NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Statistics – Class 10 Maths (Chapter 14) is an important and scoring chapter for CBSE board exams. Many students look for statistics class 10 NCERT solutions to understand concepts clearly and practise exam-oriented questions. Class 10 Maths chapter Statistics, also known as Class 10 Maths Chapter 14, focuses on data handling and analysis using methods like mean, median, and mode of grouped data.

NCERT solutions for Class 10 Maths Statistics provide step-by-step explanations for every question, making calculations easy to understand. These solutions strictly follow the latest CBSE syllabus and NCERT guidelines, ensuring students prepare in the right direction. Clear tables, formulas, and solved examples help reduce errors and improve accuracy.

Practicing class 10 maths chapter Statistics questions answers helps students develop strong numerical and analytical skills. The chapter includes practical problems based on real-life data, which enhances logical thinking and exam confidence. Regular practice also improves speed and time management during exams.

Overall, proper preparation of Class 10 Maths Chapter 14 using NCERT-based solutions helps students score better in board exams and build a strong foundation for higher-level mathematics.

Check Out: CBSE Class 10 Books

Class 10 Maths Chapter Statistics Questions Answers

Class 10 Maths Chapter 14 Exercise 14.1 Statistics

1. A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of Plants	0-2	2-4	4-6	6-8	8-10	10-12	12-14
Number of Houses	1	2	1	5	6	2	3

Which method did you use for finding the mean, and why?

Solution:

To find the mean value, we will use the direct method because the numerical value of f _i and x _i are small. Find the midpoint of the given interval using the formula. Midpoint (x _i ) = (upper limit + lower limit)/2

No. of plants (Class interval)	No. of houses Frequency (f i )	Mid-point (x i )	f i x i
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
	Sum f i = 20		Sum f i x i = 162

The formula to find the mean is: Mean = x̄ = ∑f _i x _i /∑f _i = 162/20 = 8.1 Therefore, the mean number of plants per house is 8.1.

2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs.)	500-520	520-540	540-560	560-580	580-600
Number of workers	12	14	8	6	10

Find the mean daily wages of the workers of the factory by using an appropriate method .

Solution: Find the midpoint of the given interval using the formula. Midpoint (x _i ) = (upper limit + lower limit)/2 In this case, the value of mid-point (x _i ) is very large, so let us assume the mean value, a = 550. Class interval (h) = 20 So, u _i = (x _i – a)/h u _i = (x _i – 550)/20 Substitute and find the values as follows:

Daily wages (Class interval)	Number of workers frequency (f i )	Mid-point (x i )	u i = (x i – 550)/20	f i u i
500-520	12	510	-2	-24
520-540	14	530	-1	-14
540-560	8	550 = a	0	0
560-580	6	570	1	6
580-600	10	590	2	20
Total	Sum f i = 50			Sum f i u i = -12

So, the formula to find out the mean is: Mean = x̄ = a + h(∑f _i u _i /∑f _i ) = 550 + [20 × (-12/50)] = 550 – 4.8 = 545.20 Thus, mean daily wage of the workers = Rs. 545.20

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

Daily Pocket Allowance(in c)	11-13	13-15	15-17	17-19	19-21	21-23	23-35
Number of children	7	6	9	13	f	5	4

Solution: To find out the missing frequency, use the mean formula. Given, mean x̄ = 18

Class interval	Number of children (f i )	Mid-point (x i )	f i x i
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	f i = 44+f		Sum f i x i = 752+20f

The mean formula is Mean = x̄ = ∑f _i x _i /∑f _i = (752 + 20f)/ (44 + f) Now substitute the values and equate to find the missing frequency (f) ⇒ 18 = (752 + 20f)/ (44 + f) ⇒ 18(44 + f) = (752 + 20f) ⇒ 792 + 18f = 752 + 20f ⇒ 792 + 18f = 752 + 20f ⇒ 792 – 752 = 20f – 18f ⇒ 40 = 2f ⇒ f = 20 So, the missing frequency, f = 20.

