NCERT Solutions for Class 9 Maths Chapter 14 – Statistics

Here we provide NCERT Solutions for Class 9 Maths Chapter 14 – Statistics for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Class 9 Maths solution Chapter 14 pdf. Now you will get step by step solution to each question.

Lives of batteries (in years)No. of batteries
  (Frequency)
2-2.52
2.5-36
3-3.514
3.5-411
4-4.54
4.5-53
Total40


Page No. 258

Exercise 14.3

1. A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 – 44 (in years) worldwide, found the following figures (in %):

S.No.Causes Female fatality rate (%)
1.Reproductive health conditions 31.8
2.Neuropsychiatric conditions 25.4
3.Injuries12.4
4.Cardiovascular conditions 4.3
5.Respiratory conditions 4.1
6.Other causes 22.0


(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.

Answer

(i) The data is represented below graphically.

(ii) From the above graphical data, we observe that reproductive health conditions is the major cause of women’s ill health and death worldwide.

(iii) Two factors responsible for cause in (ii)

• Lack of proper care and understanding.

• Lack of medical facilities.

2. The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below.

S.No.SectionNumber of girls per thousand boys 
1.Scheduled Caste (SC)  940
2.Scheduled Tribe (ST) 970
3.Non SC/ST 920
4.Backward districts 950
5.Non-backward districts 920
6.Rural 930
7.Urban 910


(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.

Answer

(i)

(ii) It can be observed from the above graph that the maximum number of girls per thousand boys is in ST. Also, the backward districts and rural areas have more number of girls per thousand boys than non-backward districts and urban areas.

Page No. 59

3. Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

Political party    A        B       C         D         E         F    
Seats won755537291037

(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?

Answer

(i)

(ii) The party named A has won the maximum number of seat.

4. The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

S.No.Length (in mm)Number of leaves 
1.118 – 126  3
2.127 – 135 5
3.136 – 144 9
4.145 – 15312
5.154 – 162 5
6.163 – 171  4
7.172 – 1802


(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]
(ii) Is there any other suitable graphical representation for the same data?
(iii)Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?

Answer

(i) The data is represented in a discontinuous class interval. So, first we will make continuous. The difference is 1, so we subtract 1/2 = 0.5 from lower limit and add 0.5 to the upper limit.

S.No.Length (in mm)Number of leaves
1.117.5 – 126.53
2.126.5 – 135.55
3.135.5 – 144.59
4.144.5 – 153.512
5.153.5 – 162.55
6.162.5 – 171.54
7.171.5 – 180.52

(ii) Yes, the data can also be represented by frequency polygon.

(iii) No, it is incorrect to conclude that the maximum number of leaves are 153 mm long because maximum number of leaves are lying between the length of 144.5 – 153.5

5. The following table gives the life times of 400 neon lamps:

Life Time (in hours)Number of lamps
300 – 400 14
400 – 50056
500 – 60060
600 – 70086
700 – 80074
800 – 90062
900 – 100048

(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours?

Answer

(i)

(ii) 74 + 62 + 48 = 184 lamps have a life time of more than 700 hours.

Page No. 260

6. The following table gives the distribution of students of two sections according to the marks obtained by them:

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.


Answer 

The class mark can be found by (Lower limit + Upper limit)/2.
For section A, 

MarksClass MarkFrequency
0-1053
10-20159
20-302517
30-403512
40-50459


For section B,

MarksClass MarkFrequency
0-1055
10-201519
20-302515
30-403510
40-50451


Now, we draw frequency polygon for the given data.

7. The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

Represent the data of both the teams on the same graph by frequency polygons.
[Hint : First make the class intervals continuous.]

Answer

The data is represented in a discontinuous class interval. So, first we will make continuous. The difference is 1, so we subtract 1/2 = 0.5 from lower limit and add 0.5 to the upper limit.

