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NCERT Solutions
Class 9
Maths
Chapter 12 Statistics

NCERT Solutions Class 9 Maths Chapter 12 Statistics

Class 9 chapter 12 statistics deals with information about how data can be presented graphically in the form of bar graphs, histogram and frequency polygons. The NCERT class 9 statistics enables the students to know the collection, analysis, interpretation as well as presentation of data information.

By practicing NCERT Solutions for Class 9 Maths statistics solutions, students not only strengthen the understanding of the concept being studied but also enhance their problem solving ability. Every ncert class 9 statistics question is described clearly, giving the students an understanding of the concepts used in statistics. Such an exhaustive practice makes sure that learners are prepared for their exams because statistics is an important subject in mathematics which is essential for higher classes.

1.0Download Class 9 Maths Chapter 12 NCERT Solutions PDF Online

In the below table, students can find chapter 12 statistics class 9 pdf which can be downloaded so students can access it anytime and anywhere.

NCERT Solutions for  Class 9 Maths Chapter 12 - Statistics

2.0NCERT Solutions Class 9 Maths Chapter 12 Statistics: All Exercises

In addition, answering such questions enables a student to discover his/her areas of weakness and be in a position to attend to them long before their examinations. This way the students get the confidence they need to succeed in exams through enhancing their statistics competence.

Exercises

Total Number of Questions

Statistics class 9 exercise 12.1

9

Statistics class 9 exercise 12.2

6

Statistics class 9 exercise 12.3

7

Statistics class 9 exercise 12.4

3

3.0What Will Students Learn in Chapter 12: Statistics?

  • A guide on how to gather and categorize raw information into valuable information.
  • Arithmetic mean, median and mode for describing data.
  • Methods including bar graphs, histograms and frequency polygon to display data.
  • Frequency and relative frequency, making cumulative frequency tables and cumulative frequency graphs.
  • Applying statistical methods to solve real problems that may involve data analysis and/or data interpretation.

4.0NCERT Questions with Solutions for Class 9 Maths Chapter 12 - Detailed Solutions

Exercise : 12.1

  1. A survey was conducted by a group of students as a part of their environment awareness programme, 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 plantsNumber of houses
0−21
2−42
4−61
6−85
8−106
10−122
12−143
  1. Which method did you use for finding the mean, and why?

Sol.

Number of plants) Class(Number of Houses Frequency) (fi​)Marks (xi​)fixi​​
0−2111
2−4236
4−6155
6−85735
8−106954
10−1221122
12−1431339
TotalN=20162

We have, N=Σfi​=20 and Σfi​=162. Then mean of the data is x= N1​×Σfi​Xi​=201​×162=8.1 Hence, the required mean of the data is 8.1 plants.

We find the mean of the data by direct method because the figures are small.

  • Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in Rs.)Number of workers
500−52012
520−54014
540−5608
560−5806
580−50010
  1. Find the mean daily wages of the workers of the factory by using an appropriate method.

Sol.

Daily wages (In Rs.)No. of workers (fi​)Class marks (xi​)fi​xi​
500−520125106120
520−540145307420
540−56085504400
560−58065703420
580−600105905900
TotalN=5027260

We have ∑fi​=50 and ∑fi​=27260 Mean =∑fi​∑fi​xi​​=5027260​ =545.2

  1. 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 Rs.)Number of children
11−137
13−156
15−179
17−1913
19−21f
21−235
23−254

Sol. We may prepare the table as given below :

Daily pocket allowance (in Rs.)Number of children ( fi​ )Class marks ( xi​ )di​=xi​−18​fi​di​
11-13712-6-42
13-15614-4-24
15-17916-2-18
17-191318=a00
19-21f2022f
21-23522420
23-25424624
∑fi​=44+f2f-20

It is given that mean =18. From the table, we have a=18, N=44+f and ∑fi​di​=2f−40 Now, mean =a+ N1​×Σfi​di​ Then substituting the values as given above, we have 18=18+44+f1​×(2f−40) ⇒0=44+f2f−40​⇒f=20.

  1. 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 minuteNumber of women
65−682
68−714
71−743
74−778
77−807
80−834
83−862

Sol.

