RBSE Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 9 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 9. Students can also read RBSE Class 9 Maths Important Questions for exam preparation. Students can also go through RBSE Class 9 Maths Notes to understand and remember the concepts easily. Practicing the class 9 math chapter 13 hindi medium textbook questions will help students analyse their level of preparation.

RBSE Class 9 Maths Solutions Chapter 14 Statistics Exercise 14.4

Question 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:
Using x̄ = \(\frac{x_{1}+x_{2}+\ldots .+x_{10}}{10}\), the mean is
x̄ = \(\frac{2+3+4+5+0+1+3+3+4+3}{10}=\frac{28}{10}\) = 2.8
To find the median, arrange the given data in ascending order, as follows:
0, 1, 2, 3, 3, 3, 3, 4, 4, 5.
There are 10 terms. So, there are two middle terms, i.e. the \(\left(\frac{10}{2}\right)\)th and \(\left(\frac{10}{2}+1\right)\)th, i.e. the 5th and 6th terms.
So, the median is the mean of the values of the 5th and 6th terms.
i.e. the median = \(\frac{3+3}{2}\) = 3
Again, in the data 3 occurs most frequently, i.e. 4 times. So, mode = 3.

RBSE Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4

Question 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:
Using x̄ = \(\frac{x_{1}+x_{2}+\ldots .+x_{10}}{10}\), the mean is
x̄ = \(\frac{41+39+48+52+46+62+54+40+96+52+98+40+42+52+60}{15}\)
= \(\frac{822}{15}\) = 54.8
To find the median, arrange the given data in ascending order, as follows:
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98.
Since, the number of terms is 15, an odd number, we find out the median by finding the marks obtained by \(\left(\frac{15+1}{2}\right)\)th student, i.e. the 8th student.
∴ The median marks = 52.
Again, in the data, 52 occurs most frequently, i.e. 3 times.
∴ Mode = 52.

Question 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:
The given data arranged in ascending order is as follows:
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95.
There are 10 terms. So, there are two middle terms, i.e. the \(\left(\frac{10}{2}\right)\)th and \(\left(\frac{10}{2}+1\right)\)th, i.e. the 5th and 6th terms.
So, the median is the mean of the values of the 5th and 6th terms.
i.e. median = \(\frac{x+(x+2)}{2}\) = x + 1
But median = 63
∴ x + 1 = 63
or x = 63 - 1 = 62.

Question 4.
Find the mode of 14, 25, 14, 28, 18,17, 18, 14, 23, 22, 14, 18.
Answer:
Here, 14 occurs most frequently, i.e. 4 times.
∴ Mode = 14.

RBSE Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4

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

Salary (in )

Number of workers

3000

16

4000

12

5000

10

6000

8

7000

6

8000

4

9000

3

10000

1

Total

60

Answer:
Calculation of Mean

Salary (in ) xi

Number of workers

(fi)

fixi

3000

16

48000

4000

12

48000

5000

10

50000

6000

8

48000

7000

6

42000

8000

4

32000

9000

3

27000

10000

1

10000

 

Σfi = 60

ΣfiXi = 305000

∴ Mean = \(\frac{\sum f_{i} x_{i}}{\sum f_{i}}\) = \(\frac{305000}{60}\) 5083.33
Thus, mean salary of 60 workers is ₹ 5083.33 (approx.).

Question 6.
Give one example of a situation in which :
(i) the mean is an appropriate measure of central tendency.
Answer:
Marks award to a student in 5 weekly tests are : 7, 8, 8, 9, 10 (out of 10).
Here, Median = 8, Mode = 8
But we find, Mean = \(\frac{7+8+8+9+10}{5}\) = 8.4
So, here we find that the mean value is more appropriate measure of central tendency.

RBSE Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4

(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
Answer:
Since, the scores 3, 5, 14, 18, 20 are all different and median is not affected by extreme values, therefore, median is a more suitable representative of marks.
Here, median = \(\left(\frac{5+1}{2}\right)\)th item = 14
mean = \(\frac{3+5+14+18+20}{5}\) = 12.

admin_rbse
Last Updated on April 20, 2022, 12:50 p.m.
Published April 20, 2022