Measures of Central Tendency
Reading Time: 120 mins
Solved Examples: 50
Practice Questions: 5
What are Measures of Central Tendency?
Measures of Central Tendency are statistical measures used to represent a set of data by a single value that describes its centre or typical value. The three commonly used measures of central tendency are the Mean, Median and Mode. Each measure summarises the data in a different way and is useful in different situations.
Types of Measures of Central Tendency
Measures of Central Tendency are statistical measures that describe the centre or typical value of a data set. Instead of examining every observation individually, they provide a single value that represents the entire collection of data.
The three commonly used measures of central tendency are:
Mean
Median
Mode
Each measure represents the centre of the data in a different way and is suitable for different types of data.
The three commonly used measures of central tendency are:
Mean
Median
Mode
Each measure represents the centre of the data in a different way and is suitable for different types of data.
Arithmetic Mean (Mean)
The Arithmetic Mean, commonly called Mean or Average, is the most frequently used measure of central tendency.
It is calculated by adding all observations and dividing the sum by the total number of observations.
Formula:
Mean = Sum of all observations ÷ Number of observations
Using symbols:
Mean = Σx ÷ n
where:
Σx = Sum of all observations
n = Total number of observations
Example:
The marks obtained by a student in five subjects are:
72, 65, 80, 75, 68
Mean
= (72 + 65 + 80 + 75 + 68) ÷ 5
= 360 ÷ 5
= 72
Therefore, the average marks of the student are 72.
The mean considers every observation in the data set, making it a useful measure when all values are important.
It is calculated by adding all observations and dividing the sum by the total number of observations.
Formula:
Mean = Sum of all observations ÷ Number of observations
Using symbols:
Mean = Σx ÷ n
where:
Σx = Sum of all observations
n = Total number of observations
Example:
The marks obtained by a student in five subjects are:
72, 65, 80, 75, 68
Mean
= (72 + 65 + 80 + 75 + 68) ÷ 5
= 360 ÷ 5
= 72
Therefore, the average marks of the student are 72.
The mean considers every observation in the data set, making it a useful measure when all values are important.
Mean of Frequency Distribution
Sometimes, an observation occurs more than once. Instead of writing repeated values again and again, we represent them using frequencies.
Frequency tells us how many times a particular observation occurs.
For frequency distribution, the mean is calculated using:
Mean = Σfx ÷ Σf
where:
f = Frequency of an observation
x = Observation
Σfx = Sum of the product of observation and frequency
Σf = Total frequency
Example:
Frequency tells us how many times a particular observation occurs.
For frequency distribution, the mean is calculated using:
Mean = Σfx ÷ Σf
where:
f = Frequency of an observation
x = Observation
Σfx = Sum of the product of observation and frequency
Σf = Total frequency
Example:
| Number of Books (x) | Frequency (f) |
|---|---|
| 2 | 3 |
| 4 | 5 |
| 6 | 2 |
Calculation:
Σfx
= (2 × 3) + (4 × 5) + (6 × 2)
= 6 + 20 + 12
= 38
Σf
= 3 + 5 + 2
= 10
Mean
= 38 ÷ 10
= 3.8
Therefore, the average number of books is 3.8.
Median
The Median is the middle value of a data set when the observations are arranged in ascending or descending order.
The median divides the data into two equal parts:
• Half of the observations are smaller than the median.
• Half of the observations are greater than the median.
Steps to find Median:
Arrange the observations in ascending or descending order.
Count the total number of observations.
Identify the middle value.
When the number of observations is odd:
Median = Value of the middle observation
Example:
Find the median of: 8, 3, 5, 12, 7
Arrange the data: 3, 5, 7, 8, 12
Middle value = 7
Median = 7
When the number of observations is even:
Median = Average of the two middle observations
Example:
Find the median of: 4, 8, 6, 10, 12, 14
Arrange the data: 4, 6, 8, 10, 12, 14
Middle values are 8 and 10.
Median = (8 + 10) ÷ 2
= 9
The median divides the data into two equal parts:
• Half of the observations are smaller than the median.
• Half of the observations are greater than the median.
Steps to find Median:
Arrange the observations in ascending or descending order.
Count the total number of observations.
Identify the middle value.
