Exercise 9.2 — Mean, Median and Mode

Mean, median and mode of ungrouped frequency distributions.

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Chapter 9 · Statistics

Exercise 9.2 — Measures of Central Tendency

Complete step-by-step solutions for Mean, Median and Mode — Class 9 Mathematics | CBSE, Telangana & Andhra Pradesh syllabi.

📊 Arithmetic Mean 📈 Median 🔢 Mode ✅ 11 Solved Problems 📝 CBSE · Telangana · AP

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:

Arithmetic Mean

The sum of all observations divided by the number of observations. It uses every value in the data.

M
Median

The middle value when data is arranged in order. It is not affected by very large or very small values.

Mo
Mode

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)

Sum = 4+3+5+5+4+9+7+7+8+7+4+3+7+5+8+8+8+7+4+7 = 120 n = 20 Mean = 120 ÷ 20 = 6

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.

Σfᵢxᵢ = (2×3)+(4×4)+(3×5)+(6×7)+(4×8)+(1×9) = 6+16+15+42+32+9 = 120 Σfᵢ = 2+3+4+6+4+1 = 20 Mean = 120 ÷ 20 = 6 ✓ (Same answer as raw data method)
📌 Both methods give the same mean. The frequency distribution method is faster when large datasets have repeating values.

Question 1 — Mean Weight of Parcels

Question 1
Weights of parcels in a transport office are given. Find the mean weight.
Weight xᵢ (kg)No. of Parcels fᵢfᵢxᵢ
50251250
65342210
75382850
90403600
110475170
120161920
TotalΣfᵢ = 200Σfᵢxᵢ = 17000
Mean x̄ = Σfᵢxᵢ / Σfᵢ = 17000 / 200 = 85
∴ Mean weight of parcels = 85 kg

Question 2 — Mean Number of Children per Family

Question 2
Number of families in a village by number of children are given. Find the mean number of children per family.
No. of Children xᵢNo. of Families fᵢfᵢxᵢ
0110
12525
23264
31030
4520
515
TotalΣfᵢ = 84Σfᵢxᵢ = 144
Mean x̄ = 144 / 84 = 1.71 (144 ÷ 84 = 12/7 ≈ 1.71)
∴ Mean number of children per family = 1.71
💡 Note that 0 × 11 = 0, so families with no children contribute nothing to the sum Σfᵢxᵢ. Always include zero-value rows in the table but remember they don't change the numerator.

Question 3 — Finding the Unknown Frequency 'K'

Question 3
The mean of the following frequency distribution is 7.2. Find the value of K.
xffx
248
4728
61060
816128
10K10K
12336
TotalΣf = 40 + KΣfx = 260 + 10K

Setting up the equation:

Mean = Σfx / Σf  →  7.2 = (260 + 10K) / (40 + K) 7.2 × (40 + K) = 260 + 10K 288 + 7.2K = 260 + 10K 288 − 260 = 10K − 7.2K 28 = 2.8K K = 28 / 2.8 = 10
∴ K = 10
📌 This type of "find the missing frequency given the mean" is a standard 3-mark question in Telangana and AP board exams. Always set up the equation Mean × Σf = Σfx and then solve for the unknown.

Question 4 — Average Village Population (Census 2011)

Question 4
Number of villages vs. their population (India Census 2011). Find the average population per village.
Population xᵢ (thousands)No. of Villages fᵢfᵢxᵢ
1220240
51575
3032960
2035700
1536540
8756
TotalΣfᵢ = 145Σfᵢxᵢ = 2571
Mean x̄ = 2571 / 145 = 17.7 thousands = 17.7 × 1000 = 17,700 persons
∴ Average population per village = 17,700

Question 5 — AFLATOUN School Savings Programme (Hyderabad)

Question 5
AFLATOUN mandal-wise school savings in Hyderabad. Find the school-wise mean savings per mandal and overall mean.
MandalNo. of SchoolsTotal Saved (₹)Mean per School (₹)
Amberpet62154359
Thirumalagiri62478413
Saidabad5975195
Khairathabad4912228
Secunderabad3600200
Bahadurpura97533837
All Mandals3314,652444

