|---|---|---|---|---|---|---|
| Number of women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Concept: The classes are continuous and equal in width (3), and all values are small, so the direct method is clean. The only trap is the first class mark: 65 – 68 has mid-point \((65+68)/2 = 66.5\), not 66.
Step 1: Class marks: 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5.
Build the product column.
| Class | Class mark \(x_i\) | \(f_i\) | \(f_i x_i\) |
|---|---|---|---|
| 65 – 68 | 66.5 | 2 | 133 |
| 68 – 71 | 69.5 | 4 | 278 |
| 71 – 74 | 72.5 | 3 | 217.5 |
| 74 – 77 | 75.5 | 8 | 604 |
| 77 – 80 | 78.5 | 7 | 549.5 |
| 80 – 83 | 81.5 | 4 | 326 |
| 83 – 86 | 84.5 | 2 | 169 |
| Total | 30 | 2277 |
Step 2: Apply the direct formula.
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2277}{30} = 75.9 \]
Final answer: The mean number of heartbeats per minute for these women is 75.9.
Common error: writing 66 instead of 66.5 as the first class mark. The mark is always the average of the two limits, so a half appears whenever the two limits add to an odd number.
Check: 75.9 lies between 66.5 and 84.5, and step deviation with \(a = 75.5\), \(h = 3\) gives the same value: \(\bar{x} = 75.5 + 3(4/30) = 75.9\). Two methods agreeing is the strongest sign of a correct mean.
Question 5: In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
| Number of mangoes | 50 – 52 | 53 – 55 | 56 – 58 | 59 – 61 | 62 – 64 |
|---|---|---|---|---|---|
| Number of boxes | 15 | 110 | 135 | 115 | 25 |
Concept: The intervals 50 – 52, 53 – 55 and so on are not adjacent, because the value 53 falls between 52 and 53. Gaps like this do not affect the mean, since the mean uses only the class marks, and each mark is the average of its own limits: 51, 54, 57, 60, 63.
Step 1: The marks are equally spaced 3 apart and the frequencies are large, so step deviation keeps the arithmetic small.
Take \(a = 57\), \(h = 3\) and form \(u_i = (x_i – 57)/3\).
| Class | Class mark \(x_i\) | \(f_i\) | \(u_i\) | \(f_i u_i\) |
|---|---|---|---|---|
| 50 – 52 | 51 | 15 | -2 | -30 |
| 53 – 55 | 54 | 110 | -1 | -110 |
| 56 – 58 | 57 | 135 | 0 | 0 |
| 59 – 61 | 60 | 115 | 1 | 115 |
| 62 – 64 | 63 | 25 | 2 | 50 |
| Total | 400 | 25 |
Step 2: Apply the step-deviation formula.
\[ \bar{x} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right) = 57 + 3\left(\frac{25}{400}\right) = 57 + 0.1875 = 57.1875 \]
Final answer: The mean number of mangoes kept in a packing box is 57.19 (rounded to two decimal places). Method chosen: step deviation, because the class marks are equally spaced and the frequencies are large.
Common error: assuming non-continuous intervals break the method. Continuity matters for the mode and median formulas, not for the mean. Another slip is adding the gap into a class; the mark of 50 – 52 stays 51.
Check: 57.19 lies between 51 and 63. The direct method confirms it: \(\sum f_i x_i = 22875\), and \(22875/400 = 57.1875\).
Question 6: The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.
| Daily expenditure (in ₹) | 100 – 150 | 150 – 200 | 200 – 250 | 250 – 300 | 300 – 350 |
|---|---|---|---|---|---|
| Number of households | 4 | 5 | 12 | 2 | 2 |
Concept: Expenditure values run from 125 to 325 rupees, large enough that direct products are clumsy. The assumed mean method subtracts a fixed number \(a\) from every class mark first, computes the mean of the smaller deviations \(d_i = x_i – a\), and adds \(a\) back: \(\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}\).