4. Thirty women were examined in a hospital by a doctor, and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Number of heart beats per minute	65-68	68-71	71-74	74-77	77-80	80-83	83-86
Number of women	2	4	3	8	7	4	2

Solution: From the given data, let us assume the mean as a = 75.5 x _i = (Upper limit + Lower limit)/2 Class size (h) = 3 Now, find the u _i and f _i u _i as follows:

Class Interval	Number of women (f i )	Mid-point (x i )	u i = (x i – 75.5)/h	f i u i
65-68	2	66.5	-3	-6
68-71	4	69.5	-2	-8
71-74	3	72.5	-1	-3
74-77	8	75.5 = a	0	0
77-80	7	78.5	1	7
80-83	4	81.5	2	8
83-86	2	84.5	3	6
	Sum f i = 30			Sum f i u i = 4

Mean = x̄ = a + h(∑f _i u _i /∑f _i ) = 75.5 + 3 × (4/30) = 75.5 + (4/10) = 75.5 + 0.4 = 75.9 Therefore, the mean heart beats per minute for these women is 75.9

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes	50-52	53-55	56-58	59-61	62-64
Number of boxes	15	110	135	115	25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution: The given data is not continuous, so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals is 1. Here, assumed mean (a) = 57 Class size (h) = 3 Here, the step deviation is used because the frequency values are big.

Class Interval	Number of boxes (f i )	Mid-point (x i )	u i = (x i – 57)/h	f i u i
49.5-52.5	15	51	-2	-30
52.5-55.5	110	54	-1	-110
55.5-58.5	135	57 = a	0	0
58.5-61.5	115	60	1	115
61.5-64.5	25	63	2	50
	Sum f i = 400			Sum f i u i = 25

The formula to find out the Mean is: Mean = x̄ = a + h(∑f _i u _i /∑f _i ) = 57 + 3(25/400) = 57 + 0.1875 = 57.19 Therefore, the mean number of mangoes kept in a packing box is 57.19

6. The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure(in c)	100-150	150-200	200-250	250-300	300-350
Number of households	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method.

Solution: Find the midpoint of the given interval using the formula. Midpoint (x _i ) = (upper limit + lower limit)/2 Let us assume the mean (a) = 225 Class size (h) = 50

Class Interval	Number of households (f i )	Mid-point (x i )	d i = x i – A	u i = d i /50	f i u i
100-150	4	125	-100	-2	-8
150-200	5	175	-50	-1	-5
200-250	12	225 = a	0	0	0
250-300	2	275	50	1	2
300-350	2	325	100	2	4
	Sum f i = 25				Sum f i u i = -7

Mean = x̄ = a + h(∑f _i u _i /∑f _i ) = 225 + 50(-7/25) = 225 – 14 = 211 Therefore, the mean daily expenditure on food is 211.

7. To find out the concentration of SO ₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of SO 2 ( in ppm)	Frequency
0.00 – 0.04	4
0.04 – 0.08	9
0.08 – 0.12	9
0.12 – 0.16	2
0.16 – 0.20	4
0.20 – 0.24	2

Find the mean concentration of SO ₂ in the air.

Solution: To find out the mean, first find the midpoint of the given frequencies as follows:

Concentration of SO 2 (in ppm)	Frequency (f i )	Mid-point (x i )	f i x i
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	Sum f i = 30		Sum (f i x i ) = 2.96

The formula to find out the mean is Mean = x̄ = ∑f _i x _i /∑f _i = 2.96/30 = 0.099 ppm Therefore, the mean concentration of SO ₂ in the air is 0.099 ppm.

8. 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.

Number of days	0-6	6-10	10-14	14-20	20-28	28-38	38-40
Number of students	11	10	7	4	4	3	1

Solution: Find the midpoint of the given interval using the formula. Midpoint (x _i ) = (upper limit + lower limit)/2

Class interval	Frequency (f i )	Mid-point (x i )	f i x i
0-6	11	3	33
6-10	10	8	80
10-14	7	12	84
14-20	4	17	68
20-28	4	24	96
28-38	3	33	99
38-40	1	39	39
	Sum f i = 40		Sum f i x i = 499

The mean formula is, Mean = x̄ = ∑f _i x _i /∑f _i = 499/40 = 12.48 days Therefore, the mean number of days a student was absent = 12.48.

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

literacy rate.