No. of heart beats per minNo. of (fi​)Class (xi​)
65−68266.5133
68−71469.5278
71−74372.5217.5
74−77875.5604
77−80778.5549.5
80−83481.5326
83−86284.5169
TotalN=302277

Mean =∑fi​∑fi​xi​​=302277​=75.9.

  1. 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.
No. of mangoesNumber of Boxes
50−5215
53−55110
56−58135
59−61115
62−5425
  1. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Sol.

Number of mangoesNumber of boxes fi​Mark xi​ui​=3xi​−57​fi​ui​
49.5−52.51551-2-30
52.5−55.511054-1-110
55.5−58.51355700
58.5−61.5115601115
61.5−64.52563250
TotalN=4025

a=57, h=3, N=400 and Σfi​ui​=25. By step deviation method, Mean =a+h× N1​×Σfi​ui​=57+3×4001​×25 =57.19

  • The table below shows the daily expenditure on food of 25 households in a locality.
Daily expenditure (in Rs.)No. of households
100−1504
150−2005
200−25012
250−3002
300−5002
  1. Find the mean daily expenditure on food by a suitable method.

Sol.

DailyNo. of Exp. (in Rs.) holds (fi​)Class ( xi​ )fi​xi​
100−1504125500
150−2005175875
200−250122252700
250−3002275550
300−3502325650
Total255275

Mean =∑fi​∑fii​​​=255275​=211

  1. 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
  1. Find the mean concentration of SO2​ in the air.

Sol.

Concentration of So2​ (in ppm)xi​Frequencyfi​xi​
0.00−0.040.0240.08
0.04−0.080.0690.54
0.08−0.120.1090.90
0.12−0.160.1420.28
0.16−0.200.1840.72
0.20−0.240.2220.44
Σfi​=30Σfi​Xi​=2.96

Mean =∑fi​∑fi​xi​​=302.96​=0.0986

  1. 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 daysNo. of students
0−611
6−1010
10−147
14−204
20−284
28−383
38−401

Sol.

No. of daysNo. of students (fi​)Class marks (xi​)fixi​​
0−611333
6−1010880
10−1471284
14−2041768
20−2842496
28−3833399
38−4013939
Total40499

Mean =∑fi​∑fi​xi​​=40499​=12.475 12. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %)No. of cities
45−553
55−6510
65−7511
75−858
85−953

Sol.

Literac y rate (in %)No. of cities (fi​)Class marks (xi​)fi​xi​
45−55350150
55−651060600
65−751170770
75−85880640
85−95390270
Total352430

Mean =∑fi​∑fi​xi​​=352430​=69.43

Exercise: 12.2

  1. The following table shows the ages of the patients admitted in a hospital during a year :
Age (in years)No. of patients
5−156
15−2511
25−3521
35−4023
40−4514
45−505
  1. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. Sol. From the given data, we have the modal class 35-45. {∵ It has largest frequency among the given classes of the data}

So, ℓ=35,fm​=23,f1​=21,f2​=14 and h=10

Lifetimes (in hours)Frequency
0−2010
20−4035
40−6052
60−8061
80−10038
100−12029

Mode =ℓ+{2fm​−f1​−f2​fm​−f1​​}×h =35+{46−21−1423−21​}×10=35+1120​ =36.8 years Now, let us find the mean of the data :

Age (in Years)Nuber of patient fi​Class Mark xi​ui​=10xi​−3​fi​ui​
5−15610-2-12
15−251120-1-11
25−352130=a00
35−452340123
45−551450228
55−65560315
TotalN=8043

a=30, h=10, N=80 and ∑fi​ui​=43 Mean =a+h× N1​×Σfi​ui​=30+10×801​×43 =30+5.37=35.37 years Thus, mode =36.8 years and mean =35.37 years. So, we conclude that the maximum number of patients admitted in the hospital are of the age 36.8 years (approx), whereas on an average the age of a patient admitted to the hospital is 35.37 years.

  1. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components : Determine the modal lifetimes of the components. Sol. Modal class of the given data is 60-80. Here, ℓ=60,f1​=61,f0​=52,f2​=38 and h=
  • Mode =ℓ+{2f1​−f0​−f2​f1​−f0​​}×h =60+{122−52−3861−52​}×20 =60+329×20​=60+845​ =60+5.625 =65.625 hours
  • 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.)No. of Families
1000−150024
1500−200040
2000−250033
2500−300028
3000−350030
3500−400022
4000−450016
4500−50007

Sol.