Step 1: Class marks: 125, 175, 225, 275, 325.
Take the central mark \(a = 225\).
Step 2: The key column is the deviation \(d_i = x_i – a\), computed from the class mark before anything is multiplied by the frequency.
| Class | Class mark \(x_i\) | \(f_i\) | \(d_i\) | \(f_i d_i\) |
|---|---|---|---|---|
| 100 – 150 | 125 | 4 | -100 | -400 |
| 150 – 200 | 175 | 5 | -50 | -250 |
| 200 – 250 | 225 | 12 | 0 | 0 |
| 250 – 300 | 275 | 2 | 50 | 100 |
| 300 – 350 | 325 | 2 | 100 | 200 |
| Total | 25 | -350 |
Step 3: Substitute into the formula.
\[ \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 225 + \frac{-350}{25} = 225 – 14 = 211 \]
Final answer: The mean daily expenditure on food is ₹211.
Common error: writing the deviation as \(a – x_i\), or subtracting \(a\) from the class limits instead of the class marks. Either mistake shifts every product and corrupts the mean. The mean is independent of the choice of \(a\), but the column is cleanest when \(a\) is a central class mark (NCERT, p. 176).
Check: 211 lies between 125 and 325. Direct method: \(\sum f_i x_i = 5275\), so \(5275/25 = 211\).
Question 7: To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below: Find the mean concentration of SO₂ in the air.
| Concentration of SO₂ (in ppm) | Frequency |
|---|---|
| 0.00 – 0.04 | 4 |
| 0.04 – 0.08 | 9 |
| 0.08 – 0.12 | 9 |
| 0.12 – 0.16 | 2 |
| 0.16 – 0.20 | 4 |
| 0.20 – 0.24 | 2 |
Concept: The decimal widths (0.04) look awkward, but they are equal, which is exactly what step deviation needs. Dividing each deviation by \(h = 0.04\) turns the decimal marks into the simple integers -2, -1, 0, 1, 2, 3 before any multiplication by frequency.
Step 1: Class marks: 0.02, 0.06, 0.10, 0.14, 0.18, 0.22.
Take \(a = 0.10\) (the mark of the middle class) and \(h = 0.04\).
| Class | Class mark \(x_i\) | \(f_i\) | \(u_i\) | \(f_i u_i\) |
|---|---|---|---|---|
| 0.00 – 0.04 | 0.02 | 4 | -2 | -8 |
| 0.04 – 0.08 | 0.06 | 9 | -1 | -9 |
| 0.08 – 0.12 | 0.10 | 9 | 0 | 0 |
| 0.12 – 0.16 | 0.14 | 2 | 1 | 2 |
| 0.16 – 0.20 | 0.18 | 4 | 2 | 8 |
| 0.20 – 0.24 | 0.22 | 2 | 3 | 6 |
| Total | 30 | -1 |
Step 2: Apply the step-deviation formula.
\[ \bar{x} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right) = 0.10 + 0.04\left(\frac{-1}{30}\right) = 0.10 – 0.0013 = 0.0987 \]
Final answer: The mean concentration of SO₂ in the air is 0.0987 ppm, which rounds to 0.10 ppm.
Common error: writing the first class mark as 0.2 instead of 0.02. The mark is \((0.00 + 0.04)/2 = 0.02\), so a quick place-value check prevents the mistake.
Check: round only at the final step. The unrounded value 0.0987 lies between the extremes 0.02 and 0.22, and the direct method agrees: \(\sum f_i x_i = 2.96\), so \(2.96/30 = 0.0987\).
Question 8: A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
| Number of days | 0 – 6 | 6 – 10 | 10 – 14 | 14 – 20 | 20 – 28 | 28 – 38 | 38 – 40 |
|---|---|---|---|---|---|---|---|
| Number of students | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Concept: Here the class widths are unequal: 6, 4, 4, 6, 8, 10 and 2 days. Step deviation needs one common divisor of all the deviations, and these marks (3, 8, 12, 17, 24, 33, 39) share none, so direct multiplication is the safest route. The class mark is always the average of the two limits, for example \((20+28)/2 = 24\) and \((38+40)/2 = 39\).
