CHAPTER 14 TELANGANA CLASS X MATHEMATICS - CHAPTER 14: STATISTICS (EXERCISE 14.1 TO 14.4 FULL SOLUTIONS)

BOARD OF SECONDARY EDUCATION, TELANGANA

CLASS 10 MATHEMATICS - CHAPTER 14: STATISTICS

Complete Exercise-Wise Questions & Step-by-Step Solutions (Ex 14.1 - 14.4) | Muqeem Tutorial

SUBJECT: MATHEMATICS CHAPTER 14 (FULL EXERCISES WITH OGIVE GRAPHS) WEIGHTAGE: 10 - 12 MARKS

• Mean (Direct): x̄ = Σfᵢxᵢ / Σfᵢ  |  Step-Deviation: x̄ = a + (Σfᵢuᵢ / Σfᵢ) × h
• Mode = l + [(f₁ - f₀) / (2f₁ - f₀ - f₂)] × h  |  Median = l + [(N/2 - cf) / f] × h
• Empirical Relationship: Mode = 3 Median - 2 Mean

EXERCISE 14.1 — Mean of Grouped Data (Q1 to Q9)

Q1. A survey was conducted by a group of students to collect data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

No. of Plants	0 - 2	2 - 4	4 - 6	6 - 8	8 - 10	10 - 12	12 - 14	Total
No. of Houses (fᵢ)	1	2	1	5	6	2	3	Σfᵢ = 20
Class Mark (xᵢ)	1	3	5	7	9	11	13	-
fᵢxᵢ	1	6	5	35	54	22	39	Σfᵢxᵢ = 162

Solution:
Direct Method: Mean x̄ = Σfᵢxᵢ / Σfᵢ = 162 / 20 = 8.1.
👉 Mean number of plants per house = 8.1

Q2. Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages using the step-deviation method.

Daily Wages (₹)	500 - 520	520 - 540	540 - 560	560 - 580	580 - 600	Total
Workers (fᵢ)	12	14	8	6	10	Σfᵢ = 50
xᵢ	510	530	550 (a)	570	590	-
uᵢ = (xᵢ - a)/h	-2	-1	0	1	2	-
fᵢuᵢ	-24	-14	0	6	20	Σfᵢuᵢ = -12

Solution:
Assumed Mean a = 550, Class size h = 20.
Mean x̄ = a + (Σfᵢuᵢ / Σfᵢ) × h = 550 + (-12 / 50) × 20 = 550 - 4.8 = ₹ 545.20.
👉 Mean daily wages = ₹ 545.20

Q3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.

Allowance (₹)	11-13	13-15	15-17	17-19	19-21	21-23	23-25	Total
Children (fᵢ)	7	6	9	13	f	5	4	44 + f
Class Mark (xᵢ)	12	14	16	18	20	22	24	-
fᵢxᵢ	84	84	144	234	20f	110	96	752 + 20f

Solution:
Mean = 18. Formula: Mean = Σfᵢxᵢ / Σfᵢ
18 = (752 + 20f) / (44 + f) ⇒ 18(44 + f) = 752 + 20f
792 + 18f = 752 + 20f ⇒ 2f = 40 ⇒ f = 20.
👉 Missing Frequency f = 20

Q4. Thirty women were examined in a hospital and their heart beats per minute were recorded. Find the mean heart beats per minute.

Heart Beats	65-68	68-71	71-74	74-77	77-80	80-83	83-86	Total
Women (fᵢ)	2	4	3	8	7	4	2	30
xᵢ	66.5	69.5	72.5	75.5 (a)	78.5	81.5	84.5	-
uᵢ	-3	-2	-1	0	1	2	3	-
fᵢuᵢ	-6	-8	-3	0	7	8	6	4

Solution:
a = 75.5, h = 3, Σfᵢ = 30, Σfᵢuᵢ = 4.
Mean x̄ = 75.5 + (4 / 30) × 3 = 75.5 + 0.4 = 75.9.
👉 Mean heart beats per minute = 75.9

Q5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. Find the mean number of mangoes kept in a packing box.

