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.

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.

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.

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

Answer:
Calculation of Mean

∴ 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.

(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.