These exercise 13.3 class 10 maths ncert solutions solve all seven questions of the Statistics chapter’s final exercise with complete working — not just final answers.
This is the exercise where grouped-data median, mean and mode finally come together: Question 1, Question 2 and Question 6 each mix two or three measures, Question 2 works backwards from a known median to find two missing frequencies, Question 3 hands you a less-than-type table, and Question 4 carries a printed hint about converting classes to continuous ones.
Finish this exercise and you have mastered Chapter 13.
What Exercise 13.3 Covers
These seven questions test you on the three measures of central tendency applied to grouped data — mean, median and mode. The one trick the exercise is built around is reading a frequency table correctly before you touch any formula.
- Q1, Q3, Q5, Q7 — straightforward drills: find the median (and in Q1, all three measures) directly from a grouped table.
- Q2 — the reversal: the median is given, so you work backwards to find two missing frequencies.
- Q4 — the continuous-classes trap: the classes have gaps, and the hint forces you to close them before using the formula.
- Q6 — the only question demanding median, mean and mode all together on one small table.
For the full chapter background — including how cumulative frequency tables and ogives are built — see the Class 10 Maths notes for Statistics.
Formulas You Need Before Starting This Exercise
All three formulas assume equal class sizes and continuous classes — that last point matters for Question 4 (NCERT, p. 201). Keep the formula sheet below open while you solve.
| Measure | Formula | When to use it |
|---|---|---|
| Median | \( \text{Median} = l + \left( \frac{n/2 – cf}{f} \right) \times h \) | Locate the class whose cumulative frequency first passes \( n/2 \) (NCERT, p. 194) |
| Mode | \( \text{Mode} = l + \left( \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \right) \times h \) | The modal class has the highest frequency; you need its two neighbours too (NCERT, p. 185) |
| Mean (direct) | \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \) | Small class marks and frequencies |
| Mean (assumed mean) | \( \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \) | Large values; pick a central class mark as \( a \) |
| Mean (step deviation) | \( \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \) | Large values and every deviation shares a common factor (NCERT, p. 180) |
| Empirical relation | \( 3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean} \) | Only as a quick check — it is exact only for a perfectly symmetric distribution (NCERT, p. 198) |
What every symbol means
| Symbol | Meaning |
|---|---|
| \( l \) | Lower limit of the median class (or modal class) |
| \( n \) | Total number of observations, \( \sum f_i \) |
| \( cf \) | Cumulative frequency of the class before the median class — the most common slip in this exercise |
| \( f \) | Frequency of the median class |
| \( h \) | Class size (assuming all classes are equal) |
| \( f_1 \) | Frequency of the modal class |
| \( f_0 \) | Frequency of the class preceding the modal class |
| \( f_2 \) | Frequency of the class succeeding the modal class |
| \( x_i \) | Class mark — the mid-point \( \frac{\text{upper} + \text{lower}}{2} \) of a class |
| \( d_i \) | Deviation \( x_i – a \) from the assumed mean |
| \( u_i \) | \( \frac{x_i – a}{h} \) |
Question 1: The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
| Monthly consumption (in units) | Number of consumers |
|---|---|
| 65 – 85 | 4 |
| 85 – 105 | 5 |
| 105 – 125 | 13 |
| 125 – 145 | 20 |
| 145 – 165 | 14 |
| 165 – 185 | 8 |
| 185 – 205 | 4 |
Concept: Mean, median and mode summarise the same grouped data from three angles — the balancing point, the middle value and the most frequent value respectively. Finding all three on one table and comparing them tells you the shape of the distribution.