Expenditure (in Rs.)No. of families (fi​)Class marks (xi​)fi​xi​
1000−150024125030000
1500−200040175070000
2000−250033225074250
2500−300028275077000
3000−350030325097500
3500−400022375082500
4000−450016425068000
4500−50007475033250
Total2005,32,000

Mean =∑fi​∑fi​Xi​​=200532500​=2662.5 Modal class =1500−2000 Mode =ℓ+{2f1​−f0​−f2​f1​−f0​​}×h =1500+{2×40−24−3340−24​}×500 =1500+80−5716​×500=1847.83.

  1. The following distribution gives the statewise 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 teacherNo. of states /U.T
15−203
20−258
25−309
30−3510
35−403
40−450
45−500
50−552

Sol. Modal class is (30-35) and its frequency is 10. So, ℓ=30,f1​=10,f0​=9,f2​=3, h=5. Mode =ℓ+{2f1​−f0​−f2​f1​−f0​​}×h =30+{20−9−310−9​}×5=30+85​=30.6

Number of students perNumber of states/ U.T teacherfi​Class Marks xi​ui​=xi​−35
15−20317.5-3fi​ui​
20−25822.5-2-16
25−30927.5-1-9
30−351032.5=a00
35−40337.513
40−45042.520
45−50047.530
50−55252.548
 N=35-23

a=32.5, h=5, N=35 and Σfi​ui​=−23. By step-deviation method, Mean =a+h× N1​×Σfi​ui​ =32.5+5×351​×(−23) =32.5−723​=32.5−3.3=29.2 Hence, Mode =30.6 and Mean =29.2 We conclude that most states/U.T. have a student teacher ratio of 30.6 and on an average, the ratio is 29.2 .

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

Runs SecoredNo. of batsman
3000−40004
4000−500018
5000−60009
6000−70007
7000−80006
8000−90003
9000−100001
10000−110001
  1. Find the mode of the data. Sol. Modal class =4000−5000 Mode =ℓ+{2f1​−f0​−f2​f1​−f0​​}×h =4000+{2×18−4−918−4​}×1000 =4000+{2314​}×1000 =4608.69
  2. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.
No. of carsFrequency
0−107
10−2014
20−3013
30−4012
40−5020
50−6011
60−7015
70−808

Sol. Modal class =40−50 Mode =40+{2×20−12−1120−12​}×10 =40+{40−238​}×10 =40+4.706 =44.706

Exercise: 12.3

  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.
Monthly consumption (in units)Number of consumers
65−854
85−1055
105−12513
125−14520
145−16514
165−1858
185−2054

Sol. (i)

Monthly Consumpti on (in units)Number of consumersCumulative frequency
65−8544
85−10559
105−1251322
125−1452042
145−1651456
165−185864
185−205468
TotalN=68

n=68 gives 2N​=34 So, we have the median class (125-145) ℓ=125, N=68,f=20,cf=22, h=20 Median =ℓ+{f2N​−cf​}×h =125+{2034−22​}×20=137 units. (ii) Modal class is (125 - 145) having maximum frequency f1​=20,f0​=13,f2​=14, ℓ=125 and h=20  Mode =ℓ+{2f1​−f0​−f2​f1​−f0​​}×h=125+{40−13−1420−13​}×20=125+ 137×20​ =125+13140​=125+10.76=135.76 units  (iii)

Monthly consumption (in units)Number of consumers fi​Class marks xi​ui​=xi​fi​xi​
65−85475-3-12
85−105595-2-10
105−12513115-1-13
125−14520135=a00
145−16514155114
165−1858175216
185−2054195312
TotalN=687

N=68,a=135, h=20 and Σfi​ui​=7 By step-deviation method. Mean =a+h× N1​×Σfi​ui​ =135+20×681​×7 =135+1735​=135+2.05 =137.05 units

  • If the median of the distribution given below is 28.5 , find the values of x and y.
Class intervalFrequency
0−105
10−20x
20−3020
30−4015
40−50y
50−605
Total60

Sol.

