NCERT Solutions For Class 10 Maths Chapter 14-Statistics - GMS - Learning Simply
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NCERT Solutions For Class 10 Maths Chapter 14-Statistics

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NCERT Solutions For Class 10 Maths Chapter 14-Statistics

NCERT Solutions for Class 10 Maths Chapter 14 Statistics are provided here, which can be downloaded for free, in PDF format. The Solutions for chapter 14 Statistics are prepared by our experts who have done their specialization in Maths.

All the solved questions of Statistics are with respect to the NCERT syllabus and guidelines, to help students solve each exercise question and prepare for the CBSE Board exam. Using these solutions as a reference tool will be helpful for the students to score good marks. Students can also get the exercise-wise Solutions for Class 10 Maths all chapters and practise solving the problems.


Access Answers of Maths NCERT Class 10 Chapter 14 – Statistics

Exercise 14.1 Page: 270

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 Plants0-22-44-66-88-1010-1212-14
Number of Houses1215623

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

Solution:

In order to find the mean value, we will use direct method because the numerical value of fi and xi are small.

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

No. of plants

(Class interval)

No. of houses

Frequency (fi)

Mid-point (xi)fixi
0-2111
2-4236
4-6155
6-85735
8-106954
10-1221122
12-1431339
Sum f= 20Sum fixi = 162

The formula to find the mean is:

Mean = x̄ = ∑fxi /∑f

= 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.)100-120120-140140-160160-180180-200
Number of workers12148610

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 (xi) = (upper limit + lower limit)/2

In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 150 and class interval is h = 20.

So, u= (xi – A)/h = ui  = (xi – 150)/20

Substitute and find the values as follows:

Daily wages

(Class interval)

Number of workers

frequency (fi)

Mid-point (xi)u= (xi – 150)/20fiui
100-12012110-2-24
120-14014130-1-14
140-160815000
160-180617016
180-20010190220
TotalSum f= 50Sum fiui = -12

So, the formula to find out the mean is:

Mean = x̄ = A + h∑fiui /∑f=150 + (20 × -12/50) = 150 – 4.8 = 145.20

Thus, mean daily wage of the workers = Rs. 145.20

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)11-1313-1515-1717-1919-2121-2323-35
Number of children76913f54

Solution:

To find out the missing frequency, use the mean formula.

Here, the value of mid-point (xi)  mean x̄ = 18

Class intervalNumber of children (fi)Mid-point (xi)    fixi    
11-1371284
13-1561484
15-17916144
17-191318 = A234
19-21f2020f
21-23522110
23-2542496
Totalfi = 44+fSum fixi = 752+20f

The mean formula is

Mean = x̄ = ∑fixi /∑f= (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 heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute65-6868-7171-7474-7777-8080-8383-86
Number of women2438742

Solution:

From the given data, let us assume the mean as A = 75.5

x= (Upper limit + Lower limit)/2

Class size (h) = 3

Now, find the uand fiui as follows:

Class IntervalNumber of women (fi)Mid-point (xi)ui = (xi – 75.5)/hfiui
65-68266.5-3-6
68-71469.5-2-8
71-74372.5-1-3
74-77875.500
77-80778.517
80-83481.538
83-86284.536
Sum fi= 30Sum fiu= 4

Mean = x̄ = A + h∑fiui /∑f

= 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 mangoes50-5253-5556-5859-6162-64
Number of boxes1511013511525

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

Solution:

Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals are 1

Here, assumed mean (A) = 57

Class size (h) = 3

Here, the step deviation is used because the frequency values are big.

Class IntervalNumber of boxes (fi)Mid-point (xi)di = xi – Afidi
49.5-52.51551-690
52.5-55.511054-3-330
55.5-58.513557 = A00
58.5-61.5115603345
61.5-64.525636150
Sum fi = 400Sum fidi = 75

The formula to find out the Mean is:

Mean = x̄ = A +h ∑fidi /∑f

= 57 + 3(75/400)

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

Daily expenditure(in c)100-150150-200200-250250-300300-350
Number of households451222

Solution:

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

Let is assume the mean (A) = 225

Class size (h) = 50

Class IntervalNumber of households (fi)Mid-point (xi)di = xi – Au= di/50fiui
100-1504125-100-2-8
150-2005175-50-1-5
200-25012225000
250-30022755012
300-350232510024
Sum fi = 25Sum fiui = -7

Mean = x̄ = A +h∑fiui /∑fi

 = 225+50(-7/25)

= 225-14

= 211

Therefore, the mean daily expenditure on food is 211

7. 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 and is presented below:

Concentration of SO2 ( in ppm)Frequency
0.00 – 0.044
0.04 – 0.089
0.08 – 0.129
0.12 – 0.162
0.16 – 0.204
0.20 – 0.242

Find the mean concentration of SO2 in the air.