Step 1: Build columns for the class mark \( x_i \), the product \( f_i x_i \), and the cumulative frequency \( cf \):
| Class (units) | \( f_i \) | \( x_i \) | \( f_i x_i \) | \( cf \) |
|---|---|---|---|---|
| 65 – 85 | 4 | 75 | 300 | 4 |
| 85 – 105 | 5 | 95 | 475 | 9 |
| 105 – 125 | 13 | 115 | 1495 | 22 |
| 125 – 145 | 20 | 135 | 2700 | 42 |
| 145 – 165 | 14 | 155 | 2170 | 56 |
| 165 – 185 | 8 | 175 | 1400 | 64 |
| 185 – 205 | 4 | 195 | 780 | 68 |
| Total | 68 | 9320 |
Step 2 — Mean (step deviation): The class marks are large, so take the assumed mean \( a = 135 \) with \( h = 20 \).
The deviations \( x_i – a \) give \( u_i \) values \( -3, -2, -1, 0, 1, 2, 3 \) and \( \sum f_i u_i = -12 – 10 – 13 + 0 + 14 + 16 + 12 = 7 \).
\[ \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h = 135 + \frac{7}{68} \times 20 = 135 + 2.06 = 137.06 \text{ units} \]
Step 3 — Median: With \( n = 68 \), \( n/2 = 34 \).
The cumulative frequency 42 first passes 34, so the median class is 125 – 145.
Here \( l = 125 \), \( cf = 22 \), \( f = 20 \), \( h = 20 \).
\[ \text{Median} = 125 + \frac{34 – 22}{20} \times 20 = 125 + 12 = 137 \text{ units} \]
Step 4 — Mode: The highest frequency is 20, so the modal class is 125 – 145 with \( f_1 = 20 \), \( f_0 = 13 \) (class 105 – 125) and \( f_2 = 14 \) (class 145 – 165).
\[ \text{Mode} = 125 + \frac{20 – 13}{2(20) – 13 – 14} \times 20 = 125 + \frac{7}{13} \times 20 = 125 + 10.77 = 135.77 \text{ units} \]
Step 5 — Compare: Median \( = 137 \) units, mean \( \approx 137.06 \) units and mode \( \approx 135.77 \) units all sit within about one unit of each other.
That closeness tells you the distribution is nearly symmetric — the typical consumer uses about 137 units of electricity per month, and the most common value is also close to that.
Final answer: Median = 137 units, mean ≈ 137.06 units, mode ≈ 135.77 units; all three are nearly equal, so the distribution is fairly symmetric.
Question 2: If the median of the distribution given below is 28.5, find the values of x and y.
| Class interval | Frequency |
|---|---|
| 0 – 10 | 5 |
| 10 – 20 | x |
| 20 – 30 | 20 |
| 30 – 40 | 15 |
| 40 – 50 | y |
| 50 – 60 | 5 |
| Total | 60 |
Concept: A known median pins down two independent facts. The total frequency gives you one equation, and the median value tells you which class is the median class, giving you the second. Two unknowns, two equations.
Step 1: The total is \( n = 60 \), so \( n/2 = 30 \).
Because the given median 28.5 lies inside 20 – 30, that is the median class: \( l = 20 \), \( f = 20 \), \( h = 10 \).
The cumulative frequency before this class is \( cf = 5 + x \).
Step 2: Substitute into the median formula:
\[ 28.5 = 20 + \frac{30 – (5 + x)}{20} \times 10 \]
\[ 8.5 = \frac{25 – x}{20} \times 10 \;\;\Rightarrow\;\; 8.5 = \frac{25 – x}{2} \;\;\Rightarrow\;\; 17 = 25 – x \;\;\Rightarrow\;\; x = 8 \]
Step 3: Now use the total frequency: \( 5 + x + 20 + 15 + y + 5 = 60 \), so \( x + y = 15 \).
With \( x = 8 \), we get \( y = 7 \).
Final answer: \( x = 8 \) and \( y = 7 \).