Step 1: Build the product column.
| Class | Class mark \(x_i\) | \(f_i\) | \(f_i x_i\) |
|---|---|---|---|
| 0 – 6 | 3 | 11 | 33 |
| 6 – 10 | 8 | 10 | 80 |
| 10 – 14 | 12 | 7 | 84 |
| 14 – 20 | 17 | 4 | 68 |
| 20 – 28 | 24 | 4 | 96 |
| 28 – 38 | 33 | 3 | 99 |
| 38 – 40 | 39 | 1 | 39 |
| Total | 40 | 499 |
Step 2: Apply the direct formula.
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{499}{40} = 12.475 \]
Final answer: The mean number of days a student was absent is 12.48 days (rounded to two decimal places).
Common error: guessing the mark of 20 – 28 by eye; students often take 26. The correct mark is the average \((20+28)/2 = 24\). Every mark follows the same rule, whatever the class width.
Check: 12.48 lies between the smallest mark (3) and the largest (39). An assumed mean with \(a = 12\) gives \(\sum f_i d_i = 19\), so \(\bar{x} = 12 + 19/40 = 12.475\), the same answer.
Question 9: The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
| Literacy rate (in %) | 45 – 55 | 55 – 65 | 65 – 75 | 75 – 85 | 85 – 95 |
|---|---|---|---|---|---|
| Number of cities | 3 | 10 | 11 | 8 | 3 |
Concept: The class size is a constant 10 and the class marks are round values (50, 60, 70, 80, 90), so both the direct and step-deviation methods are tidy. Direct method is chosen here because the marks and the frequencies are small; step deviation would give the identical mean.
Step 1: The mark of 45 – 55 is \((45+55)/2 = 50\), not 55.
Complete the product column.
| Class | Class mark \(x_i\) | \(f_i\) | \(f_i x_i\) |
|---|---|---|---|
| 45 – 55 | 50 | 3 | 150 |
| 55 – 65 | 60 | 10 | 600 |
| 65 – 75 | 70 | 11 | 770 |
| 75 – 85 | 80 | 8 | 640 |
| 85 – 95 | 90 | 3 | 270 |
| Total | 35 | 2430 |
Step 2: Apply the direct formula.
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2430}{35} = 69.43 \]
Final answer: The mean literacy rate of the 35 cities is 69.43%.
Common error: using 55 as the first class mark by reading the upper limit of 45 – 55 instead of averaging the limits. The mark is 50, and taking 55 pulls the whole mean up.
Check: 69.43 lies between 50 and 90. Keeping the unit in the final line, 69.43%, completes the answer.
Method Recap: How to Choose Between the Three Mean Methods
The textbook’s Remark (NCERT, p. 180) is the rule to remember: all three methods always give the same mean, so the choice depends on the numbers in front of you. This table maps every question in Exercise 13.1 to the method used here and the reason for it.
| Question | Method used | Reason |
|---|---|---|
| Q1 Plants per house | Direct | Class marks 1-13 and frequencies are small |
| Q2 Daily wages | Step deviation | Wages in hundreds; common class size \(h = 20\) |
| Q3 Missing frequency \(f\) | Direct formula, solved backwards | Mean is given; form one equation in \(f\) |
| Q4 Heartbeats | Direct (step deviation also works) | Small marks and frequencies; equal width 3 |
| Q5 Mangoes per box | Step deviation | Marks equally spaced 3 apart; large frequencies; works despite non-continuous classes |
| Q6 Daily expenditure | Assumed mean | Values in hundreds; central mark \(a = 225\) |
| Q7 SO₂ concentration | Step deviation | Decimal width 0.04; dividing removes the decimals |
| Q8 Days absent | Direct | Unequal widths; the deviations have no common divisor |
| Q9 Literacy rate | Direct | Round marks (50-90) and small frequencies |
Decision guide in three lines
- Small \(x_i\) and \(f_i\) go to the direct method (Q1, Q4, Q9).