Class intervalFrequencyCumulative frequency
0−1055
10−20x5+x
20−302025+x
30−401540+x
40−50y40+x+y
50−60545+x+y
Total60

Median =28.5 lies in the class-interval (20-30). Then median class is (20-30). So, we have ℓ=20,f=20,cf=5+x,h=10, N=60 Median =ℓ+{f2N​−cf​}×h=28.5 28.5=20+{2030−(5+x)​}×10 ⇒8.5=225−x​⇒17=25−x⇒x=8 Find the given table, we have i.e., x+y+45=60 or x+y=15 ⇒y=15−x=15−8=7, i.e., y=7

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

Age (in years)No. of policy holders
Below 202
Below 256
Below 3024
Below 3545
Below 4078
Below 4589
Below 5092
Below 5598
Below 60100

Sol.

Age (in years)Number of policy holders fi​Cumulative frequency
Below 202=22
20−25(6−2) =46
Median
class25−30(24−6) =18
30−35(45−24) =2124
40−40(78−45) =2178
40−45(89−78) =1189
45−50(92−89) =392
50−55(98−92) =698
55−60(100−98) =2100
TotalN=100

Here, ℓ=35, N=100,f=33,cf=45, h=5 Median =ℓ+{f2N​−cf​}×h =35+{3350−45​}×5 =35+3325​=35+0.76 =35.76 years.

  1. The length of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table.
Length (in mm)No. of leaves
118−1263
127−1355
136−1449
145−15312
154−1625
163−1714
172−1802
  1. Find the median length of the leaves.

Sol. The given series is in inclusive form. We may prepare the table in exclusive form and prepare the cumulative frequency table as given below :

Length (in mm)No. of leaves (fi​)Cumulative 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
N=40

Here, N=40 ⇒2N​=20 The cumulative frequency just greater than 20 is 29 and the corresponding class is 144.5-153.5. So, the median class is 144.5-153.5. ∴ℓ=144.5, N=40,cf=17,f=12 and h=9 Therefore, median =ℓ+{f2N​−cf​}×h =144.5+12(20−17)​×9 =144.5+123×9​ =144.5+2.25=146.75 Hence, median length of leaves is 146.75 mm .

The following table gives the distribution of the life time of 400 neon lamps :

Life time (in hours)No. of lamps
1500−200014
2000−250056
2500−300060
3000−350086
3500−400074
4000−450062
4500−500048
  1. Find the median life time of a lamp.

Sol.

Life time (in hours)No. of lampscf
1500−20001414
2000−25005670
2500−300060130
3000−350086216
3500−400074290
4000−450062352
4500−500048400

2N​=2400​=200 Median class =3000−3500 Median =ℓ+{f2N​−cf​}×h =3000+{86200−130​}×500=3406.98 hours Hence, median life time of a lamp 3406.98 hours.

  • 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
No. of lettersNo. of surnames
1−46
4−730
7−1040
10−1316
13−164
16−194
  1. Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Sol.

Median classNo. of lettersNo. of Surnames fi​Cumulative frequency
1-466=6
4-7306+30=36​
7-104036+40=76​50=2N​
10-131676+16=92​
13-16492+4=96​
16-194​96+4=100​
TotalN=100

(i) Here,

ℓ=7, N=100,f=40,cf=36, h=3

Median =ℓ+{f2N​−cf​}×h =7+{4050−36​}×3 =7+2021​=8.05 (ii) Modal class is (7-10).

ℓ=7,f1​=40,f0​=30,f2​=16, h=3 Mode =ℓ+{2f1​−f0​−f2​f1​−f0​​}×h=7+{80−30−1640−30​}×3=7+3430​=7.88 (iii) Here, a=8.5, h=3,n=100 and Σfi​ui​=−6.

Number of lettersfi​mark xi​ui​=3xi​−8​3fi​ui​
1−462.5-2-12
4−7305.5-1-30
7−10408.5=a00
10−131611.5116
13−16414.528
16−19417.5312
TotalN=100−6

Mean =a+h× N1​×Σfi​ui​ =8.5+3×1001​×(−6). =8.5−10018​=8.5−0.18=8.32

  • The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Weight (in kg)No. of students
40−452
45−503
50−558
55−606
60−656
65−703
70−75

Sol.

