Samacheer Kalvi Class 9 Maths Solutions Chapter 8 Statistics Exercise 8.3

 

Question 1. The monthly salary of 10 employees in a factory are given below: Rs 5000, Rs 7000, Rs 5000, Rs 7000, Rs 8000, Rs 7000, Rs 7000, Rs 8000, Rs 7000, Rs 5000. Find the mean, median and mode.
Answer: First, we list the salaries in increasing order to make calculations easier: Rs 5000, Rs 5000, Rs 5000, Rs 7000, Rs 7000, Rs 7000, Rs 7000, Rs 7000, Rs 8000, Rs 8000.

**1. Mean (Average):**
The mean is found by adding all the salaries together and then dividing by the total number of employees.
Sum of salaries \( = 3 \times 5000 + 5 \times 7000 + 2 \times 8000 \)
\( = 15000 + 35000 + 16000 \)
\( = 66000 \)
Total number of employees \( = 10 \)
Mean \( = \frac{66000}{10} = 6600 \)
So, the average salary is Rs 6600.

**2. Median (Middle Value):**
Since there are 10 salaries (an even number), the median is the average of the two middle values. These are the \( \left(\frac{10}{2}\right)^{th} \) and \( \left(\frac{10}{2}+1\right)^{th} \) values, which means the \( 5^{th} \) and \( 6^{th} \) values in the ordered list.
The \( 5^{th} \) value is Rs 7000.
The \( 6^{th} \) value is Rs 7000.
Median \( = \frac{7000+7000}{2} = 7000 \)
The median salary is Rs 7000, which divides the data into two equal halves.

**3. Mode (Most Frequent Value):**
The mode is the salary that appears most often in the list. By looking at the ordered list, we can see:
Rs 5000 appears 3 times.
Rs 7000 appears 5 times.
Rs 8000 appears 2 times.
Since Rs 7000 appears 5 times, more than any other salary, it is the mode.
Mode \( = 7000 \)
Therefore, the mode of the salaries is Rs 7000.
In simple words: The mean (average) salary is Rs 6600. The median (middle) salary is Rs 7000. The mode (most common) salary is also Rs 7000, as it appeared most often.

🎯 Exam Tip: Remember to arrange data in ascending order before calculating the median. For mode, simply count the frequency of each value to find the most common one.

 

Question 2. Find the mode of the given data: 3.1, 3.2, 3.3, 2.1, 1.3, 3.3, 3.1
Answer: To find the mode, we need to see which number appears most frequently in the given data set. Let's list the numbers and count how many times each one appears:
1.3: 1 time
2.1: 1 time
3.1: 2 times
3.2: 1 time
3.3: 2 times
Both 3.1 and 3.3 appear two times, which is more than any other number. When a data set has two modes, it is called bimodal.
Therefore, the modes of the given data are 3.1 and 3.3.
In simple words: The mode is the number that shows up most often. In this list, both 3.1 and 3.3 show up twice, which is more than any other number, so they are both modes.

🎯 Exam Tip: A data set can have one mode, multiple modes (bimodal or multimodal), or no mode if all values appear with the same frequency.

 

Question 3. For the data 11, 15, 17, x + 1, 19, x − 2, 3 if the mean is 14, find the value of x. Also find the mode of the data.
Answer: First, let's find the value of x using the given mean.
The given data values are 11, 15, 17, x + 1, 19, x - 2, 3.
The total number of observations (n) is 7.
The mean is given as 14.

**1. Find x:**
The formula for mean is: Mean \( = \frac{\text{Sum of observations}}{\text{Number of observations}} \)
Sum of observations \( = 11 + 15 + 17 + (x+1) + 19 + (x-2) + 3 \)
\( = (11+15+17+1+19-2+3) + (x+x) \)
\( = 64 + 2x \)
Now, substitute this into the mean formula:
\( 14 = \frac{2x+64}{7} \)
Multiply both sides by 7:
\( 14 \times 7 = 2x+64 \)
\( 98 = 2x+64 \)
Subtract 64 from both sides:
\( 98 - 64 = 2x \)
\( 34 = 2x \)
Divide by 2:
\( x = \frac{34}{2} \)
\( x = 17 \)

**2. Find the mode of the data:**
Now that we have the value of x, let's substitute it back into the data points that contained x:
x + 1 \( = 17 + 1 = 18 \)
x - 2 \( = 17 - 2 = 15 \)
So, the complete data set is: 11, 15, 17, 18, 19, 15, 3.
To find the mode, we count how many times each number appears:
3: 1 time
11: 1 time
15: 2 times
17: 1 time
18: 1 time
19: 1 time
The number 15 appears two times, which is more than any other number. Therefore, 15 is the mode.
Final Answer: The value of x is 17 and the mode of the data is 15.
In simple words: We first added up all the numbers, including the 'x' terms, and used the given mean to find that 'x' is 17. Then, we put 17 back into the list of numbers and found that 15 was the number that appeared most often, making it the mode.

