Chapter 14 : Statistics  Ncert Solutions for Class 9 Maths CBSE
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Chapter 14  Statistics Excercise Ex. 14.1
1. Number of females per 1000 males in various states of our country.
2. Height and Weights of students of our class.
3. Temperature of past 10 days in our city.
4. Number of plants in our locality.
5. Rain fall in our city.
6. Marks obtained by students of the class in a test.
7. Date of birth of students.
8. Subjects taught in various schools in class X.
Chapter 14  Statistics Excercise Ex. 14.2
So, the table representing the data is as follows:
Blood group 
Number of students 
A 
9 
B 
6 
AB 
3 
O 
12 
Total 
30 
As 12 students have the blood group O and 3 have their blood group as AB, clearly that the most common blood group and the rarest blood group among these students is O and AB respectively.
Required grouped frequency distribution table as following 
Distance (in km) 
Tally marks 
Number of engineers 
0  5 

5 
5  10 

11 
10 15 

11 
15  20 

9 
20  25 

1 
25  30 

1 
30  35 

2 
Total 

40 
Most of the engineers are having their workplace up to 15 km distance, from their homes.
Class intervals will be as follows 84  86, 86  88, and 88  90... ...
Relative humidity (in %) 
Number of days (frequency ) 
84  86 
1 
86  88 
1 
88  90 
2 
90  92 
2 
92  94 
7 
94  96 
6 
96  98 
7 
98  100 
4 
Total 
30 
(iii) Range of data = maximum value  minimum value
= 99.2  84.9
Heights (in cm) 
Number of students (frequency ) 
150  155 
12 
155  160 
9 
160  165 
14 
165  170 
10 
170  175 
5 
Total 
50 
(ii) From the table we can see that 50% of students are shorter than 165 cm.
Concentration of SO_{2} (in ppm) 
Number of days (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 
Total 
30 
Number of days for which concentration SO2 is more than 0.11 is number of days for which concentration is in between 0.12  0.16, 0.16  0.20, 0.20  0.24.
So, required number of days = 2 + 4 + 2 = 8
Number of heads 
Number of times (frequency) 
0 
6 
1 
10 
2 
9 
3 
5 
Total 
30 
Digit 
Frequency 
0 
2 
1 
5 
2 
5 
3 
8 
4 
4 
5 
5 
6 
4 
7 
4 
8 
5 
9 
8 
Total 
50 
(ii) From the above table the least frequency is 2 of digit 0, and the maximum frequency is 8 of digit 3 and 9. So, the most frequently occurring digits are 3 and 9 and the least occurring digit is 0.
The grouped frequency distribution table is as follows:
Hours 
Number of children 
0  5 
10 
5  10 
13 
10  15 
5 
15  20 
2 
Total 
30 
(ii) The number of children, who watched TV for 15 or more hours a week
is 2 (i.e. number of children in class interval 15  20).
Lives of batteries (in hours) 
Number of batteries 
2  2.5 
2 
2.5  3.0 
6 
3.0  3.5 
14 
3.5  4.0 
11 
4.0  4.5 
4 
4.5  5.0 
3 
Total 
40 
Chapter 14  Statistics Excercise Ex. 14.3
All the rectangle bars are of same width and having equal spacing between them.
(ii) Reproductive health condition is the major cause of women's ill health and death worldwide as 31.8% of women are affected by it.
Here all the rectangle bars are of same width and have equal spacing in between them.
Here rectangle bars are of same width and have equal spacing in between them.
(ii). We may find that political party 'A' won maximum number of seats.
Length (in mm) 
Number of leaves 
117.5  126.5 
3 
126.5  135.5 
5 
135.5  144.5 
9 
144.5  153.5 
12 
153.5  162.5 
5 
162.5  171.5 
4 
171.5  180.5 
2 
Now taking length of leaves on x axis and number of leaves on y axis we can draw the histogram of this information as below 
Here 1 unit on y axis represents 2 leaves.
(ii). Other suitable graphical representation of this data could be frequency polygon.
(iii). No as maximum number of leaves (i.e. 12) have their length in between of 144.5 mm and 153.5 mm. It is not necessary that all have their lengths as 153 mm.
Here 1 unit on y axis represents 10 lamps.
(ii). Number of neon lamps having their lifetime more than 700 are sum of number of neon lamps having their lifetime as 700  800, 800  900, and 900  1000.
So number of neon lamps having their lifetime more than 700 hours is 184. (74 + 62 + 48 = 184)
Class mark
Section A 
Section B 