Solution:

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

Concentration of SO(in ppm)Frequency (fi)Mid-point (xi)fixi
0.00-0.0440.020.08
0.04-0.0890.060.54
0.08-0.1290.100.90
0.12-0.1620.140.28
0.16-0.2040.180.72
0.20-0.2420.200.40
TotalSum fi = 30Sum (fixi) = 2.96

The formula to find out the mean is

Mean = x̄ = ∑fixi /∑fi

= 2.96/30

= 0.099 ppm

Therefore, the mean concentration of SO2 in 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 days0-66-1010-1414-2020-2828-3838-40
Number of students111074431

Solution:

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

Class intervalFrequency (fi)Mid-point (xi)fixi
0-611333
6-1010880
10-1471284
14-2041768
20-2842496
28-3833399
38-4013939
Sum fi = 40Sum fixi = 499

The mean formula is,

Mean = x̄ = ∑fixi /∑fi

= 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 %)45-5555-6565-7575-8585-98
Number of cities3101183

Solution:

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 70 and class interval is h = 10.

So, u= (xi-A)/h = u= (xi-70)/10

Substitute and find the values as follows:

Class IntervalFrequency (fi)(xi)di = xi – aui = di/hfiui
45-55350-20-2-6
55-651060-10-1-10
65-751170000
75-858801018
85-953902026
Sum fi  = 35Sum fiui  = -2

So, Mean = x̄ = A+(∑fiui /∑fi)×h

= 70+(-2/35)×10

= 69.42

Therefore, the mean literacy part = 69.42


Exercise 14.2 Page: 275

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

Age (in years)5-1515-2525-3535-4545-5555-65
Number of patients6112123145

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,

l = 35,

class width (h) = 10,

fm = 23,

f1 = 21 and f2 = 14

The formula to find the mode is

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values in the formula, we get

Mode = 35+[(23-21)/(46-21-14)]×10

Mode = 35+(20/11) = 35+1.8

Mode = 36.8 year

So the mode of the given data = 36.8 year

Calculation of Mean:

First find the midpoint using the formula, x= (upper limit +lower limit)/2

Class IntervalFrequency (fi)Mid-point (xi)fixi
5-1561060
15-251120220
25-352130630
35-452340920
45-551450700
55-65560300
Sum fi = 80Sum fixi = 2830

The mean formula is

Mean = x̄ = ∑fixi /∑fi

= 2830/80

= 35.37 years

Therefore, the mean of the given data = 35.37 years

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

Lifetime (in hours)0-2020-4040-6060-8080-100100-120
Frequency103552613829

Determine the modal lifetimes of the components.

Solution:

From the given data the modal class is 60–80.

l = 60,

The frequencies are:

fm = 61, f1 = 52, f2 = 38 and h = 20

The formula to find the mode is

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values in the formula, we get

Mode =60+[(61-52)/(122-52-38)]×20

Mode = 60+((9 x 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:

ExpenditureNumber of families
1000-150024
1500-200040
2000-250033
2500-300028
3000-350030
3500-400022
4000-450016
4500-50007

Solution:

Given data:

Modal class = 1500-2000,

l = 1500,

Frequencies:

fm = 40 f1 = 24, f2 = 33 and

h = 500

Mode formula:

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×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=(upper limit +lower limit)/2

Let us assume a mean, A be 2750

Class Intervalfixidi = xi – aui = di/hfiui
1000-1500241250-1500-3-72
1500-2000401750-1000-2-80
2000-2500332250-500-1-33
2500-3000282750000
3000-3500303250500130
3500-40002237501000244
4000-45001642501500348
4500-5000747502000428
fi = 200fiui = -35

The formula to calculate the mean,

Mean = x̄ = a +(∑fiui /∑fi)×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 = Rupees 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 teacherNumber of states / U.T
15-203
20-258
25-309
30-3510
35-403
40-450
45-500
50-552

Solution:

Given data:

Modal class = 30 – 35,

l = 30,

Class width (h) = 5,

fm = 10, f1 = 9 and f2 = 3

Mode Formula:

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values in the given formula

Mode = 30+((10-9)/(20-9-3))×5

Mode = 30+(5/8) = 30+0.625

Mode = 30.625

Therefore, the mode of the given data = 30.625

Calculation of mean:

Find the midpoint using the formula, x=(upper limit +lower limit)/2

Class IntervalFrequency (fi)Mid-point (xi)fixi
15-20317.552.5
20-25822.5180.0
25-30927.5247.5
30-351032.5325.0
35-40337.5112.5
40-45042.50
45-50047.50
50-55252.5105.5
Sum fi = 35Sum fixi = 1022.5

Mean = x̄ = ∑fixi /∑fi

= 1022.5/35

= 29.2

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 ScoredNumber of Batsman
3000-40004
4000-500018
5000-60009
6000-70007
7000-80006
8000-90003
9000-100001
10000-110001

Find the mode of the data.

Solution:

Given data:

Modal class = 4000 – 5000,

l = 4000,

class width (h) = 1000,

fm = 18, f1 = 4 and f2 = 9

Mode Formula:

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values

Mode = 4000+((18-4)/(36-4-9))×1000

Mode = 4000+(14000/23) = 4000+608.695

Mode = 4608.695

Mode = 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 carsFrequency
0-107
10-2014
20-3013
30-4012
40-5020
50-6011
60-7015
70-808

Solution:

Given Data:

Modal class = 40 – 50, l = 40,

Class width (h) = 10, fm = 20, f1 = 12 and f2 = 11

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values

Mode = 40+((20-12)/(40-12-11))×10

Mode = 40 + (80/17) = 40 + 4.7 = 44.7

Thus, the mode of the given data is 44.7 cars


Exercise 14.3 Page: 287

1. The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption(in units)No. of customers
65-854
85-1055
105-12513
125-14520
145-16514
165-1858
185-2054

Solution:

Find the cumulative frequency of the given data as follows:

Class IntervalFrequencyCumulative frequency
65-8544
85-10559
105-1251322
125-1452042
145-1651456
165-185864
185-205468
N=68

From the table, it is observed that, n = 68 and hence n/2=34

Hence, the median class is 125-145 with cumulative frequency = 42

Where, l = 125, n = 68, C= 22, f = 20, h = 20

Median is calculated as follows:

Ncert solutions class 10 chapter 14-1

=125+((34−22)/20) × 20

=125+12 = 137

Therefore, median = 137

To calculate the mode:

Modal class = 125-145,

f1=20, f0=13, f2=14 & h = 20

Mode formula:

Mode = l+ [(f1-f0)/(2f1-f0-f2)]×h

Mode = 125 + ((20-13)/(40-13-14))×20

=125+(140/13)

=125+10.77

=135.77

Therefore, mode = 135.77

Calculate the Mean:

Class Intervalfixidi=xi-aui=di/hfiui
65-85475-60-3-12
85-105595-40-2-10
105-12513115-20-1-13
125-14520135000
145-1651415520114
165-185817540216
185-205419560312
Sum fi= 68Sum fiui= 7

x̄ =a+h ∑fiui/∑fi =135+20(7/68)

Mean=137.05

In this case, mean, median and mode are more/less equal in this distribution.

2. If the median of a distribution given below is 28.5 then, find the value of x & y.

Class IntervalFrequency
0-105
10-20x
20-3020
30-4015
40-50y
50-605
Total60

Solution:

Given data, n = 60

Median of the given data = 28.5

Where, n/2 = 30

Median class is 20 – 30 with a cumulative frequency = 25+x

Lower limit of median class, = 20,

Cf = 5+x,

f = 20 & h = 10

Ncert solutions class 10 chapter 14-2

Substitute the values

28.5=20+((30−5−x)/20) × 10

8.5 = (25 – x)/2

17 = 25-x

Therefore, x =8

Now, from cumulative frequency, we can identify the value of x + y as follows:

Since,

60=5+20+15+5+x+y

Now, substitute the value of x, to find y

60 = 5+20+15+5+8+y

y = 60-53

y = 7

Therefore, the value of x = 8 and y = 7.

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

Age (in years)Number of policy holder
Below 202
Below 256
Below 3024
Below 3545
Below 4078
Below 4589
Below 5092
Below 5598
Below 60100

Solution:

Class intervalFrequencyCumulative frequency
15-2022
20-2546
25-301824
30-352145
35-403378
40-451189
45-50392
50-55698
55-602100

Given data: n = 100 and n/2 = 50

Median class = 35-45

Then, l = 35, cf = 45, f = 33 & h = 5

Ncert solutions class 10 chapter 14-3

Median = 35+((50-45)/33) × 5

= 35 + (5/33)5

= 35.75

Therefore, the median age = 35.75 years.