When the number of observations is odd:
Median = Value of the middle observation
Example:
Find the median of: 8, 3, 5, 12, 7
Arrange the data: 3, 5, 7, 8, 12
Middle value = 7
Median = 7
When the number of observations is even:
Median = Average of the two middle observations
Example:
Find the median of: 4, 8, 6, 10, 12, 14
Arrange the data: 4, 6, 8, 10, 12, 14
Middle values are 8 and 10.
Median = (8 + 10) ÷ 2
= 9
Mode
The Mode is the value that occurs most frequently in a data set.
It represents the most common observation.
Example:
Find the mode of:
5, 7, 8, 5, 9, 5, 10
Frequency of 5 = 3
Other values occur only once.
Therefore:
Mode = 5
A data set can have:
• One mode → When one value occurs most frequently.
• Two modes → When two values have the same highest frequency.
• No mode → When no value repeats.
Mode is especially useful when finding the most popular choice, most common score, or most frequently occurring category.
It represents the most common observation.
Example:
Find the mode of:
5, 7, 8, 5, 9, 5, 10
Frequency of 5 = 3
Other values occur only once.
Therefore:
Mode = 5
A data set can have:
• One mode → When one value occurs most frequently.
• Two modes → When two values have the same highest frequency.
• No mode → When no value repeats.
Mode is especially useful when finding the most popular choice, most common score, or most frequently occurring category.
Comparison of Mean, Median and Mode
Although all three measures represent the centre of a data set, they are calculated differently and are suitable for different situations.
| Measure | Based On | Affected by Extreme Values | Best Used When |
|---|---|---|---|
| Mean | All observations in the data set. | Yes | The data does not contain extremely large or small values. |
| Median | Middle value after arranging the data. | No | The data contains extreme values or unusual observations. |
| Mode | Most frequently occurring observation. | No | Finding the most common value or category is required. |
Choosing the appropriate measure depends on the nature of the data and the purpose of the analysis.
Relationship Between Mean, Median and Mode
Mean, Median and Mode are related to each other in a moderately distributed data set.
The empirical relationship is:
Mode = 3 × Median − 2 × Mean
This relationship helps us find one measure when the other two measures are known.
Example:
If Mean = 20 and Median = 25, find Mode.
Mode = 3 × 25 − 2 × 20
= 75 − 40
= 35
Therefore, Mode = 35.
This formula is an approximate relationship and is mainly used for solving problems where one value is missing.
The empirical relationship is:
Mode = 3 × Median − 2 × Mean
This relationship helps us find one measure when the other two measures are known.
Example:
If Mean = 20 and Median = 25, find Mode.
Mode = 3 × 25 − 2 × 20
= 75 − 40
= 35
Therefore, Mode = 35.
This formula is an approximate relationship and is mainly used for solving problems where one value is missing.
Effect of Change in Data on Mean
The mean changes when every observation in a data set is changed.
If the same value is added to every observation:
New Mean = Old Mean + Added Value
If the same value is subtracted from every observation:
New Mean = Old Mean − Subtracted Value
If every observation is multiplied by a number:
New Mean = Old Mean × Multiplier
Example:
The average age of five students is 14 years. If each student’s age increases by 2 years, what will be the new average age?
New Mean
= 14 + 2
= 16 years
The change in every observation produces the same change in the mean.
If the same value is added to every observation:
New Mean = Old Mean + Added Value
If the same value is subtracted from every observation:
New Mean = Old Mean − Subtracted Value
If every observation is multiplied by a number:
New Mean = Old Mean × Multiplier
Example:
The average age of five students is 14 years. If each student’s age increases by 2 years, what will be the new average age?
New Mean
= 14 + 2
= 16 years
The change in every observation produces the same change in the mean.
Choosing the Appropriate Measure
The choice of measure depends on the type of data and the purpose of analysis.
Mean is preferred when all observations are important and extreme values are not present.
Example:
Average marks of students in a class.
Median is preferred when extreme values can affect the result.
Example:
Average income of people where a few individuals have very high incomes.
Mode is preferred when the most common value or category is required.
Example:
Most popular shoe size sold in a store.
Choosing the correct measure helps in understanding the data more accurately.
Mean is preferred when all observations are important and extreme values are not present.
Example:
Average marks of students in a class.
Median is preferred when extreme values can affect the result.
Example:
Average income of people where a few individuals have very high incomes.
Mode is preferred when the most common value or category is required.
Example:
Most popular shoe size sold in a store.
Choosing the correct measure helps in understanding the data more accurately.