Question 3: A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
| Age (in years) | Number of policy holders |
|---|---|
| Below 20 | 2 |
| Below 25 | 6 |
| Below 30 | 24 |
| Below 35 | 45 |
| Below 40 | 78 |
| Below 45 | 89 |
| Below 50 | 92 |
| Below 55 | 98 |
| Below 60 | 100 |
Concept: A “less than” table lists cumulative frequencies at each upper limit, not the class frequencies themselves. So the frequency of each class is found by subtracting consecutive cumulative totals — the numbers 2, 6, 24, … are running totals, not per-class frequencies.
Step 1: Rebuild the ordinary classes.
The upper limits 20, 25, 30, …, 60 give the classes below 20, 20 – 25, 25 – 30, …, 55 – 60, and each frequency is a successive difference:
| Class | Frequency \( f \) | \( cf \) |
|---|---|---|
| Below 20 | 2 | 2 |
| 20 – 25 | \( 6 – 2 = 4 \) | 6 |
| 25 – 30 | \( 24 – 6 = 18 \) | 24 |
| 30 – 35 | \( 45 – 24 = 21 \) | 45 |
| 35 – 40 | \( 78 – 45 = 33 \) | 78 |
| 40 – 45 | \( 89 – 78 = 11 \) | 89 |
| 45 – 50 | \( 92 – 89 = 3 \) | 92 |
| 50 – 55 | \( 98 – 92 = 6 \) | 98 |
| 55 – 60 | \( 100 – 98 = 2 \) | 100 |
Step 2: With \( n = 100 \), \( n/2 = 50 \).
The cumulative frequency 78 first passes 50, so the median class is 35 – 40: \( l = 35 \), \( cf = 45 \), \( f = 33 \), \( h = 5 \).
\[ \text{Median} = 35 + \frac{50 – 45}{33} \times 5 = 35 + \frac{25}{33} = 35 + 0.76 = 35.76 \text{ years} \]
Final answer: The median age is 35.76 years. About half the policy holders are younger than roughly 35.8 years and half are older.
Question 4: The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :
| Length (in mm) | Number of leaves |
|---|---|
| 118 – 126 | 3 |
| 127 – 135 | 5 |
| 136 – 144 | 9 |
| 145 – 153 | 12 |
| 154 – 162 | 5 |
| 163 – 171 | 4 |
| 172 – 180 | 2 |
Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, …, 171.5 – 180.5.
Concept: The median formula assumes there is no gap between the end of one class and the start of the next. Here 126 and 127 are separate values with a gap of 1 unit, so every boundary shifts by half the gap — 0.5 — to close the classes (NCERT, p. 201).
Step 1: Convert the classes as the hint directs (each lower limit drops by 0.5, each upper limit rises by 0.5).
Frequencies stay unchanged.
The cumulative frequencies are 3, 8, 17, 29, 34, 38, 40.
Step 2: With \( n = 40 \), \( n/2 = 20 \).
The cumulative frequency 29 first passes 20, so the median class is 144.5 – 153.5: \( l = 144.5 \), \( cf = 17 \), \( f = 12 \), \( h = 153.5 – 144.5 = 9 \).
\[ \text{Median} = 144.5 + \frac{20 – 17}{12} \times 9 = 144.5 + \frac{27}{12} = 144.5 + 2.25 = 146.75 \text{ mm} \]
Final answer: The median length of the leaves is 146.75 mm.
Question 5: The following table gives the distribution of the life time of 400 neon lamps :
| Life time (in hours) | Number of lamps |
|---|---|
| 1500 – 2000 | 14 |
| 2000 – 2500 | 56 |
| 2500 – 3000 | 60 |
| 3000 – 3500 | 86 |
| 3500 – 4000 | 74 |
| 4000 – 4500 | 62 |
| 4500 – 5000 | 48 |
Concept: The median is the value that splits the data into two equal halves — here, the life time that divides 400 lamps into two equal groups of 200. You simply locate the class containing the 200th lamp.
Step 1: Build the cumulative frequency column: 14, 70, 130, 216, 290, 352, 400.
With \( n = 400 \), \( n/2 = 200 \).