Literacy rate (in %)	45-55	55-65	65-75	75-85	85-98
Number of cities	3	10	11	8	3

Solution: Find the midpoint of the given interval using the formula. Midpoint (x _i ) = (upper limit + lower limit)/2 In this case, the value of mid-point (x _i ) is very large, so let us assume the mean value, a = 70. Class interval (h) = 10 So, u _i = (x _i – a)/h u _i = (x _i – 70)/10 Substitute and find the values as follows:

Class Interval	Frequency (f i )	(x i )	u i = (x i – 70)/10	f i u i
45-55	3	50	-2	-6
55-65	10	60	-1	-10
65-75	11	70 = a	0	0
75-85	8	80	1	8
85-95	3	90	2	6
	Sum f i = 35			Sum f i u i = -2

So, Mean = x̄ = a + (∑f _i u _i /∑f _i ) × h = 70 + (-2/35) × 10 = 69.43 Therefore, the mean literacy part = 69.43%

Read More: NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Class 10 Maths Chapter 14 Exercise 14.2 Statistics

1. The following table shows the ages of the patients admitted to a hospital during a year:

Age (in years)	5-15	15-25	25-35	35-45	45-55	55-65
Number of patients	6	11	21	23	14	5

Find the mode and the mean of the data given above. Compare and interpret the two

measures of central tendency.

Solution: To find out the modal class, let us the consider the class interval with high frequency. Here, the greatest frequency = 23, so the modal class = 35 – 45, Lower limit of modal class = l = 35, class width (h) = 10, f _m = 23, f ₁ = 21 and f ₂ = 14 The formula to find the mode is Mode = l + [(f _m – f ₁ )/ (2f _m – f ₁ – f ₂ )] × h Substitute the values in the formula, we get Mode = 35+[(23-21)/(46-21-14)]×10 = 35 + (20/11) = 35 + 1.8 = 36.8 years So the mode of the given data = 36.8 years Calculation of Mean: First find the midpoint using the formula, x _i = (upper limit +lower limit)/2

Class Interval	Frequency (f i )	Mid-point (x i )	f i x i
5-15	6	10	60
15-25	11	20	220
25-35	21	30	630
35-45	23	40	920
45-55	14	50	700
55-65	5	60	300
	Sum f i = 80		Sum f i x i = 2830

The mean formula is Mean = x̄ = ∑f _i x _i /∑f _i = 2830/80 = 35.375 years Therefore, the mean of the given data = 35.375 years

2. The following data gives the information on the observed lifetimes (in hours) of 225

electrical components:

Lifetime (in hours)	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	10	35	52	61	38	29

Determine the modal lifetimes of the components.

Solution: From the given data the modal class is 60–80. Lower limit of modal class = l = 60, The frequencies are: f _m = 61, f ₁ = 52, f ₂ = 38 and h = 20 The formula to find the mode is Mode = l+ [(f _m – f ₁ )/(2f _m – f ₁ – f ₂ )] × h Substitute the values in the formula, we get Mode = 60 + [(61 – 52)/ (122 – 52 – 38)] × 20 Mode = 60 + [(9 × 20)/32] Mode = 60 + (45/8) = 60 + 5.625 Therefore, modal lifetime of the components = 65.625 hours.

3. The following data gives the distribution of total monthly household expenditure of 200

families of a village. Find the modal monthly expenditure of the families. Also, find the

mean monthly expenditure:

Expenditure (in Rs.)	Number of families
1000-1500	24
1500-2000	40
2000-2500	33
2500-3000	28
3000-3500	30
3500-4000	22
4000-4500	16
4500-5000	7

Solution: Given data: Modal class = 1500-2000, l = 1500, Frequencies: f _m = 40 f ₁ = 24, f ₂ = 33 and h = 500 Mode formula: Mode = l + [(f _m – f ₁ )/ (2f _m – f ₁ – f ₂ )] × h Substitute the values in the formula, we get Mode = 1500 + [(40 – 24)/ (80 – 24 – 33)] × 500 Mode = 1500 + [(16 × 500)/23] Mode = 1500 + (8000/23) = 1500 + 347.83 Therefore, modal monthly expenditure of the families = Rupees 1847.83 Calculation for mean: First find the midpoint using the formula, x _i =(upper limit +lower limit)/2 Let us assume a mean, (a) be 2750.