Calculations:

Amberpet: 2154 ÷ 6 = ₹359 Thirumalagiri: 2478 ÷ 6 = ₹413 Saidabad: 975 ÷ 5 = ₹195 Khairathabad: 912 ÷ 4 = ₹228 Secunderabad: 600 ÷ 3 = ₹200 Bahadurpura: 7533 ÷ 9 = ₹837   Overall mean = (2154+2478+975+912+600+7533) ÷ (6+6+5+4+3+9) = 14652 ÷ 33 = ₹444
∴ Overall mean savings per school = ₹444
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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.

n is ODD
Median = ((n+1)/2)th observation

Example: n=7 → Median = 4th value

n is EVEN
Median = avg of (n/2)th and (n/2+1)th

Example: 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

n = 7 (odd) → Median = ((7+1)/2)th = 4th observation 4th value in sorted list = 42

Median Example 2 (Even n):

Data: 75, 21, 56, 36, 81, 05, 42, 68 → Arranged: 5, 21, 36, 42, 56, 68, 75, 81

n = 8 (even) → Average of (8/2)th and (8/2+1)th = avg of 4th & 5th 4th = 42, 5th = 56 → Median = (42+56)/2 = 98/2 = 49

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.

🔑 Step-by-step method: (1) Create the cf column. (2) Find N/2. (3) Look for the first cf value ≥ N/2. (4) The corresponding observation is the Median.

Worked example (wages): N = 100, N/2 = 50. The cf first reaches 52 at wage ₹8500.

Wages (₹)Employees (f)Cumulative Frequency (cf)
750044
80001822
85003052 ← cf first ≥ 50 = N/2
90002072
95001587
10000895
110005100
TotalN = 100
Median = ₹8500

Question 6 — Comparing Heights of Boys and Girls

Question 6
Heights of boys and girls of Class IX are given. Compare the heights using Median.
📐 Median Height of Boys
Height (cm)Boys (f)Cumulative Frequency (cf)
13522
14057
1471219 ← cf first ≥ 18.5
1521029
155736
160137
TotalN = 37N/2 = 18.5
Median height of Boys = 147 cm
📐 Median Height of Girls
Height (cm)Girls (f)Cumulative Frequency (cf)
13511
14023
1471013
152518 ← cf first ≥ 14.5
155624
160529
TotalN = 29N/2 = 14.5
Median height of Girls = 152 cm
Comparison: The median height of girls (152 cm) is greater than that of boys (147 cm) in this Class IX sample. Median is used here instead of mean because it is not affected by extreme values at the ends of the distribution.

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
🛍️ Real-world use: A shoe shop uses mode to decide which shoe size to stock the most. If size 7 sells most often (mode = 7), the shop orders more of size 7.

Question 7 — Centuries Scored by Cricketers (Mean + Median + Mode)

Question 7
Centuries scored by world cricketers: find the Mean, Median, and Mode.
📊 Mean
No. of Centuries xᵢNo. of Cricketers fᵢfᵢxᵢ
556280
1023230
1539585
2013260
258200
TotalΣfᵢ = 139Σfᵢxᵢ = 1555
Mean = 1555 / 139 = 11.18
📈 Median
CenturiesCricketers (f)Cumulative Frequency (cf)
55656
102379 ← cf first ≥ 69.5
1539118
2013131
258139
TotalN = 139N/2 = 69.5
Median = 10 centuries
🔢 Mode
Maximum frequency = 56 (for 5 centuries) Mode = 5 centuries
MeasureValue
Mean11.18 centuries
Median10 centuries
Mode5 centuries

Question 8 — Sweet Packets on New Year's Day (Mean + Median + Mode)

