Exercise 9.2 — Mean, Median and Mode
Mean, median and mode of ungrouped frequency distributions.
What are Measures of Central Tendency?
When we collect a large set of data — marks, heights, salaries, weights — we need a single value that best represents the entire dataset. This representative value is called a measure of central tendency. It tells us where the "centre" of the data lies. In Class 9 Statistics, three such measures are studied:
The sum of all observations divided by the number of observations. It uses every value in the data.
The middle value when data is arranged in order. It is not affected by very large or very small values.
The value that appears most frequently in the data. A dataset can have one, two, or no mode.
1. Arithmetic Mean — Formulas and Concepts
Mean of Raw (Ungrouped) Data
When observations are listed directly without frequencies, the mean is simply the total divided by the count.
Mean (x̄) = (x₁ + x₂ + x₃ + … + xₙ) / n = (Sum of all observations) / (Number of observations)
Example from the textbook: Data: 4, 3, 5, 5, 4, 9, 7, 7, 8, 7, 4, 3, 7, 5, 8, 8, 8, 7, 4, 7 (20 observations)
Mean of Frequency Distribution
When the same values repeat, we use a frequency table. Each value xᵢ appears fᵢ times, so we multiply them to get the weighted sum:
Mean (x̄) = (Σfᵢxᵢ) / (Σfᵢ) = (f₁x₁ + f₂x₂ + … + fₙxₙ) / (f₁ + f₂ + … + fₙ)
Using the same data grouped by frequency: values 3,4,5,7,8,9 appear 2,4,3,6,4,1 times respectively.
Question 1 — Mean Weight of Parcels
| Weight xᵢ (kg) | No. of Parcels fᵢ | fᵢxᵢ |
|---|---|---|
| 50 | 25 | 1250 |
| 65 | 34 | 2210 |
| 75 | 38 | 2850 |
| 90 | 40 | 3600 |
| 110 | 47 | 5170 |
| 120 | 16 | 1920 |
| Total | Σfᵢ = 200 | Σfᵢxᵢ = 17000 |
Question 2 — Mean Number of Children per Family
| No. of Children xᵢ | No. of Families fᵢ | fᵢxᵢ |
|---|---|---|
| 0 | 11 | 0 |
| 1 | 25 | 25 |
| 2 | 32 | 64 |
| 3 | 10 | 30 |
| 4 | 5 | 20 |
| 5 | 1 | 5 |
| Total | Σfᵢ = 84 | Σfᵢxᵢ = 144 |
Question 3 — Finding the Unknown Frequency 'K'
| x | f | fx |
|---|---|---|
| 2 | 4 | 8 |
| 4 | 7 | 28 |
| 6 | 10 | 60 |
| 8 | 16 | 128 |
| 10 | K | 10K |
| 12 | 3 | 36 |
| Total | Σf = 40 + K | Σfx = 260 + 10K |
Setting up the equation:
Question 4 — Average Village Population (Census 2011)
| Population xᵢ (thousands) | No. of Villages fᵢ | fᵢxᵢ |
|---|---|---|
| 12 | 20 | 240 |
| 5 | 15 | 75 |
| 30 | 32 | 960 |
| 20 | 35 | 700 |
| 15 | 36 | 540 |
| 8 | 7 | 56 |
| Total | Σfᵢ = 145 | Σfᵢxᵢ = 2571 |
Question 5 — AFLATOUN School Savings Programme (Hyderabad)
| Mandal | No. of Schools | Total Saved (₹) | Mean per School (₹) |
|---|---|---|---|
| Amberpet | 6 | 2154 | 359 |
| Thirumalagiri | 6 | 2478 | 413 |
| Saidabad | 5 | 975 | 195 |
| Khairathabad | 4 | 912 | 228 |
| Secunderabad | 3 | 600 | 200 |
| Bahadurpura | 9 | 7533 | 837 |
| All Mandals | 33 | 14,652 | 444 |
Calculations:
2. Median — Finding the Middle Value
The median is the middle observation of a dataset arranged in ascending (or descending) order. It divides the data into two equal halves — one half has values below the median and the other half has values above it.