No. of Mangoes	50 - 52	53 - 55	56 - 58	59 - 61	62 - 64	Total
No. of Boxes (fᵢ)	15	110	135	115	25	400
Class Mark (xᵢ)	51	54	57 (a)	60	63	-
uᵢ = (xᵢ - a)/h	-2	-1	0	1	2	-
fᵢuᵢ	-30	-110	0	115	50	25

Solution:
Assumed mean a = 57, class size h = 3, Σfᵢ = 400, Σfᵢuᵢ = 25.
Mean x̄ = a + (Σfᵢuᵢ / Σfᵢ) × h = 57 + (25 / 400) × 3 = 57 + 0.1875 = 57.19.
👉 Mean number of mangoes = 57.19

Q6. 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 Exp. (₹)	100 - 150	150 - 200	200 - 250	250 - 300	300 - 350	Total
Households (fᵢ)	4	5	12	2	2	25
xᵢ	125	175	225 (a)	275	325	-
uᵢ	-2	-1	0	1	2	-
fᵢuᵢ	-8	-5	0	2	4	-7

Solution:
a = 225, h = 50, Σfᵢ = 25, Σfᵢuᵢ = -7.
Mean x̄ = 225 + (-7 / 25) × 50 = 225 - 14 = ₹ 211.
👉 Mean daily expenditure = ₹ 211

Q7. 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. Find the mean concentration of SO₂ in the air.

Conc. of SO₂ (ppm)	0.00-0.04	0.04-0.08	0.08-0.12	0.12-0.16	0.16-0.20	0.20-0.24	Total
Frequency (fᵢ)	4	9	9	2	4	2	30
Class Mark (xᵢ)	0.02	0.06	0.10	0.14	0.18	0.22	-
fᵢxᵢ	0.08	0.54	0.90	0.28	0.72	0.44	2.96

Solution:
Mean x̄ = Σfᵢxᵢ / Σfᵢ = 2.96 / 30 = 0.09867 ppm.
👉 Mean concentration of SO₂ = 0.099 ppm

Q8. 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.

No. of Days	0 - 6	6 - 10	10 - 14	14 - 20	20 - 28	28 - 38	38 - 40	Total
Students (fᵢ)	11	10	7	4	4	3	1	40
xᵢ	3	8	12	17	24	33	39	-
fᵢxᵢ	33	80	84	68	96	99	39	499

Solution:
Direct Method (since class intervals are unequal):
Mean x̄ = Σfᵢxᵢ / Σfᵢ = 499 / 40 = 12.475 days.
👉 Mean number of absent days = 12.48 days

Q9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy Rate (%)	45 - 55	55 - 65	65 - 75	75 - 85	85 - 95	Total
No. of Cities (fᵢ)	3	10	11	8	3	35
xᵢ	50	60	70 (a)	80	90	-
uᵢ	-2	-1	0	1	2	-
fᵢuᵢ	-6	-10	0	8	6	-2

Solution:
a = 70, h = 10, Σfᵢ = 35, Σfᵢuᵢ = -2.
Mean x̄ = 70 + (-2 / 35) × 10 = 70 - 0.57 = 69.43%.
👉 Mean literacy rate = 69.43%

EXERCISE 14.2 — Mode of Grouped Data (Q1 to Q6)

Q1. The following table shows the ages of patients admitted in a hospital during a year. Find the mode and the mean of the data:

Age (Years)	5 - 15	15 - 25	25 - 35	35 - 45	45 - 55	55 - 65
Patients (fᵢ)	6	11	21	23	14	5

Solution:
Max frequency = 23 ⇒ Modal class = 35 - 45.
l = 35, f₁ = 23, f₀ = 21, f₂ = 14, h = 10.
Mode = 35 + [(23 - 21) / (2(23) - 21 - 14)] × 10 = 35 + [2 / 11] × 10 = 36.81 years.
👉 Mode = 36.81 years  |  Mean = 35.37 years

Q2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components. Determine the modal lifetimes of the components:

Lifetimes (hrs)	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100	100 - 120
Frequency (fᵢ)	10	35	52	61	38	29

Solution:
Max frequency = 61 ⇒ Modal class = 60 - 80.
l = 60, f₁ = 61, f₀ = 52, f₂ = 38, h = 20.
Mode = 60 + [(61 - 52) / (122 - 52 - 38)] × 20 = 60 + [9 / 32] × 20 = 65.625 hours.
👉 Modal Lifetime = 65.625 hours

Q3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure.