Marks 
Class marks 
Frequency 
Marks 
Class marks 
Frequency 
0  10 
5 
3 
0  10 
5 
5 
10  20 
15 
9 
10  20 
15 
19 
20  30 
25 
17 
20  30 
25 
15 
30  40 
35 
12 
30  40 
35 
10 
40  50 
45 
9 
40  50 
45 
1 
From the graph we can see performance of students of section 'A' is better than the students of section 'B' as for good marks.
Also class mark of each interval can be found by using formula 
Number of balls 
Class mark 
Team A 
Team B 
0.5  6.5 
3.5 
2 
5 
6.5  12.5 
9.5 
1 
6 
12.5  18.5 
15.5 
8 
2 
18.5  24.5 
21.5 
9 
10 
24.5  30.5 
27.5 
4 
5 
30.5  36.5 
33.5 
5 
6 
36.5  42.5 
39.5 
6 
3 
42.5  48.5 
45.5 
10 
4 
48.5  54.5 
51.5 
6 
8 
54.5  60.5 
57.5 
2 
10 
Now by taking class marks on x axis and runs scored on y axis we can construct frequency polygon as following 
Age (in years) 
Frequency (Number of children) 
Width of class 
Length of rectangle 
1  2

5

1 

2  3 
3 
1 

3  5 
6 
2 

5  7 
12 
2 

7  10 
9 
3 

10  15 
10 
5 

15  17 
4 
2 

Now taking age of children on x axis and proportion of children per 1 year interval on y axis we may draw histogram as below 
Number of letters 
Frequency (Number of surnames) 
Width of class 
Length of rectangle 
1  4 
6 
3 

4  6 
30 
2 

6  8 
44 
2 

8 12 
16 
4 

12  20 
4 
8 

Now by taking number of letters on x axis and proportion of number of surnames per 2 letters interval on y axis and choosing an appropriate scale (1 unit = 4 students for y axis) we will construct the histogram as below
(ii). The class interval in which the maximum number of surname lie is
6  8 as there are 44 number of surnames in it i.e. maximum for this data.
Chapter 14  Statistics Excercise Ex. 14.4
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
0, 1, 2, 3, 3, 3, 3, 4, 4, 5
As the number of observations is 10. 10 is an even number. So, median score will be
Mode of data is the observation with the maximum frequency in data.
So, mode score of data is 3 as it is having maximum frequency as 4 in the data.
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
As the number of observations is 15 that is odd so, median of data will be = 8^{th} observation while data is arranged in an ascending or descending order
So, median score of data = 52
Mode of data is the observation with the maximum frequency in data. So mode of this data is 52 having the highest frequency in data as 3.
14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28
Here observation 14 is having the highest frequency i.e. 4 in given data. So, mode of given data is 14.
Salary (in Rs) (x_{i}) 
Number of workers (f_{i}) 
f_{i}x_{i} 
3000 
16 
3000 * 16 = 48000 
4000 
12 
4000 * 12 = 48000 
5000 
10 
5000 * 10 = 50000 
6000 
8 
6000 * 8 = 48000 
7000 
6 
7000 * 6 = 42000 
8000 
4 
8000 * 4 = 32000 
9000 
3 
9000 * 3 = 27000 
10000 
1 
10000 * 1 = 10000 
Total 


(ii) Mean is not suitable in cases where there are very high and low values for example salary in a company.
CBSE Class 9 Maths Homework Help
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