GSEB 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:
1. Mean:
Mean \( = \frac{\text{Sum of all the observation}}{\text{Total number of observation}} \)
\( = \frac{2+3+4+5+0+1+3+3+4+3}{10} \)
\( = \frac{28}{10} = 2.8 \)

2. Median:
To find the median, first arrange the given data in increasing order:
0, 1, 2, 3, 3, 3, 3, 4, 4, 5.
The total number of observations (n) is 10, which is an even number.
For even 'n', the median is the average of the \( \left( \frac{n}{2} \right)^{th} \) observation and the \( \left( \frac{n}{2}+1 \right)^{th} \) observation.
Median \( = \frac{\left( \frac{10}{2} \right)^{th} \text{ observation} + \left( \frac{10}{2}+1 \right)^{th} \text{ observation}}{2} \)
\( = \frac{5^{th} \text{ observation} + 6^{th} \text{ observation}}{2} \)
\( = \frac{3+3}{2} = \frac{6}{2} = 3 \)

3. Mode:
Arranging the data in ascending order, we have:
0, 1, 2, 3, 3, 3, 3, 4, 4, 5.
Here, the number 3 appears most frequently (4 times).
So, the mode is 3.
In simple words: We calculated the average score (mean), the middle score after ordering (median), and the most frequent score (mode). The mean score is 2.8, the median score is 3, and the mode is also 3.

Exam Tip: Remember to always arrange data in ascending or descending order before finding the median. For mode, count the frequency of each observation carefully.

 

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:
1. Mean:
Mean \( = \frac{\text{Sum of all the observations}}{\text{Total number of observations}} \)
\( = \frac{41+39+48+52+46+62+54+40+96+52+98+40+42+52+60}{15} \)
\( = \frac{822}{15} = 54.8 \)

2. Median:
First, arrange the given data in increasing order:
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98.
The total number of observations (n) is 15, which is an odd number.
For odd 'n', the median is the \( \left( \frac{n+1}{2} \right)^{th} \) observation.
Median \( = \left( \frac{15+1}{2} \right)^{th} \) observation
\( = 8^{th} \) observation \( = 52 \)

3. Mode:
Arranging the data in increasing order, we have:
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98.
Here, the number 52 occurs most frequently (3 times).
So, the mode is 52.
In simple words: We calculated the mean, median, and mode for the given test marks. The average score (mean) is 54.8, the middle score (median) is 52, and the most common score (mode) is also 52.

Exam Tip: For test scores, it's common for the mean, median, and mode to be close to each other, indicating a relatively normal distribution of marks.

 

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 observations in ascending order are:
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95.
The total number of observations (n) is 10, which is an even number.
When 'n' is even, the median is the average of the \( \left( \frac{n}{2} \right)^{th} \) observation and the \( \left( \frac{n}{2}+1 \right)^{th} \) observation.
Median \( = \frac{\left( \frac{10}{2} \right)^{th} \text{ observation} + \left( \frac{10}{2}+1 \right)^{th} \text{ observation}}{2} \)
\( = \frac{5^{th} \text{ observation} + 6^{th} \text{ observation}}{2} \)
From the given data, the 5th observation is \( x \) and the 6th observation is \( x+2 \).
So, Median \( = \frac{x+(x+2)}{2} = \frac{2x+2}{2} = x+1 \)
According to the question, the median is 63.
Therefore, \( x+1 = 63 \)
\( \implies x = 63 - 1 \)
\( \implies x = 62 \)
Hence, the value of x is 62.
In simple words: Since we know the observations are already ordered and the total count is even, we found the median by averaging the two middle values. We set this average equal to the given median of 63 and solved for x, which turned out to be 62.

Exam Tip: When finding the median for an even number of observations, make sure to average the two middle values. Don't pick just one of them.

 

Question 4. Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
Answer:
First, arrange the given data in ascending order:
14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28.
Now, let's count the frequency of each number:
- 14 appears 4 times.
- 17 appears 1 time.
- 18 appears 3 times.
- 22 appears 1 time.
- 23 appears 1 time.
- 25 appears 1 time.
- 28 appears 1 time.
Here, the number 14 occurs most frequently (4 times).
Therefore, the mode is 14.
In simple words: We listed all the numbers in order and then counted how many times each number appeared. The number 14 showed up more than any other number, so 14 is the mode.

Exam Tip: The mode is simply the value that appears most often in a data set. Listing the data in order can help you spot the most frequent value easily.

 

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

Answer:
To find the mean salary, we create a frequency distribution table and calculate \( f_i x_i \) for each class.

The formula for the mean \( \bar{x} \) is:
\( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \)
\( \bar{x} = \frac{305000}{60} \)
\( \bar{x} = 5083.33 \)
Hence, the mean salary is Rs. 5083.33.
In simple words: To find the average salary, we multiplied each salary by the number of workers earning it, then added all those products together. We divided this total by the total number of workers. The average salary for the factory workers is Rs. 5083.33.

Exam Tip: Remember to calculate \( f_i x_i \) correctly for each row and sum both \( f_i \) and \( f_i x_i \) accurately before applying the mean formula.

 

Question 6. Give an example of a situation in which:
• the mean is an appropriate measure of central tendency.
• the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
Answer:
Here are examples for the given situations:
• When the mean is an appropriate measure of central tendency: Mean marks in a test in mathematics. (Since marks usually don't have extreme outliers that would skew the average significantly.)
• When the mean is not an appropriate measure of central tendency but the median is appropriate: Average beauty scores given by judges. (Beauty scores can be very subjective, and extreme ratings by a few judges might pull the mean drastically, making the median a better representation of the typical score.)
In simple words: The mean works well for things like test scores because they are usually grouped together. But for things like beauty scores, where a few very high or very low ratings could make the average seem wrong, the median is a better choice because it finds the middle score and ignores those extreme numbers.

Exam Tip: The mean is suitable for data without extreme values (outliers), while the median is preferred when data might have outliers, as it is less affected by them.