Summary of Measures of Central Tendency
| Measure | Definition | Formula / Method | Important Points |
|---|---|---|---|
| Mean | The average value obtained by dividing the sum of all observations by the total number of observations. | Mean = Σx ÷ n | Uses every observation and is affected by extremely large or small values. |
| Mean of Frequency Distribution | The average value calculated when observations occur with different frequencies. | Mean = Σfx ÷ Σf | Frequency represents the number of times an observation occurs. |
| Median | The middle value of a data set after arranging the observations in ascending or descending order. | Odd observations: Median = Value of (n + 1) ÷ 2 th observation
Even observations: Median = Average of two middle observations |
Not affected by extreme values and divides the data into two equal parts. |
| Mode | The observation that occurs most frequently in a data set. | Mode = Value with the highest frequency | Useful for finding the most common value or category. |
| Mean, Median and Mode Relationship | Shows the approximate relationship between the three measures for moderately distributed data. | Mode ≈ 3 × Median − 2 × Mean | Helps find one measure when the other two measures are known. |
| Effect of Change on Mean | Explains how the mean changes when the same value is added, subtracted, multiplied or divided from every observation. | Add a value → New Mean = Old Mean + Value
Subtract a value → New Mean = Old Mean − Value |
Any equal change in all observations produces the same change in the mean. |
| Choosing the Appropriate Measure | The choice of measure depends on the type of data and the purpose of analysis. | Mean → General average
Median → Data with extreme values Mode → Most common value |
Selecting the correct measure gives a more meaningful representation of data. |
Solved Examples
Question 1: Find the mean of the following numbers 12, 15, 18, 20, 25
Sum of observations
= 12 + 15 + 18 + 20 + 25
= 90
Number of observations = 5
Mean = 90 ÷ 5
= 18
= 12 + 15 + 18 + 20 + 25
= 90
Number of observations = 5
Mean = 90 ÷ 5
= 18
Question 2: Find the mean of the first five consecutive natural numbers.
The first five consecutive natural numbers are:
1, 2, 3, 4, 5
Mean
= (1 + 2 + 3 + 4 + 5) ÷ 5
= 15 ÷ 5
= 3
1, 2, 3, 4, 5
Mean
= (1 + 2 + 3 + 4 + 5) ÷ 5
= 15 ÷ 5
= 3
Question 3: The daily number of customers visiting a shop for six days was:
45, 52, 48, 60, 55, 40
Find the average number of customers per day.
45, 52, 48, 60, 55, 40
Find the average number of customers per day.
Total customers
= 45 + 52 + 48 + 60 + 55 + 40
= 300
Number of days = 6
Average customers per day
= 300 ÷ 6
= 50
= 45 + 52 + 48 + 60 + 55 + 40
= 300
Number of days = 6
Average customers per day
= 300 ÷ 6
= 50
Question 4: The mean of five numbers is 24. Four of the numbers are 18, 22, 25 and 30. Find the fifth number.
Mean = Sum of Observations ÷ Number of Observations
Total sum of five numbers
= 24 × 5
= 120
Sum of known numbers
= 18 + 22 + 25 + 30
= 95
Missing number
= 120 − 95
= 25
Total sum of five numbers
= 24 × 5
= 120
Sum of known numbers
= 18 + 22 + 25 + 30
= 95
Missing number
= 120 − 95
= 25
Question 5: Find which group has a higher average.
(a) Group A: 15, 20, 25, 30, 35
(b) Group B: 10, 25, 30, 35, 40
(a) Group A: 15, 20, 25, 30, 35
(b) Group B: 10, 25, 30, 35, 40
Mean of Group A:
= (15 + 20 + 25 + 30 + 35) ÷ 5
= 125 ÷ 5
= 25
Mean of Group B:
= (10 + 25 + 30 + 35 + 40) ÷ 5
= 140 ÷ 5
= 28
So Group B has higher average
= (15 + 20 + 25 + 30 + 35) ÷ 5
= 125 ÷ 5
= 25
Mean of Group B:
= (10 + 25 + 30 + 35 + 40) ÷ 5
= 140 ÷ 5
= 28
So Group B has higher average
Question 6: The mean of 7 numbers is 24. If six of the numbers are 18, 22, 25, 27, 30 and 35, find the seventh number.