- Large \(x_i\) or \(f_i\) go to the assumed mean or step-deviation method (Q2, Q5, Q6).
- Equal widths or decimals suit step deviation with a common \(h\) (Q2, Q5, Q7); unequal widths with no common divisor stay with the direct method (Q8).
Common mistakes and how to catch them
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Using class limits (2, 4, 6) instead of class marks | \(x_i = \frac{\text{upper limit} + \text{lower limit}}{2}\): always average the two limits | Recompute one mark: \((0+2)/2 = 1\), so a mark is never a limit itself |
| Forgetting the missing \(f\) belongs in \(\sum f_i\) too | Both numerator and denominator carry \(f\): \(\bar{x} = \frac{752 + 20f}{44 + f}\) | Substitute \(f = 20\) back: \(1152/64 = 18\) |
| Multiplying by frequency before subtracting the assumed mean | First form \(d_i = x_i – a\) from the class mark, then compute \(f_i d_i\) | The \(d_i\) column must be 0 at the chosen class mark |
| Writing 66 instead of 66.5, or 0.2 instead of 0.02, for a class mark | The mark is the average of the two limits; decimals and place values follow from the limits | The final mean must lie between the smallest and largest class marks |
For revision, the Class 10 Maths notes on this site cover every chapter, including Chapter 12 Surface Areas and Volumes and Chapter 14 Probability. From the Class 10 notes hub you can reach every subject, and the complete notes index lists material for all classes.
The chapter continues in section 13.3 with the mode of grouped data. For Exercise 13.1, the checking habit to carry forward is the one used throughout: every mean must lie between the smallest and largest class marks.
Frequently Asked Questions on Exercise 13.1
Why did the textbook example give 59.3 for the ungrouped data but 62 for the grouped data?
Because the grouped calculation replaces every actual observation by the class mark of its class. The 30 real marks in the textbook’s Example 1 give exactly 59.3; grouping the same marks and using mid-points gives 62. The textbook calls 59.3 the exact mean and 62 the approximate mean (NCERT, p. 174).
The mid-point assumption (NCERT, p. 173) is the cause, and 62 is the best estimate available when the data exists only in grouped form.
Can the step deviation method be used when the class sizes are unequal?
Yes. The textbook’s Remark (NCERT, p. 180) allows step deviation for unequal class sizes as long as \(h\) is a suitable divisor of all the deviations \(d_i\). Question 8 shows the other side: the marks 3, 8, 12, 17, 24, 33, 39 share no common divisor, so the direct method is simpler there.
How do I find a missing frequency when the mean is already given?
Write the mean formula with \(f\) in both places: \(\bar{x} = \frac{\text{known sum of } f_i x_i + f \times \text{class mark}}{\text{known } \sum f_i + f}\). This gives one linear equation, as in question 3: \(18 = \frac{752 + 20f}{44 + f}\). Solve for \(f\), then substitute the value back into the formula to confirm the mean.
Which method of finding the mean should I choose in the exam?
Follow the textbook’s Remark (NCERT, p. 180): direct when \(x_i\) and \(f_i\) are small; assumed mean or step deviation when the numbers are large; step deviation whenever all deviations share a common factor.
All three give the same mean, so the method is your choice, but questions 1 and 5 ask for the reason, so state it in terms of the size of the numbers.
Do non-continuous class intervals like 50-52 and 53-55 affect the mean?
No. The mean uses only the class marks, and each mark is the average of its own limits: \((50+52)/2 = 51\), \((53+55)/2 = 54\), and so on. The gap between 52 and 53 changes nothing. Continuity must be ensured for the mode and median formulas, but the mean methods of this exercise work on any grouped table.
Reference: NCERT Class 10 Mathematics textbook, chapter Statistics.
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