Median = ((n+1)/2)th observationExample: n=7 → Median = 4th value
Median = avg of (n/2)th and (n/2+1)thExample: n=8 → avg of 4th & 5th values
Median Example 1 (Odd n):
Data: 75, 21, 56, 36, 81, 05, 42 → Arranged: 5, 21, 36, 42, 56, 75, 81
Median Example 2 (Even n):
Data: 75, 21, 56, 36, 81, 05, 42, 68 → Arranged: 5, 21, 36, 42, 56, 68, 75, 81
Median of a Frequency Distribution
Build a cumulative frequency (cf) column. The median is the observation whose cumulative frequency first reaches or exceeds N/2, where N = Σfᵢ.
Worked example (wages): N = 100, N/2 = 50. The cf first reaches 52 at wage ₹8500.
| Wages (₹) | Employees (f) | Cumulative Frequency (cf) |
|---|---|---|
| 7500 | 4 | 4 |
| 8000 | 18 | 22 |
| 8500 | 30 | 52 ← cf first ≥ 50 = N/2 |
| 9000 | 20 | 72 |
| 9500 | 15 | 87 |
| 10000 | 8 | 95 |
| 11000 | 5 | 100 |
| Total | N = 100 | — |
Question 6 — Comparing Heights of Boys and Girls
| Height (cm) | Boys (f) | Cumulative Frequency (cf) |
|---|---|---|
| 135 | 2 | 2 |
| 140 | 5 | 7 |
| 147 | 12 | 19 ← cf first ≥ 18.5 |
| 152 | 10 | 29 |
| 155 | 7 | 36 |
| 160 | 1 | 37 |
| Total | N = 37 | N/2 = 18.5 |
| Height (cm) | Girls (f) | Cumulative Frequency (cf) |
|---|---|---|
| 135 | 1 | 1 |
| 140 | 2 | 3 |
| 147 | 10 | 13 |
| 152 | 5 | 18 ← cf first ≥ 14.5 |
| 155 | 6 | 24 |
| 160 | 5 | 29 |
| Total | N = 29 | N/2 = 14.5 |
3. Mode — The Most Frequent Value
The mode is the observation that occurs most often in a dataset. In a frequency table, the mode is simply the value with the highest frequency.
- Unimodal data — one value has the highest frequency (e.g., mode = 7 appearing 5 times)
- Bimodal data — two values tie for the highest frequency (e.g., both 4 and 6 appear 5 times each)
- Trimodal data — three values have the same highest frequency
- No mode — all values appear equally often; no single observation dominates
Question 7 — Centuries Scored by Cricketers (Mean + Median + Mode)
| No. of Centuries xᵢ | No. of Cricketers fᵢ | fᵢxᵢ |
|---|---|---|
| 5 | 56 | 280 |
| 10 | 23 | 230 |
| 15 | 39 | 585 |
| 20 | 13 | 260 |
| 25 | 8 | 200 |
| Total | Σfᵢ = 139 | Σfᵢxᵢ = 1555 |
| Centuries | Cricketers (f) | Cumulative Frequency (cf) |
|---|---|---|
| 5 | 56 | 56 |
| 10 | 23 | 79 ← cf first ≥ 69.5 |
| 15 | 39 | 118 |
| 20 | 13 | 131 |
| 25 | 8 | 139 |
| Total | N = 139 | N/2 = 69.5 |
| Measure | Value |
|---|---|
| Mean | 11.18 centuries |
| Median | 10 centuries |
| Mode | 5 centuries |
Question 8 — Sweet Packets on New Year's Day (Mean + Median + Mode)
| Cost xᵢ (₹) | Packets fᵢ | fᵢxᵢ |
|---|---|---|
| 25 | 20 | 500 |
| 50 | 36 | 1800 |
| 75 | 32 | 2400 |
| 100 | 29 | 2900 |
| 125 | 22 | 2750 |
| 150 | 11 | 1650 |
| Total | Σfᵢ = 150 | Σfᵢxᵢ = 12000 |
| Cost (₹) | Packets (f) | Cumulative Frequency |
|---|---|---|
| 25 | 20 | 20 |
| 50 | 36 | 56 |
| 75 | 32 | 88 ← cf first ≥ 75 |
| 100 | 29 | 117 |
| 125 | 22 | 139 |
| 150 | 11 | 150 |
| Total | N = 150 | N/2 = 75 |
| Measure | Value |
|---|---|
| Mean | ₹80 |
| Median | ₹75 |
| Mode | ₹50 |
Question 9 — Finding an Unknown Weight Using Mean
Question 10 — School Donations to Orphanage (Mean + Median + Mode)
| Class | Donation xᵢ (₹) | Students fᵢ | fᵢxᵢ |