4. The lengths of 40 leaves in a plant are measured correctly to the nearest millimeter, and the data obtained is represented as in the following table:

Length (in mm)Number of leaves
118-1263
127-1355
136-1449
145-15312
154-1625
163-1714
172-1802

Find the median length of leaves.             

Solution:

Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.

Class IntervalFrequencyCumulative frequency
117.5-126.533
126.5-135.558
135.5-144.5917
144.5-153.51229
153.5-162.5534
162.5-171.5438
171.5-180.5240

So, the data obtained are:

n = 40 and n/2 = 20

Median class = 144.5-153.5

then, l = 144.5,

cf = 17, f = 12 & h = 9

Ncert solutions class 10 chapter 14-4

Median = 144.5+((20-17)/12)×9

= 144.5+(9/4)

= 146.75 mm

Therefore, the median length of the leaves = 146.75 mm.

5. The following table gives the distribution of a life time of 400 neon lamps.

Lifetime (in hours)Number of lamps
1500-200014
2000-250056
2500-300060
3000-350086
3500-400074
4000-450062
4500-500048

Find the median lifetime of a lamp.

Solution:

Class IntervalFrequencyCumulative
1500-20001414
2000-25005670
2500-300060130
3000-350086216
3500-400074290
4000-450062352
4500-500048400

Data:

n = 400 &n/2 = 200

Median class = 3000 – 3500

Therefore, l = 3000, C= 130,

f = 86 & h = 500

Ncert solutions class 10 chapter 14-5

Median = 3000 + ((200-130)/86) × 500

= 3000 + (35000/86)

= 3000 + 406.97

= 3406.97

Therefore, the median life time of the lamps = 3406.97 hours

6. In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:

Number of letters1-44-77-1010-1313-1616-19
Number of surnames630401644

Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

Solution:

To calculate median:

Class IntervalFrequencyCumulative Frequency
1-466
4-73036
7-104076
10-131692
13-16496
16-194100

Given:

n = 100 &n/2 = 50

Median class = 7-10

Therefore, l = 7, Cf = 36, f = 40 & h = 3

Ncert solutions class 10 chapter 14-6

Median = 7+((50-36)/40) × 3

Median = 7+42/40

Median=8.05

Calculate the Mode:

Modal class = 7-10,

Where, l = 7, f1 = 40, f0 = 30, f2 = 16 & h = 3

Ncert solutions class 10 chapter 14-7

Mode = 7+((40-30)/(2×40-30-16)) × 3

= 7+(30/34)

= 7.88

Therefore mode = 7.88

Calculate the Mean:

Class Intervalfixifixi
1-462.515
4-7305.5165
7-10408.5340
10-131611.5184
13-16414.551
16-19417.570
Sum fi = 100Sum fixi = 825

Mean = x̄ = ∑fxi /∑f

Mean = 825/100 = 8.25

Therefore, mean = 8.25

7. The distributions of below give a weight of 30 students of a class. Find the median weight of a student.

Weight(in kg)40-4545-5050-5555-6060-6565-7070-75
Number of students2386632

Solution:

Class IntervalFrequencyCumulative frequency
40-4522
45-5035
50-55813
55-60619
60-65625
65-70328
70-75230

Given: n = 30 and n/2= 15

Median class = 55-60

l = 55, Cf = 13, f = 6 & h = 5

Ncert solutions class 10 chapter 14-8

Median = 55+((15-13)/6)×5

Median=55 + (10/6) = 55+1.666

Median =56.67

Therefore, the median weight of the students = 56.67


Exercise 14.4 Page: 293

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

Daily income in Rupees100-120120-140140-160160-180180-200
Number of workers12148610

Solution

Convert the given distribution table to a less than type cumulative frequency distribution, and we get

Daily incomeFrequencyCumulative Frequency
Less than 1201212
Less than 1401426
Less than 160834
Less than 180640
Less than 2001050

From the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve

Ncert solutions class 10 chapter 14-9

2.During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight in kgNumber of students
Less than 380
Less than 403
Less than 425
Less than 449
Less than 4614
Less than 4828
Less than 5032
Less than 5235

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Solution:

From the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in x-axis and frequencies on y-axis by choosing the convenient scale. Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper an join them to get a smooth curve. The curve obtained is known as less than type ogive.

Ncert solutions class 10 chapter 14-10

Locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From the point, draw a perpendicular line to the x-axis. The intersection point perpendicular to x-axis is the median of the given data. Now, to find the mode by making a table.