Number of ballsTeam ATeam B
0.5-6.525
6.5-12.516
12.5-18.582
18.5-24.5910
24.5-30.545
30.5-36.556
36.5-42.563
42.5-48.5104
48.5-54.568
54.5-60.5210


Now, we draw frequency polygon for the given data.


Page No. 261

8. A random survey of the number of children of various age groups playing in a park was found as follows:

Draw a histogram to represent the data above.

Answer

The class intervals in the data is having varying width. We know that the area of rectangle is proportional to the frequencies in the histogram. The class interval with minimum class size 1 is selected and the length of the rectangle is proportionate to it.

Age (in years)Number of children (frequency)Width of classLength of rectangle
1-251(5/1)×1 = 5
2-331(3/1)×1 = 3
3-562(6/2)×1 = 3
5-7122(12/2)×1 = 6
7-1093(9/3)×1 = 3
10-15105(10/5)×1 = 2
15-1742(4/2)×1 = 2


Taking the age of children on x-axis and proportion of children per 1 year interval on y-axis, the histogram can be drawn

9. 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

(i) Draw a histogram to depict the given information.

(ii) Write the class interval in which the maximum number of surnames lie.

Answer

(i) The class intervals in the data is having varying width. We know that the area of rectangle is proportional to the frequencies in the histogram. The class interval with minimum class size 2 is selected and the length of the rectangle is proportionate to it.

The proportion of the surnames per 2 letters interval can be calculated as:

Number of lettersNumber of surnamesWidth of classLength of rectangle
1-463(6/3)×2 = 4
4-6302(30/2)×2 = 30
6-8442(44/2)×2 = 44
8-12164(16/4)×2 = 8
12-2048(4/8)×2 = 1

(ii) The class interval in which the maximum number of surnames lie is 6-8.

Page No. 269

Exercise 14.4

1. The following number of goals were scored by a team in a series of 10 matches:
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.

Answer

Mean = Sum of all the observations/Total number of observations
= (2+3+4+5+0+1+3+3+4+3)/10 = 28/10 = 2.8

For Median, we will arrange the given data in ascending order,
0, 1, 2, 3, 3, 3, 3, 4, 4, 5
Number of observations (n) = 10
Number of observations are even so we will calculate median as,

= (3+3)/2 = 6/2 = 3

For Mode, we will arrange the given data in ascending order, we have

0, 1, 2, 3, 3, 3, 3, 4, 4, 5.

Here, 3 occurs most frequently (4 times)

∴ Mode = 3


2. In a mathematics test given to 15 students, the following marks (out of 100) are recorded:
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.

Answer

Mean = Sum of all the observations/Total number of observations
= (41+39+48+52+46+62+54+40+96+52+98+40+42+52+60)/15 = 822/15 = 54.8

For Median, we will arrange the given data in ascending order,
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Number of observations (n) = 15
Number of observations are odd so we will calculate median as,

For Mode, we will arrange the given data in ascending order, we have

39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98

Here, 52 occurs most frequently (3 times)

∴ Mode = 52

3. The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.

29, 32, 48, 50, x, x+2, 72, 78, 84, 95

Answer

Number of observations (n) = 10 (even)

According to question, Median = 63

∴ x + 1 = 63

⇒ x = 63−1 = 62

Hence, the value of x is 62.

4. Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.

Answer

The given data is,

14,25,14,28,18,17,18,14,23,22,14,18 

Arranging the data in ascending order, 

14,14,14,14,17,18,18,18,22,23,25,28 

Here, 14 occurs most frequently (4 times). Mode = 14

5. Find the mean salary of 60 workers of a factory from the following table:

Answer

Salary (xi)Number of workers (fi)fixi
30001648000
4000 1248000
50001050000
6000848000
7000642000
8000432000
9000327000
10000110000
TotalΣfi = 60Σfixi = 305000

Hence, the mean salary is ₹5083.33

6. Give one example of a situation in which

(i) the mean is an appropriate measure of central tendency.

(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.

(i) Mean marks in a test in mathematics.

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