Question 8
Cost and count of sweet packets prepared by a stall. Find Mean, Median, and Mode.
📊 Mean
Cost xᵢ (₹)Packets fᵢfᵢxᵢ
2520500
50361800
75322400
100292900
125222750
150111650
TotalΣfᵢ = 150Σfᵢxᵢ = 12000
Mean = 12000 / 150 = ₹80
📈 Median
Cost (₹)Packets (f)Cumulative Frequency
252020
503656
753288 ← cf first ≥ 75
10029117
12522139
15011150
TotalN = 150N/2 = 75
Median = ₹75
🔢 Mode
Maximum frequency = 36 (for cost ₹50) Mode = ₹50
MeasureValue
Mean₹80
Median₹75
Mode₹50

Question 9 — Finding an Unknown Weight Using Mean

Question 9
The mean weight of three students is 40 kg. Ranga weighs 46 kg. Rahim and Reshma have the same weight. Find Rahim's weight.
Let Rahim's weight = Reshma's weight = x kg Mean = (x + x + 46) / 3 = 40 2x + 46 = 40 × 3 = 120 2x = 120 − 46 = 74 x = 74 / 2 = 37
∴ Rahim's weight = 37 kg
💡 The formula "Mean × n = Sum" is the key rearrangement here. If you know the mean and how many values there are, you can find the total sum and then solve for any unknown.

Question 10 — School Donations to Orphanage (Mean + Median + Mode)

Question 10
Donations given by students of different classes to an orphanage. Find Mean, Median, and Mode.
📊 Mean
ClassDonation xᵢ (₹)Students fᵢfᵢxᵢ
VI51575
VII715105
VIII1020200
IX1516240
X2014280
TotalΣfᵢ = 80Σfᵢxᵢ = 900
Mean = 900 / 80 = ₹11.25
📈 Median
Donation (₹)Students (f)Cumulative Frequency
51515
71530
102050 ← cf first ≥ 40
151666
201480
TotalN = 80N/2 = 40
Median = ₹10
🔢 Mode
Maximum frequency = 20 (for donation ₹10) Mode = ₹10
MeasureValue
Mean₹11.25
Median₹10
Mode₹10

Question 11 — Four Unknown Numbers Using Mean Equations

Question 11
Mean of first two numbers = 4. Mean of first three = 9. Mean of all four = 15. One number is 2. Find the others.
Step 1 — Find the 2nd number (a): (2 + a) / 2 = 4  →  2 + a = 8  →  a = 6   Step 2 — Find the 3rd number (b): (2 + a + b) / 3 = 9  →  2 + 6 + b = 27  →  b = 19   Step 3 — Find the 4th number (c): (2 + a + b + c) / 4 = 15  →  2 + 6 + 19 + c = 60  →  c = 33
∴ The four numbers are: 2, 6, 19, 33
📌 Key strategy: Work sequentially — use Mean×n = Sum to find the sum at each step, then subtract the known values to get the unknown. This chained approach is widely tested in CBSE and Telangana board exams as a 4-mark problem.

Quick Summary — All Questions at a Glance

QTopicMeasure(s)Answer
Q1Parcel weightsMean85 kg
Q2Children per familyMean1.71
Q3Find K given mean = 7.2Mean equationK = 10
Q4Village population (Census)Mean17,700
Q5AFLATOUN savingsMean (per mandal)Overall ₹444
Q6Heights: boys vs girlsMedianBoys 147 cm; Girls 152 cm
Q7Cricketer centuriesMean, Median, Mode11.18 / 10 / 5
Q8Sweet packet costsMean, Median, Mode₹80 / ₹75 / ₹50
Q9Rahim's weightMean (reverse)37 kg
Q10Orphanage donationsMean, Median, Mode₹11.25 / ₹10 / ₹10
Q11Four unknown numbersMean (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.
Most common board exam error (Median): Finding N/2 = 50 and then writing "the 50th observation is the median" — this is wrong for grouped data. You look for the first cumulative frequency ≥ N/2 and the corresponding observation is the median. The cumulative frequency reaching 50 or beyond is what matters.

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.

Board Exam Tip (CBSE, Telangana & AP): For any question asking all three measures (Mean, Median, Mode), always present them in a final summary table — it makes the answer visually clear and examiners can verify each measure at a glance. This presentation habit alone can improve your score on 5-mark questions.
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