🎯 Exam Tip: Always double-check your value of x by plugging it back into the original data and recalculating the mean to ensure it matches the given value. This helps catch calculation errors.

 

Question 4. The demand of track suit of different sizes as obtained by a survey is given below:

Which size is demanded more?
Answer: To find which size is demanded more, we need to identify the size with the highest number of persons (frequency). This is essentially finding the mode of the data.
Looking at the "No. of Persons" row in the table:
Size 38 is demanded by 36 persons.
Size 39 is demanded by 15 persons.
Size 40 is demanded by 37 persons.
Size 41 is demanded by 13 persons.
Size 42 is demanded by 26 persons.
Size 43 is demanded by 8 persons.
Size 44 is demanded by 6 persons.
Size 45 is demanded by 2 persons.
The highest frequency is 37, which corresponds to Size 40. This means Size 40 is the mode because it has the highest demand.
Therefore, Size 40 is demanded more.
In simple words: The mode tells us the most popular item. By looking at the table, we see that 37 people want size 40, which is more than any other size, so size 40 is the most demanded.

🎯 Exam Tip: When given frequency distribution data, the mode is simply the value or class interval that has the highest frequency. No complex calculations are needed, just careful observation.

 

Question 5. Find the mode of the following data:

Answer: To find the mode for grouped data, we first identify the modal class, which is the class interval with the highest frequency. Then, we use the mode formula.

From the table, the highest frequency is 46, which corresponds to the class interval 20-30. So, the modal class is 20-30.
Now, we identify the values needed for the mode formula:
\( l \) = lower limit of the modal class = 20
\( f \) = frequency of the modal class = 46
\( f_1 \) = frequency of the class preceding the modal class = 38
\( f_2 \) = frequency of the class succeeding the modal class = 34
\( c \) = class size (upper limit - lower limit) = \( 30 - 20 = 10 \)

The formula for mode of grouped data is:
\( Mode = l + \left(\frac{f-f_1}{2f-f_1-f_2}\right) \times c \)
Substitute the values into the formula:
\( Mode = 20 + \left(\frac{46-38}{2 \times 46 - 38 - 34}\right) \times 10 \)
\( Mode = 20 + \left(\frac{8}{92 - 72}\right) \times 10 \)
\( Mode = 20 + \left(\frac{8}{20}\right) \times 10 \)
\( Mode = 20 + \frac{80}{20} \)
\( Mode = 20 + 4 \)
\( Mode = 24 \)
Therefore, the mode of the given data is 24.
In simple words: For grouped data, the mode is found by using a special formula. First, find the class with the most students (that's the modal class). Then, use its lower value, its number of students, and the number of students from the classes just before and after it in the formula to calculate the exact mode.

🎯 Exam Tip: Be careful with the class frequencies \( f_1 \) (preceding) and \( f_2 \) (succeeding) in the mode formula; a common mistake is to swap them or use the wrong frequency. Always confirm the modal class before proceeding.

 

Question 6. Find the mode of the following distribution

Answer: The given class intervals are in inclusive form (e.g., 25-34, 35-44). Before calculating the mode, it's good practice to convert them to exclusive form. We do this by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class. This creates continuous intervals for calculations.

The converted table with exclusive class intervals is:

From the updated table, the highest frequency is 14, which corresponds to the class interval 54.5-64.5. This is our modal class.
Now, let's identify the values for the mode formula:
\( l \) = lower limit of the modal class = 54.5
\( f \) = frequency of the modal class = 14
\( f_1 \) = frequency of the class preceding the modal class (44.5-54.5) = 10
\( f_2 \) = frequency of the class succeeding the modal class (64.5-74.5) = 8
\( c \) = class size = \( 64.5 - 54.5 = 10 \)

Using the formula for mode of grouped data:
\( Mode = l + \left(\frac{f-f_1}{2f-f_1-f_2}\right) \times c \)
Substitute the values:
\( Mode = 54.5 + \left(\frac{14-10}{2 \times 14 - 10 - 8}\right) \times 10 \)
\( Mode = 54.5 + \left(\frac{4}{28 - 18}\right) \times 10 \)
\( Mode = 54.5 + \left(\frac{4}{10}\right) \times 10 \)
\( Mode = 54.5 + 4 \)
\( Mode = 58.5 \)
Therefore, the mode of the given distribution is 58.5.
In simple words: We first adjusted the weight ranges to be continuous. Then, we found the range with the most students (the modal class). Using a formula with the details of this class and the classes next to it, we calculated the most common weight, which is 58.5 kgs.

🎯 Exam Tip: Always check if the class intervals are inclusive or exclusive. If they are inclusive, convert them to exclusive form by adjusting the limits by 0.5 before applying the mode formula for accurate results.