Expenditure (₹)	1000-1500	1500-2000	2000-2500	2500-3000	3000-3500	3500-4000	4000-4500	4500-5000
Families (fᵢ)	24	40	33	28	30	22	16	7

Solution:
Max frequency = 40 ⇒ Modal class = 1500 - 2000.
l = 1500, f₁ = 40, f₀ = 24, f₂ = 33, h = 500.
Mode = 1500 + [(40 - 24) / (80 - 24 - 33)] × 500 = 1500 + [16 / 23] × 500 = ₹ 1847.83.
👉 Modal Monthly Expenditure = ₹ 1847.83

Q4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data.

Students/Teacher	15-20	20-25	25-30	30-35	35-40	40-45	45-50	50-55
States/UTs (fᵢ)	3	8	9	10	3	0	0	2

Solution:
Max frequency = 10 ⇒ Modal class = 30 - 35.
l = 30, f₁ = 10, f₀ = 9, f₂ = 3, h = 5.
Mode = 30 + [(10 - 9) / (20 - 9 - 3)] × 5 = 30 + [1 / 8] × 5 = 30.625.
👉 Mode = 30.63  |  Mean = 29.22

Q5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Find the mode.

Runs Scored	3000-4000	4000-5000	5000-6000	6000-7000	7000-8000	8000-9000	9000-10000	10000-11000
Batsmen (fᵢ)	4	18	9	7	6	3	1	1

Solution:
Max frequency = 18 ⇒ Modal class = 4000 - 5000.
l = 4000, f₁ = 18, f₀ = 4, f₂ = 9, h = 1000.
Mode = 4000 + [(18 - 4) / (36 - 4 - 9)] × 1000 = 4000 + [14 / 23] × 1000 = 4608.7 runs.
👉 Mode = 4608.7 runs

Q6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table below. Find the mode of the data:

No. of Cars	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80
Frequency (fᵢ)	7	14	13	12	20	11	15	8

Solution:
Max frequency = 20 ⇒ Modal class = 40 - 50.
l = 40, f₁ = 20, f₀ = 12, f₂ = 11, h = 10.
Mode = 40 + [(20 - 12) / (40 - 12 - 11)] × 10 = 40 + [8 / 17] × 10 = 44.71 cars.
👉 Mode = 44.71 cars

EXERCISE 14.3 — Median of Grouped Data & Missing Frequencies (Q1 to Q7)

Q1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median of the data.

Monthly Consumption (Units)	65-85	85-105	105-125	125-145	145-165	165-185	185-205
Consumers (fᵢ)	4	5	13	20	14	8	4
Cumulative Frequency (cf)	4	9	22	42	56	64	68

Solution:
N = 68 ⇒ N / 2 = 34. Median class = 125 - 145.
l = 125, cf = 22, f = 20, h = 20.
Median = l + [(N/2 - cf) / f] × h = 125 + [(34 - 22) / 20] × 20 = 137 units.
👉 Median = 137 units

Q2. If the median of the distribution given below is 28.5, find the values of x and y (Total frequency = 60):

Class Interval	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	Total
Frequency (fᵢ)	5	x	20	15	y	5	60
cf	5	5 + x	25 + x	40 + x	40 + x + y	45 + x + y	-

Solution:
1. 45 + x + y = 60 ⇒ x + y = 15 --- (Eq 1)
2. Median = 28.5 lies in class 20 - 30 ⇒ l = 20, f = 20, cf = 5 + x, h = 10.
28.5 = 20 + [(30 - (5 + x)) / 20] × 10 ⇒ 8.5 = (25 - x) / 2 ⇒ x = 8.
From Eq 1: 8 + y = 15 ⇒ y = 7.
👉 x = 8,   y = 7

Q3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age:

Age (Years)	15-20	20-25	25-30	30-35	35-40	40-45	45-50	50-55	55-60
fᵢ	2	4	18	21	33	11	3	6	2
cf	2	6	24	45	78	89	92	98	100

Solution:
N = 100 ⇒ N / 2 = 50. Median class = 35 - 40.
l = 35, cf = 45, f = 33, h = 5.
Median = 35 + [(50 - 45) / 33] × 5 = 35 + [25 / 33] = 35.76 years.
👉 Median Age = 35.76 years

Q4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre. Find the median length of the leaves.

