Exercise 14.1 — Finding the Mean

Problems based on finding the mean of grouped data.

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Class 10 » Mathematics » Statistics » Exercise 14.1

Exercise 14.1 — Statistics: Mean of Grouped Data

All 9 Questions Fully Solved — Direct Method, Assumed Mean Method & Step Deviation Method | CBSE, Telangana & Andhra Pradesh

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Exercise 14.1 — Finding the Mean of Grouped Data

Exercise 14.1 is the first exercise of Chapter 14 — Statistics in Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus). All nine questions in this exercise test your ability to find the mean of grouped data presented as a frequency distribution table. Across these questions, each of the three textbook methods is used at least twice — so by the end you will have solid practice with all three.

Q1 · Direct Method Q2 · Assumed Mean Q3 · Missing Frequency Q4 · Assumed Mean Q5 · Step Deviation Q6 · Step Deviation Q7 · Direct Method Q8 · Assumed Mean Q9 · Assumed Mean

💡 How to choose the right method: Use the Direct Method when class marks are small. Use the Assumed Mean Method when class marks are large (e.g. three-digit wages). Use the Step Deviation Method when class marks are large and all class widths are equal — it gives the smallest numbers to work with.

Question 1 Plants per House — Direct Method

A survey was conducted to find the number of plants in 20 houses in a locality. Find the mean number of plants per house using the Direct Method.

Plants (C.I.)	Frequency (fᵢ)	Class Mark (xᵢ)	fᵢ × xᵢ
0 – 2	1	1	1
2 – 4	2	3	6
4 – 6	1	5	5
6 – 8	5	7	35
8 – 10	6	9	54
10 – 12	2	11	22
12 – 14	3	13	39
Total	Σfᵢ = 20	—	Σfᵢxᵢ = 162

Mean = Σfᵢxᵢ ÷ Σfᵢ = 162 ÷ 20 = 8.1 plants per house

✅ Mean number of plants per house = 8.1

📌 Why Direct Method here? The class marks (1, 3, 5, 7, 9, 11, 13) are small single- and double-digit numbers, so direct multiplication is quick and error-free. No shortcut is needed.

Question 2 Daily Wages of 50 Workers — Assumed Mean Method

The daily wages (in Rs.) of 50 factory workers are distributed as shown. Find the mean daily wages using an appropriate method.

Wages (Rs.) C.I.	fᵢ	xᵢ	dᵢ = xᵢ − 325	fᵢ × dᵢ
200 – 250	12	225	−100	−1200
250 – 300	14	275	−50	−700
300 – 350	8	325	0 (a)	0
350 – 400	6	375	+50	+300
400 – 450	10	425	+100	+1000
Total	Σfᵢ = 50	—	—	Σfᵢdᵢ = −600

Assumed mean a = 325 (middle class mark) Mean = a + Σfᵢdᵢ ÷ Σfᵢ = 325 + (−600 ÷ 50) = 325 + (−12) = 325 − 12 = Rs. 313

✅ Mean daily wages of workers = Rs. 313

🔍 Why Assumed Mean? The class marks (225, 275, 325, 375, 425) are large three-digit numbers — multiplying them directly by frequencies gives very large products. Subtracting the assumed mean 325 reduces every value to −100, −50, 0, +50, +100, making arithmetic far simpler. Notice the negative Σfᵢdᵢ means the true mean lies below the assumed mean.

Question 3 Missing Frequency — Pocket Allowance — Direct Method + Algebra

The mean pocket allowance of children in a locality is Rs. 18. One frequency value f is missing from the distribution. Find the value of f.

Pocket Allowance (Rs.)	fᵢ	xᵢ	fᵢ × xᵢ
11 – 13	7	12	84
13 – 15	6	14	84
15 – 17	9	16	144
17 – 19	13	18	234
19 – 21	f	20	20f
21 – 23	5	22	110
23 – 25	4	24	96
Total	Σfᵢ = 44 + f	—	Σfᵢxᵢ = 752 + 20f

Since the mean is given as Rs. 18, substitute into the mean formula and solve for f:

Mean = Σfᵢxᵢ ÷ Σfᵢ ⟹ 18 = (752 + 20f) ÷ (44 + f) ⟹ 18 × (44 + f) = 752 + 20f ⟹ 792 + 18f = 752 + 20f ⟹ 792 − 752 = 20f − 18f ⟹ 40 = 2f ⟹ f = 20

✅ Missing frequency f = 20

📌 Strategy for missing-frequency questions: Write the unknown as f in the table, build Σfᵢ and Σfᵢxᵢ both in terms of f, plug the known mean into the formula, and solve the resulting linear equation. This type of question appears regularly in board exams and typically carries 3–4 marks.

Question 4 Heart Beats of 30 Women — Assumed Mean Method

Thirty women had their heart beats per minute recorded and summarised below. Find the mean heart beats per minute using a suitable method.