Class Interval	f i	x i	d i = x i – a	u i = d i /h	f i u i
1000-1500	24	1250	-1500	-3	-72
1500-2000	40	1750	-1000	-2	-80
2000-2500	33	2250	-500	-1	-33
2500-3000	28	2750 = a	0	0	0
3000-3500	30	3250	500	1	30
3500-4000	22	3750	1000	2	44
4000-4500	16	4250	1500	3	48
4500-5000	7	4750	2000	4	28
	f i = 200				f i u i = -35

The formula to calculate the mean, Mean = x̄ = a +(∑f _i u _i /∑f _i ) × h Substitute the values in the given formula = 2750 + (-35/200) × 500 = 2750 – 87.50 = 2662.50 So, the mean monthly expenditure of the families = Rs. 2662.50

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

No of students per teacher	Number of states / U.T
15-20	3
20-25	8
25-30	9
30-35	10
35-40	3
40-45	0
45-50	0
50-55	2

Solution: Given data: Modal class = 30 – 35, l = 30, Class width (h) = 5, f _m = 10, f ₁ = 9 and f ₂ = 3 Mode Formula: Mode = l + [(f _m – f ₁ )/ (2f _m – f ₁ – f ₂ )] × h Substitute the values in the given formula Mode = 30 + [(10 – 9)/ (20 – 9 – 3)] × 5 = 30 + (5/8) = 30 + 0.625 = 30.625 Therefore, the mode of the given data = 30.625 Calculation of mean: Find the midpoint using the formula, x _i =(upper limit +lower limit)/2

Class Interval	Frequency (f i )	Mid-point (x i )	f i x i
15-20	3	17.5	52.5
20-25	8	22.5	180.0
25-30	9	27.5	247.5
30-35	10	32.5	325.0
35-40	3	37.5	112.5
40-45	0	42.5	0
45-50	0	47.5	0
50-55	2	52.5	105.0
	Sum f i = 35		Sum f i x i = 1022.5

Mean = x̄ = ∑f _i x _i /∑f _i = 1022.5/35 = 29.2 (approx) Therefore, mean = 29.2

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

Run Scored	Number of Batsman
3000-4000	4
4000-5000	18
5000-6000	9
6000-7000	7
7000-8000	6
8000-9000	3
9000-10000	1
10000-11000	1

Find the mode of the data.

Solution: Given data: Modal class = 4000 – 5000, l = 4000, class width (h) = 1000, f _m = 18, f ₁ = 4 and f ₂ = 9 Mode Formula: Mode = l + [(f _m – f ₁ )/ (2f _m – f ₁ – f ₂ )] × h Substitute the values Mode = 4000 + [(18 – 4)/ (36 – 4 – 9)] × 1000 = 4000 + (14000/23) = 4000 + 608.695 = 4608.695 = 4608.7 (approximately) Thus, the mode of the given data is 4608.7 runs.

6. 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 given below. Find the mode of the data:

Number of cars	Frequency
0-10	7
10-20	14
20-30	13
30-40	12
40-50	20
50-60	11
60-70	15
70-80	8

Solution: Given Data: Modal class = 40 – 50, l = 40, Class width (h) = 10, f _m = 20, f ₁ = 12 and f ₂ = 11 Mode = l + [(f _m – f ₁ )/(2f _m – f ₁ – f ₂ )] × h Substitute the values Mode = 40 + [(20 – 12)/ (40 – 12 – 11)] × 10 = 40 + (80/17) = 40 + 4.7 = 44.7 Thus, the mode of the given data is 44.7 cars.

Read More: NCERT Solutions for Class 10 Maths Chapter 3

Class 10 Maths Chapter 14 Exercise 14.3 Statistics

Solve the followings Questions.

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.

Answer:

2. If the median of the distribution given below is 28.5, then find the values of x and y.

Answer:

3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onwards but less than 60 years.

Answer:

4. The lengths of 40 leaves of a plant are measured correct to the nearest millimeter and data obtained is represented in the following table. Find the median length of the leaves.

Answer:

Read More: NCERT Solutions for Class 10 Maths Chapter 4

Understanding Class Marks and Grouped Data

Before diving into formulas, students must understand how to handle grouped frequency distributions. Unlike raw data, grouped data is organized into class intervals (e.g., 10-20, 20-30). To perform calculations, we need a single value to represent each group. This is called the Class Mark (xi).

To find the Class Mark, use this formula: Class Mark (xi) = (Upper Class Limit + Lower Class Limit) / 2

This number is the "midpoint" and is the starting point for all Mean computations. Understanding this is the first step in answering any issue in chapter 14 maths class 10 exercise 14.1. There are three ways to find the mean of grouped data.

The "Mean" is the middle point of the data. In chapter 14 of math class 10, the curriculum teaches three different ways to do things. It depends on how complicated the numbers are which one you choose:

Method of Direct Action: This is the easiest way to do it when xi and fi are both tiny.