Length (mm)	117.5-126.5	126.5-135.5	135.5-144.5	144.5-153.5	153.5-162.5	162.5-171.5	171.5-180.5
Leaves (fᵢ)	3	5	9	12	5	4	2
cf	3	8	17	29	34	38	40

Solution:
Converting to continuous boundaries: Subtract 0.5 from lower limits and add 0.5 to upper limits.
N = 40 ⇒ N / 2 = 20. Median class = 144.5 - 153.5.
l = 144.5, cf = 17, f = 12, h = 9.
Median = 144.5 + [(20 - 17) / 12] × 9 = 144.5 + [27 / 12] = 144.5 + 2.25 = 146.75 mm.
👉 Median Length = 146.75 mm

Q5. The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp:

Life Time (hrs)	1500-2000	2000-2500	2500-3000	3000-3500	3500-4000	4000-4500	4500-5000
Lamps (fᵢ)	14	56	60	86	74	62	48
cf	14	70	130	216	290	352	400

Solution:
N = 400 ⇒ N / 2 = 200. Median class = 3000 - 3500.
l = 3000, cf = 130, f = 86, h = 500.
Median = 3000 + [(200 - 130) / 86] × 500 = 3000 + 406.98 = 3406.98 hours.
👉 Median Lifetime = 3406.98 hours

Q6. 100 surnames were randomly picked up from a local telephone directory. Determine the median number of letters in the surnames.

No. of Letters	1 - 4	4 - 7	7 - 10	10 - 13	13 - 16	16 - 19
Surnames (fᵢ)	6	30	40	16	4	4
cf	6	36	76	92	96	100

Solution:
N = 100 ⇒ N / 2 = 50. Median class = 7 - 10.
l = 7, cf = 36, f = 40, h = 3.
Median = 7 + [(50 - 36) / 40] × 3 = 7 + [42 / 40] = 7 + 1.05 = 8.05.
👉 Median Number of Letters = 8.05

Q7. The distribution below gives the weight of 30 students of a class. Find the median weight of the students.

Weight (kg)	40 - 45	45 - 50	50 - 55	55 - 60	60 - 65	65 - 70	70 - 75
Students (fᵢ)	2	3	8	6	6	3	2
cf	2	5	13	19	25	28	30

Solution:
N = 30 ⇒ N / 2 = 15. Median class = 55 - 60.
l = 55, cf = 13, f = 6, h = 5.
Median = 55 + [(15 - 13) / 6] × 5 = 55 + 1.67 = 56.67 kg.
👉 Median Weight = 56.67 kg

EXERCISE 14.4 — Cumulative Frequency Curves (Ogives with Vector Graphs)

Q1. The following distribution gives the daily income of 50 workers of a factory. Convert the distribution to a 'less than type' cumulative frequency distribution, and draw its ogive.

Upper Limit (Income less than ₹)	Less than 120	Less than 140	Less than 160	Less than 180	Less than 200
Cumulative Frequency (cf)	12	26	34	40	50
Plotting Points (X, Y)	(120, 12)	(140, 26)	(160, 34)	(180, 40)	(200, 50)

Figure 14.1: Less Than Type Ogive for Daily Income Distribution

👉 Plot upper limits on X-axis and cumulative frequencies on Y-axis to obtain the smooth 'Less Than Ogive' curve.

Q2. During the medical check-up of 35 students of a class, their weights were recorded. Draw a 'less than type ogive' for the given data and obtain the median weight from the graph.

Weight (in kg)	< 38	< 40	< 42	< 44	< 46	< 48	< 50	< 52
No. of Students (cf)	0	3	5	9	14	28	32	35

Figure 14.2: Less Than Ogive Curve and Graphical Determination of Median (46.5 kg)

Solution:
N = 35 ⇒ N / 2 = 17.5.
Locate y = 17.5 on Y-axis and draw a horizontal line to intersect the ogive. Drop a perpendicular from the intersection point to the X-axis.
👉 Median Weight from Ogive Graph ≈ 46.5 kg

Q3. The following table gives production yield per hectare of wheat of 100 farms of a village. Change the distribution to a 'more than type' distribution, and draw its ogive.

Lower Limit (Yield ≥)	≥ 50	≥ 55	≥ 60	≥ 65	≥ 70	≥ 75
Cumulative Frequency (cf)	100	98	90	78	54	16
Plotting Points (X, Y)	(50, 100)	(55, 98)	(60, 90)	(65, 78)	(70, 54)	(75, 16)

Figure 14.3: More Than Type Ogive Curve for Wheat Yield

👉 Plot lower limits on X-axis and cumulative frequencies on Y-axis to obtain the 'More Than Ogive' curve.