Beats/min (C.I.)	fᵢ	xᵢ	dᵢ = xᵢ − 75.5	fᵢ × dᵢ
65 – 68	2	66.5	−9	−18
68 – 71	4	69.5	−6	−24
71 – 74	3	72.5	−3	−9
74 – 77	8	75.5	0 (a)	0
77 – 80	7	78.5	+3	+21
80 – 83	4	81.5	+6	+24
83 – 86	2	84.5	+9	+18
Total	Σfᵢ = 30	—	—	Σfᵢdᵢ = +12

Assumed mean a = 75.5 (class mark of highest-frequency class) Mean = 75.5 + (12 ÷ 30) = 75.5 + 0.4 = 75.9 beats per minute

✅ Mean heart beats per minute = 75.9 beats/min

📌 Note on class marks with decimals: Because each class width is 3 (e.g. 65–68), the midpoints are all x.5 values (66.5, 69.5 …). The Assumed Mean Method still works perfectly — the deviations (−9, −6, −3, 0, +3, +6, +9) are neat whole numbers, which keeps the arithmetic clean.

Question 5 Oranges per Basket — Step Deviation Method

Fruit vendors kept oranges in 400 baskets. The number of oranges per basket is distributed as shown. Find the mean number of oranges per basket. Which method did you choose?

Oranges (C.I.)	fᵢ	xᵢ	uᵢ = (xᵢ−22)/5	fᵢ × uᵢ
10 – 14	15	12	−2	−30
15 – 19	110	17	−1	−110
20 – 24	135	22	0 (a)	0
25 – 29	115	27	+1	+115
30 – 34	25	32	+2	+50
Total	Σfᵢ = 400	—	—	Σfᵢuᵢ = +25

a = 22, h = 5 (class width = 14−10 = 5, note: these are inclusive classes) Mean = a + (Σfᵢuᵢ ÷ Σfᵢ) × h = 22 + (25 ÷ 400) × 5 = 22 + (25 ÷ 80) = 22 + 0.3125 = 22.3125 oranges

✅ Mean oranges per basket = 22.3125 | Method chosen: Step Deviation Method

📌 Inclusive class intervals: The classes here (10–14, 15–19 …) are inclusive (the upper limit of one class and the lower limit of the next are different). The class mark is still (lower + upper) ÷ 2: e.g. (10+14)÷2 = 12. Do not confuse these with exclusive classes — the method is unchanged.

Question 6 Daily Food Expenditure of 25 Households — Step Deviation Method

The daily expenditure on food of 25 households is shown below. Find the mean daily expenditure using a suitable method.

Expenditure (Rs.)	fᵢ	xᵢ	uᵢ = (xᵢ−225)/50	fᵢ × uᵢ
100 – 150	4	125	−2	−8
150 – 200	5	175	−1	−5
200 – 250	12	225	0 (a)	0
250 – 300	2	275	+1	+2
300 – 350	2	325	+2	+4
Total	Σfᵢ = 25	—	—	Σfᵢuᵢ = −7

a = 225, h = 50 Mean = 225 + (−7 ÷ 25) × 50 = 225 + (−7 × 2) = 225 − 14 = Rs. 211

✅ Mean daily expenditure on food = Rs. 211

🔍 Simplification trick used here: Instead of computing (−7 ÷ 25) × 50, notice that 50 ÷ 25 = 2, so the expression reduces to −7 × 2 = −14. Always look for this simplification in the final step — it avoids decimals and saves time in an exam.

Question 7 SO₂ Concentration in Air (30 Localities) — Direct Method

Concentration of SO₂ in the air (in ppm) was recorded for 30 localities. Find the mean concentration using the Direct Method.

SO₂ Conc. (ppm)	fᵢ	xᵢ (Class Mark)	fᵢ × xᵢ
0.00 – 0.04	4	0.02	0.08
0.04 – 0.08	9	0.06	0.54
0.08 – 0.12	9	0.10	0.90
0.12 – 0.16	2	0.14	0.28
0.16 – 0.20	4	0.18	0.72
0.20 – 0.24	2	0.22	0.44
Total	Σfᵢ = 30	—	Σfᵢxᵢ = 2.96

Mean = Σfᵢxᵢ ÷ Σfᵢ = 2.96 ÷ 30 = 0.099 ppm

✅ Mean concentration of SO₂ in the air = 0.099 ppm

📌 Decimal class marks — handle with care: When class marks are small decimals (0.02, 0.06 …) the Direct Method is actually convenient because the products are tiny. Working step-by-step and carrying all decimal places until the very end avoids rounding errors.

Question 8 Student Attendance (40 Students, 56-Day Term) — Assumed Mean Method

A class teacher recorded attendance of 40 students over a 56-day term. Find the mean number of days a student was present.