Step 2: The cumulative frequency 216 first passes 200, so the median class is 3000 – 3500: \( l = 3000 \), \( cf = 130 \), \( f = 86 \), \( h = 500 \).
\[ \text{Median} = 3000 + \frac{200 – 130}{86} \times 500 = 3000 + \frac{70}{86} \times 500 = 3000 + 406.98 = 3406.98 \text{ hours} \]
Final answer: The median life time of a lamp is 3406.98 hours — about half the lamps last less than this and half last longer.
Question 6: 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
| Number of letters | 1 – 4 | 4 – 7 | 7 – 10 | 10 – 13 | 13 – 16 | 16 – 19 |
|---|---|---|---|---|---|---|
| Number of surnames | 6 | 30 | 40 | 16 | 4 | 4 |
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Concept: One table, three questions — three different tools. The mean uses class marks and products, the median uses cumulative frequency, and the mode uses the modal class together with its two neighbouring frequencies. Keep three separate columns.
Step 1 — Mean: Class marks are 2.5, 5.5, 8.5, 11.5, 14.5, 17.5.
Then \( \sum f_i x_i = 6(2.5) + 30(5.5) + 40(8.5) + 16(11.5) + 4(14.5) + 4(17.5) = 15 + 165 + 340 + 184 + 58 + 70 = 832 \).
\[ \bar{x} = \frac{832}{100} = 8.32 \text{ letters} \]
Step 2 — Median: Cumulative frequencies are 6, 36, 76, 92, 96, 100.
With \( n/2 = 50 \), the median class is 7 – 10: \( l = 7 \), \( cf = 36 \), \( f = 40 \), \( h = 3 \).
\[ \text{Median} = 7 + \frac{50 – 36}{40} \times 3 = 7 + \frac{42}{40} = 7 + 1.05 = 8.05 \text{ letters} \]
Step 3 — Mode: The highest frequency is 40, so the modal class is 7 – 10 with \( f_1 = 40 \), \( f_0 = 30 \), \( f_2 = 16 \).
\[ \text{Mode} = 7 + \frac{40 – 30}{2(40) – 30 – 16} \times 3 = 7 + \frac{10}{34} \times 3 = 7 + 0.88 = 7.88 \text{ letters} \]
Final answer: Mean = 8.32 letters, median = 8.05 letters, modal size = 7.88 letters. Typical surnames have about 8 letters.
Question 7: The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
| Weight (in kg) | 40 – 45 | 45 – 50 | 50 – 55 | 55 – 60 | 60 – 65 | 65 – 70 | 70 – 75 |
|---|---|---|---|---|---|---|---|
| Number of students | 2 | 3 | 8 | 6 | 6 | 3 | 2 |
Concept: With 30 observations, the median falls between the 15th and 16th student’s weight, so you locate the class holding the 15th value. The cumulative frequency column does the locating for you.
Step 1: Build the cumulative frequencies: 2, 5, 13, 19, 25, 28, 30.
With \( n = 30 \), \( n/2 = 15 \).
Step 2: The cumulative frequency 19 first passes 15, so the median class is 55 – 60: \( l = 55 \), \( cf = 13 \), \( f = 6 \), \( h = 5 \).
\[ \text{Median} = 55 + \frac{15 – 13}{6} \times 5 = 55 + \frac{10}{6} = 55 + 1.67 = 56.67 \text{ kg} \]
Final answer: The median weight of the students is 56.67 kg.
Choosing the Right Measure: Mean, Median or Mode
The chapter closes by asking when each measure is the best representative of a data set (NCERT, pp. 197–198). Exam answers should state the property and the situation together.
- Mean uses every observation, so it is the natural choice for fairly even data, but extreme values drag it away from the majority — so it fails when one class is tiny and the others are large.
- Median gives the “typical” middle value and shrugs off outliers and skewed frequencies — best for typical productivity, average wage, or anything with extreme values.