Class intervalNumber of students(Frequency)Cumulative Frequency
Less than 3800
Less than 403-0=33
Less than 425-3=28
Less than 449-5=49
Less than 4614-9=514
Less than 4828-14=1428
Less than 5032-28=432
Less than 5235-22=335

The class 46 – 48 has the maximum frequency, therefore, this is modal class

Here, = 46, h = 2, f1= 14, f0= 5 and f2 = 4

The mode formula is given as:

Now, Mode =

Ncert solutions class 10 chapter 14-11

= 46 + 0.95 = 46.95

Thus, mode is verified.

3. The following tables gives production yield per hectare of wheat of 100 farms of a village.

Production Yield50-5555-6060-6565-7070-7575-80
Number of farms2812243816

Change the distribution to a more than type distribution and draw its ogive.

Solution:

Converting the given distribution to a more than type distribution, we get

Production Yield (kg/ha)Number of farms
More than or equal to 50100
More than or equal to 55100-2 = 98
More than or equal to 6098-8= 90
More than or equal to 6590-12=78
More than or equal to 7078-24=54
More than or equal to 7554-38 =16

From the table obtained draw the ogive by plotting the corresponding points where the upper limits in x-axis and the frequencies obtained in the y-axis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on

this graph paper. The graph obtained is known as more than type ogive curve.

Ncert solutions class 10 chapter 14-12


Class 10 Maths Chapter 14, Statistics, is one of the most important of all the chapter present in the textbook. The weightage of this chapter in the final exam is around 11 to 12 marks. On average, there will be 3 questions which could be asked from this chapter and marks will be distributed in a manner of 3+4+4( it could vary as per question).

Topics covered in Chapter 14, Statistics are;

  • Mean of Grouped Data
  • Mode of Grouped Data
  • Median of Grouped Data
  • Graphical Representation of Cumulative Frequency Distribution

List of Exercises in class 10 Maths Chapter 14 :
Exercise 14.1 Solutions 9 Question ( 9 long)
Exercise 14.2 Solutions 6 Question ( 6 long)
Exercise 14.3 Solutions 7 Question ( 7 long)
Exercise 14.4 Solutions 3 Question ( 3 long)

NCERT solutions for Class 10 Maths Chapter 14- Statistics are made available for students who want to obtain good marks in this chapter. The methods and procedure to solve the questions have been explained clearly in these NCERT Solutions, such that, students find it easy to understand the fundamentals quickly.

The world is highly data-oriented, in fact, each and every field has a group of data, which represents the relevant information. Statistics is the branch of mathematics which deals with the representation of data in a meaningful way.

You will face many real-life scenarios where the fundamentals of statistics are used to represent a set of data in tabular form or in graphs or in pie charts. There are a number of methods you will learn from this chapter such as, step deviation methods, finding mode and median of grouped data, converting frequency distribution and the relation between mode, mean and median methods, etc. 10th Class NCERT solutions are the best study materials to prepare for the board exam.

Key Features of NCERT Solutions for Class 10 Maths Chapter 14- Statistics

  • The solutions for the statistics chapter works as a reference for the students.
  • It will help students to score marks against the questions asked from the statistics chapter.
  • Students can prepare and do the revision for chapter 14 with this source.
  • The questions of statistics are solved by subject experts.
  • The content of the material is as per the CBSE syllabus and guidelines.

Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 14

Does NCERT Solutions for Class 10 Maths Chapter 14 help you to clear board exams?

Yes, NCERT Solutions for Class 10 Maths Chapter 14 is one of the important chapters of Class 10 Maths of NCERT Solutions. These solutions are focused on learning various Mathematics tricks and shortcuts for quick and easy calculations. This makes them learn and clear the Maths subject in board exams.

How NCERT Solutions for Class 10 Maths Chapter 14 is helps in learning areas and volumes of geometrical shapes?

NCERT Solutions for Class 10 Maths Chapter 14 is created by our faculty in an interactive manner to make it easy for the students to understand the concepts. Students can refer to the PDF of solutions as a major study material to improve their speed in solving problems accurately.

How many questions present in NCERT Solutions for Class 10 Maths Chapter 14 ?

14th Chapter of NCERT Solutions for Class 10 Maths has 4 exercises. In that first exercise contains 9 questions, second exercise contains 6 questions, third exercise has 7 questions and last or fourth exercise has 3 questions. So there are a total of 25 long questions that are present in NCERT Solutions for Class 10 Maths Chapter 14.

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At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s visio…

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