Mean = (Sum of fi * xi) / (Sum of fi)

Assumed Method of the Mean: By taking a center value (a) away from all class marks, this approach makes the numbers smaller.

Mean = a + (Sum of fi * di) / (Sum of fi) (Where di = xi - a)

Step-Deviation Method: This is the best and most advanced way to work with really large datasets. It makes the math much easier by using the class size (h).

Mean = a + h * [(Sum of fi * ui) / (Sum of fi)]

(Where ui = (xi - a) / h)

How to Find the Mode: Finding the Data Peaks

The Mode is the number that shows up the most often. When we have grouped data, we don't look for one number at first. Instead, we look for the Modal Class, which is the range with the most occurrences. People often ask this in chapter 14 math tests for 10th grade.

To find the Mode, use this formula:

Mode = l + [(f1 - f0) / (2*f1 - f0 - f2)] * h Where:

l: The lowest number in the modal class.

f1: How often the modal class happens.

f0: How often the class before the modal class happens.

f2: The number of times the class that comes after the modal class.

h: The width of the class.

This computation is important for figuring out market patterns, like what the most common shoe size is or what the most common range of rainfall is in a certain area.

The Middle: The Middle Value and the Total Frequency

The Median is the observation that is in the middle. Students need to understand the idea of Cumulative Frequency (cf) in order to locate it. This is a running total of the frequencies. The Median Class is the one where the cumulative frequency is just above n/2, where n is the overall frequency.

The Median is calculated using the following formula:

The median is equal to l plus [(n/2) - cf] / f. * h

Where: l is the lower limit of the median class.

n: The total number of times (Sum of fi).

cf: The total frequency of the class that comes before the median class.

f: The number of times the median class appears.

h: The number of students in a class.

The chapter also talks about the Empirical Relationship, which is a rule of thumb for making sure your responses are consistent:

3 * Median equals Mode + 2 * Mean

Read More: NCERT Solutions for Class 10 Maths Chapter 10

Comparison of Measures of Central Tendency

Original Framing: Statistics as a "Data Compass"

A unique way to view the concepts in chapter 14 maths class 10 is to see Statistics as a "Data Compass." Often, students see Mean, Median, and Mode as three separate, tiring math problems. However, this framing suggests that they are three different needles on a compass pointing toward the "center" of a story. The Mean shows the "balance point," the Median shows the "midpoint," and the Mode shows the "peak." Depending on the terrain of your data (whether it is skewed or balanced), you choose the needle that guides you most accurately.

Common Pitfalls to Avoid in Chapter 14

Students often lose marks not because they don't know the formulas, but due to minor calculation errors. Keep these in mind:

• Subscripts in Mode: Be very careful with f0, f1, and f2. Mixing these up will completely change the result.

• The "cf" in Median: Remember that 'cf' in the formula refers to the cumulative frequency of the previousclass, not the median class itself.

• Class Intervals: Ensure class intervals are continuous (e.g., 10-20, 20-30). If they are discontinuous (e.g., 10-19, 20-29), subtract 0.5 from the lower limits and add 0.5 to the upper limits before starting.

Check Out: CBSE Class 10 Sample Papers

PW CBSE Class 10 Maths Study Material

• Designed as per the latest CBSE and NCERT syllabus

• Explains all chapters with simple and clear concepts

• Includes step-by-step solutions for better understanding

• Provides solved examples and practice questions

• Covers important formulas and shortcuts

• Includes previous years’ questions for exam practice

• Offers chapter-wise revision notes

• Helps improve accuracy and problem-solving speed

• Builds confidence for Class 10 board exams

Check Out: CBSE Class 10 Question Banks

NCERT Solutions for Class 10 Maths Chapter 14 FAQs

1. Which method for Mean is best for exams?
   If numbers are large, the Step-Deviation method is best to avoid calculation errors.

2. Can Mean, Median, and Mode be the same?
   Yes, in a perfectly symmetrical distribution, all three values will coincide.

3. What is cumulative frequency?
   It is the running total of frequencies up to a certain class interval.

4. Why use the Empirical Formula?
   It is useful for finding one measure when the other two are known or for checking answer consistency.

5. How do you find the Median Class? Find n/2?
   The class whose cumulative frequency is just greater than n/2 is the Median Class.

6. What does a "Skewed" distribution mean?
   It means the data is not symmetrical, and the Mean, Median, and Mode will be at different points.