Days Present	fᵢ	xᵢ	dᵢ = xᵢ − 45.5	fᵢ × dᵢ
35 – 38	1	36.5	−9	−9
38 – 41	3	39.5	−6	−18
41 – 44	4	42.5	−3	−12
44 – 47	4	45.5	0 (a)	0
47 – 50	7	48.5	+3	+21
50 – 53	10	51.5	+6	+60
53 – 56	11	54.5	+9	+99
Total	Σfᵢ = 40	—	—	Σfᵢdᵢ = +141

a = 45.5 Mean = 45.5 + (141 ÷ 40) = 45.5 + 3.525 = 49.025 days

✅ Mean number of days a student was present = 49 days (textbook rounds to 49)

💡 Observe the skew: The highest frequencies (10 and 11) are in the last two classes (50–53 and 53–56), meaning most students attended more than 50 days out of 56. The mean of 49 days reflects this — well above the midpoint of the range (≈45.5), confirming the data is left-skewed.

Question 9 Literacy Rate of 35 Cities — Assumed Mean Method

The literacy rate (in %) of 35 cities is given. Find the mean literacy rate.

Literacy Rate (%)	fᵢ	xᵢ	dᵢ = xᵢ − 70	fᵢ × dᵢ
45 – 55	3	50	−20	−60
55 – 65	10	60	−10	−100
65 – 75	11	70	0 (a)	0
75 – 85	8	80	+10	+80
85 – 95	3	90	+20	+60
Total	Σfᵢ = 35	—	—	Σfᵢdᵢ = −20

a = 70 Mean = 70 + (−20 ÷ 35) = 70 − 0.571... ≈ 69.43%

✅ Mean literacy rate = 69.43%

Quick Summary — All 9 Questions at a Glance

Q	Topic	n	Method Used	Answer
1	Plants per house	20	Direct	8.1 plants
2	Daily wages of workers	50	Assumed Mean	Rs. 313
3	Pocket allowance (find f)	44+f	Missing Freq.	f = 20
4	Heart beats of women	30	Assumed Mean	75.9 beats/min
5	Oranges per basket	400	Step Deviation	22.3125
6	Food expenditure	25	Step Deviation	Rs. 211
7	SO₂ concentration	30	Direct	0.099 ppm
8	Student attendance (56 days)	40	Assumed Mean	49 days
9	Literacy rate of cities	35	Assumed Mean	69.43%

Common Mistakes to Avoid in Exercise 14.1

Computing the class mark incorrectly: The class mark is always (lower limit + upper limit) ÷ 2. For inclusive classes like 10–14, this gives (10+14)÷2 = 12, not 10 or 14 alone.
Sign errors in the dᵢ column: Classes below the assumed mean give negative dᵢ values; classes above give positive ones. A sign mistake in one row can flip the final answer significantly.
Forgetting to multiply by h in Step Deviation: The formula is a + (Σfᵢuᵢ / Σfᵢ) × h. Omitting the final ×h is the single most common error in this method (Questions 5 and 6).
Summing fᵢ instead of fᵢxᵢ: In the Direct Method, make sure you sum the product column (fᵢxᵢ), not just the frequency column. Double-check your total row.
Wrong assumed mean in Question 3 type: In missing-frequency questions, the mean formula gives an equation in f — cross-multiplying is mandatory. Forgetting to cross-multiply (treating 18 = (752+20f)/44 and just adding 18 to 44) is a very common error.
Rounding mid-calculation: Carry all decimal places through every step and round only at the very end. Early rounding (e.g. rounding 0.571 to 0.6 in Question 9) changes the final answer.

⛔ Board exam alert: Exercise 14.1 questions are almost guaranteed in Telangana and Andhra Pradesh SSC and CBSE Class 10 board papers. Questions 2, 5, and 8 (or similar datasets) are the most frequently repeated. Always set out the complete frequency table with all columns, write the formula before substituting, and show the cross-multiplication step clearly in missing-frequency problems.

What This Exercise Prepares You For

Having mastered the three methods for finding the mean of grouped data, you are ready to move on to the remaining measures of central tendency in this chapter. The same frequency table format is used again in Exercise 14.2 (Median), where cumulative frequencies are added to find the median class, and in Exercise 14.3 (Mode), where the modal class is identified by the highest frequency. The mean method reviewed here in Exercise 14.1 also connects directly with the earlier concepts from the Statistics Introduction lesson.

📐 Board Exam Tip (CBSE, Telangana & AP): A full 5-mark statistics question in the board exam typically asks for the mean using either the Assumed Mean or Step Deviation method, always with 5–7 class intervals. Set out the table cleanly with all column headings (C.I., fᵢ, xᵢ, dᵢ or uᵢ, fᵢdᵢ or fᵢuᵢ), compute each row carefully, total correctly, and write the substitution step clearly. Examiners award step marks — a wrong final answer due to an arithmetic slip can still earn 3 out of 5 marks if your method and table are correct.