|---|---|---|---|
| VI | 5 | 15 | 75 |
| VII | 7 | 15 | 105 |
| VIII | 10 | 20 | 200 |
| IX | 15 | 16 | 240 |
| X | 20 | 14 | 280 |
| Total | Σfᵢ = 80 | Σfᵢxᵢ = 900 | |
| Donation (₹) | Students (f) | Cumulative Frequency |
|---|---|---|
| 5 | 15 | 15 |
| 7 | 15 | 30 |
| 10 | 20 | 50 ← cf first ≥ 40 |
| 15 | 16 | 66 |
| 20 | 14 | 80 |
| Total | N = 80 | N/2 = 40 |
| Measure | Value |
|---|---|
| Mean | ₹11.25 |
| Median | ₹10 |
| Mode | ₹10 |
Question 11 — Four Unknown Numbers Using Mean Equations
Quick Summary — All Questions at a Glance
| Q | Topic | Measure(s) | Answer |
|---|---|---|---|
| Q1 | Parcel weights | Mean | 85 kg |
| Q2 | Children per family | Mean | 1.71 |
| Q3 | Find K given mean = 7.2 | Mean equation | K = 10 |
| Q4 | Village population (Census) | Mean | 17,700 |
| Q5 | AFLATOUN savings | Mean (per mandal) | Overall ₹444 |
| Q6 | Heights: boys vs girls | Median | Boys 147 cm; Girls 152 cm |
| Q7 | Cricketer centuries | Mean, Median, Mode | 11.18 / 10 / 5 |
| Q8 | Sweet packet costs | Mean, Median, Mode | ₹80 / ₹75 / ₹50 |
| Q9 | Rahim's weight | Mean (reverse) | 37 kg |
| Q10 | Orphanage donations | Mean, Median, Mode | ₹11.25 / ₹10 / ₹10 |
| Q11 | Four unknown numbers | Mean (chained) | 2, 6, 19, 33 |
Common Mistakes to Avoid in Board Exams
- Forgetting to arrange data before finding median: The median formula only works on sorted data. Always arrange values in ascending order first — skipping this step gives a completely wrong answer.
- Using N/2 instead of (N+1)/2 for odd n in raw data: For ungrouped data with odd n, use (n+1)/2. The N/2 shortcut applies to finding the median class in a frequency distribution, not raw data.
- Confusing mode with median: Mode = most frequent value. Median = middle value. These are different. In Q10, both happen to be ₹10, but that is a coincidence.
- Not writing units in the answer: If xᵢ is in thousands (Q4), remember to convert: 17.7 thousand = 17,700 persons. Always state units clearly.
- In Q3 type problems — not forming the equation first: Many students substitute wrongly. Always write Mean × Σf = Σfx and then substitute the given mean before expanding.
What This Exercise Prepares You For
The concepts in Exercise 9.2 are the starting point for all statistical analysis you will encounter in higher classes. The weighted mean formula (Σfᵢxᵢ / Σfᵢ) reappears in Class 10 Statistics Exercise 14.1, where it is extended to grouped data with class intervals using the direct method, assumed mean method, and step-deviation method.
The cumulative frequency technique used for median in this exercise is also the foundation for constructing ogives (cumulative frequency curves) in Class 10 Exercise 14.4. Bahadurpura's high mean in Q5 vs. other mandals is a classic example of why median can sometimes better represent a skewed dataset — a concept explored in detail in Class 11 and beyond.
For Telangana and Andhra Pradesh board examinations, Questions 3 (find K), 7 and 8 (all three measures), and 11 (chained mean) are high-frequency question types. Mastering these fully earns you 4–5 marks per problem in the annual examination.