Weight (in kg)No. of studentsCumulative frequency
40−4522
45−5035
50−55813
55−60619
60−65625
65−70328
70−75230

2N​=230​=15 Median class =55−60

 Median =ℓ+{f2N​−cf​}×h=55+{615−13​}×5=56.67 kg

Exercise: 12.4

  1. The following distribution gives the daily income of 50 workers of a factory.
Daily income (in Rs.)No. of workers
100−12012
120−14014
140−1608
160−1806
180−20010

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.

Sol.

Daily income (in Rs.)No. of workersCumulative frequency less than type
100−12012Less than 12012=12
120−14014Less than 140(12+14) =26
140−1608Less than 160(26+8) =34
160−1806Less than 180(34+6) =40
180−20010Less than 200(40+10) =50
TotalN=50

N=50 gives 2N​=25 On the graph, we will plot the points (120,12),(140,26),(160,34),(180,40), (200,50).

  • During the medial check up of 35 students of a class, their weights were recorded as follows:
Weight (in Kg.)No. 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
  1. 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.

Sol.

Weight (in Kg.)No. of studentsCumulative frequency less than type
36−380=0Less than 380
38−40(3−0) =3Less than 403
40−42(5−3) =2Less than 425
42−44(9−5) =4Less than 449
44−46(14−9) =5Less than 4614
Median Class46−48(25−14) =14Less than 4828
48−50(32−28) =4Less than 50322
50−5235−32) =3Less than 5235

To draw the 'less than' type ogive, we plot the points (38,0),(40,3),(42,5),(44,9), (46,14),(48,28),(50,32) and (52,35) on the graph.

Median from the graph =46.5 kg. median class is (46-48). (See in the table) We have ℓ=46,f=14,cf=14, N=35 and h=2. Median =ℓ+{f2N​−cf​}×h =46+{14235​−14​}×2=46+21​=46.5 kg Hence, the median is same as we have noticed from the graph

  1. The following table gives production yield per hectare of wheat of 100 farms of a village.
Production yield (in kg/ha)No. of farms
50−552
55−608
60−6512
65−7024
70−7538
75−8016

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

Production yield (in kg/ha)Number of (Frequency)Cumulative frequency less than type
50-55250 or more than 50100=100
55-60855 or more than 55(100−2)=98​
60-651260 or more than 60(98−8)=90​
65-702465 or  more  than 65​(90−12)=78​
70-753870 or more than 70(78−24)=54​
75-801675 or more than 75(54−38)=16​
N=100

Now, we will draw the ogive by plotting the points (50,100),(55,98),(60,90), (65,78),(70,54) and (75,16). Join these points by a freehand to get an ogive of 'more than' type.

NCERT Solutions for Class 9 Maths Other Chapters:-

Chapter 1: Number Systems

Chapter 2: Polynomials

Chapter 3: Coordinate Geometry

Chapter 4: Linear Equations in Two Variables

Chapter 5: Introduction to Euclid’s Geometry

Chapter 6: Lines and Angles

Chapter 7: Triangles

Chapter 8: Quadrilaterals

Chapter 9: Circles

Chapter 10: Heron’s Formula

Chapter 11: Surface Areas and Volumes

Chapter 12: Statistics


CBSE Notes for Class 9 Maths - All Chapters:-

Class 9 Maths Chapter 1 - Number Systems Notes

Class 9 Maths Chapter 2 - Polynomial Notes

Class 9 Maths Chapter 3 - Coordinate Geometry Notes

Class 9 Maths Chapter 4 - Linear Equation In Two Variables Notes

Class 9 Maths Chapter 5 - Introduction To Euclids Geometry Notes

Class 9 Maths Chapter 6 - Lines and Angles Notes

Class 9 Maths Chapter 7 - Triangles Notes

Class 9 Maths Chapter 8 - Quadrilaterals Notes

Class 9 Maths Chapter 9 - Circles Notes

Class 9 Maths Chapter 10 - Herons Formula Notes

Class 9 Maths Chapter 11 - Surface Areas and Volumes Notes

Class 9 Maths Chapter 12 - Statistics Notes

Frequently Asked Questions

Grouped data is categorized into intervals (like class intervals in histograms), while ungrouped data is raw data that hasn’t been organized into intervals.

Cumulative frequency is the addition of all frequencies up to the end point of the grouped data. It comes in handy to estimate the proportion of observations that are less than a certain value in a given dataset.

The mean for grouped data is calculated using the formula: Mean = Σfixi / Σfi

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