- Mode finds the most frequent value — best for popularity questions like the most-watched TV programme or the most demanded consumer item.
The empirical relation \( 3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean} \) links the three, and you can use it to check whether your three answers are consistent (NCERT, p. 198).
Everything in the chosen chapter feeds the rest of your course: you met the Surface Areas and Volumes solutions just before this chapter, and you will go on to Probability solutions after it. For every subject, see the Class 10 notes and the main notes hub.
Common Mistakes in Exercise 13.3 (and How to Catch Them)
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Using \( n/2 = 28.5 \) (the median value) in Question 2 | \( n/2 \) is half the total frequency — here \( 60/2 = 30 \) — while 28.5 is the median value you place inside the 20 – 30 class | \( n/2 \) must be a number between 0 and \( n \), matching the class you picked |
| Feeding the cumulative numbers 2, 6, 24, … of the less-than table in as class frequencies (Q3) | Real frequencies are successive differences: 6 − 2 = 4, 24 − 6 = 18, and so on | The frequency column must add back up to \( n = 100 \) |
| Using 118 – 126 without converting to continuous classes (Q4) | Shift every boundary by half the gap (0.5) — classes become 117.5 – 126.5 and so on | Use \( l = 144.5 \) and \( h = 9 \); the median must sit inside 144.5 – 153.5 |
| Using the median class’s own cumulative frequency as \( cf \) | \( cf \) is the cumulative frequency of the class before the median class | In Q1, for median class 125 – 145, \( cf = 22 \), not 42 |
| Reusing one column for both mean and median (Q6) | Mean uses class marks and products \( f_i x_i \); median uses cumulative frequency | Label the columns — they answer different questions |
Frequently Asked Questions
Which mean method should I choose — direct, assumed mean or step deviation?
If the class marks and frequencies are small, use the direct method. If they are large, use the assumed mean or step-deviation method (NCERT, p. 180). All three give the same result — they are just different amounts of arithmetic.
The assumed mean \( a \) is any convenient class mark (usually a central one), and the step-deviation method adds the bonus of dividing every deviation by the class size when the deviations all share a common factor. For Question 1, the class marks are large, so step deviation keeps the multiplication small.
Why does question 4 ask me to convert the classes to continuous ones before finding the median?
Because the median formula assumes continuous classes — there must be no gap between the upper limit of one class and the lower limit of the next (NCERT, p. 201). In Question 4 the class 118 – 126 is followed by 127 – 135, a gap of 1 unit, so the formula would use the wrong boundaries.
Moving every boundary by half the gap, 0.5, closes the classes into 117.5 – 126.5, 126.5 – 135.5, and so on. Then the median comes out as 146.75 mm.
In question 2, how do I use the fact that the median is 28.5 to find the values of x and y?
The given median pins down two equations. First, 28.5 lies inside the 20 – 30 class, so that is the median class with \( l = 20 \), \( f = 20 \), \( cf = 5 + x \), \( h = 10 \). Substituting 28.5 into the median formula gives \( x = 8 \). Second, the total frequency is 60, so \( 5 + x + 20 + 15 + y + 5 = 60 \), which gives \( x + y = 15 \) and hence \( y = 7 \).
Can I use 3 Median = Mode + 2 Mean to check my answers in this exercise?
Yes, but treat it as a rough check, not a proof. The relation \( 3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean} \) holds exactly only for a perfectly symmetric distribution; for nearly symmetric ones such as Question 1 and Question 6 it comes out very close (NCERT, p. 198).
If your three measures satisfy it closely, your arithmetic is almost certainly right; if they are far apart, one of the three is wrong.
Every question and table on this page is reproduced from the official NCERT Class 10 Mathematics textbook, and you can verify any table, formula or the exact wording of this exercise in the Statistics chapter PDF hosted by NCERT on the ncert.nic.in website.
Reference: NCERT Class 10 Mathematics textbook, chapter 13 Statistics.
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