Mean = Sum of observations ÷ Number of observations
24 = Sum of observations ÷ 7
Sum of observations = 24 × 7
= 168
Sum of given numbers:
= 18 + 22 + 25 + 27 + 30 + 35
= 157
Missing number
= 168 − 157
= 11
Therefore, the missing number is 11.
24 = Sum of observations ÷ 7
Sum of observations = 24 × 7
= 168
Sum of given numbers:
= 18 + 22 + 25 + 27 + 30 + 35
= 157
Missing number
= 168 − 157
= 11
Therefore, the missing number is 11.
Question 7: The average marks of 8 students is 72. When the marks of one student are removed, the average of the remaining 7 students becomes 70. Find the marks of the student who was removed.
Total marks of 8 students:
= 72 × 8
= 576
Total marks of remaining 7 students:
= 70 × 7
= 490
Marks of removed student:
= 576 − 490
= 86
Therefore, the marks of the removed student are 86.
= 72 × 8
= 576
Total marks of remaining 7 students:
= 70 × 7
= 490
Marks of removed student:
= 576 − 490
= 86
Therefore, the marks of the removed student are 86.
Question 8: The mean of five numbers is 36. If two of the numbers are 24 and 48, and the remaining three numbers are equal, find the value of each of the remaining numbers.
Total sum:
= 36 × 5
= 180
Sum of known numbers:
= 24 + 48
= 72
Sum of remaining three numbers:
= 180 − 72
= 108
Let each remaining number be x.
x + x + x = 108
3x = 108
x = 36
Therefore, each of the remaining three numbers is 36.
= 36 × 5
= 180
Sum of known numbers:
= 24 + 48
= 72
Sum of remaining three numbers:
= 180 − 72
= 108
Let each remaining number be x.
x + x + x = 108
3x = 108
x = 36
Therefore, each of the remaining three numbers is 36.
Question 9: The following table shows the number of books read by students in a month. The mean number of books read is 5. Find the missing frequency.
| Number of Books (x) | 2 | 4 | 6 | 8 |
|---|---|---|---|---|
| Frequency (f) | 3 | 5 | x | 2 |
Mean = Σfx ÷ Σf
5 = [(2×3) + (4×5) + (6×x) + (8×2)] ÷ (3 + 5 + x + 2)
5 = (6 + 20 + 6x + 16) ÷ (10 + x)
5 = (42 + 6x) ÷ (10 + x)
5(10 + x) = 42 + 6x
50 + 5x = 42 + 6x
x = 8
Therefore, the missing frequency is 8.
5 = [(2×3) + (4×5) + (6×x) + (8×2)] ÷ (3 + 5 + x + 2)
5 = (6 + 20 + 6x + 16) ÷ (10 + x)
5 = (42 + 6x) ÷ (10 + x)
5(10 + x) = 42 + 6x
50 + 5x = 42 + 6x
x = 8
Therefore, the missing frequency is 8.
Question 10: The average of 6 numbers is 18. If one number is replaced by 30, the average becomes 20. Find the original number that was replaced.
Original total: = 18 × 6
= 108
New total: = 20 × 6
= 120
Increase in total: = 120 − 108
= 12
The new number is 30.
Original number: = 30 − 12
= 18
Therefore, the original number was 18.
= 108
New total: = 20 × 6
= 120
Increase in total: = 120 − 108
= 12
The new number is 30.
Original number: = 30 − 12
= 18
Therefore, the original number was 18.
Question 11: The average age of 6 players in a team is 24 years. When the captain joins the team, the average age increases to 25 years. Find the age of the captain.
For divisibility by 11, the difference between the sum of the digits in alternate positions must be 0 or a multiple of 11.
Sum of digits in odd positions: = 4 + 3 + 3 + 8 = 18
Sum of digits in even positions:
= 6 + 9 + Y
= 15 + Y
Now, because: The maximum possible difference here is between 18 and 24 (since Y is at most 9), the difference can only lie between −6 and 3.
It can never be ±11.
Therefore, the only possible value is 0.
Therefore, 18 − (15 + Y) = 0
3 − Y = 0
∴ Y = 3
Sum of digits in odd positions: = 4 + 3 + 3 + 8 = 18
Sum of digits in even positions:
= 6 + 9 + Y
= 15 + Y
Now, because: The maximum possible difference here is between 18 and 24 (since Y is at most 9), the difference can only lie between −6 and 3.
It can never be ±11.
Therefore, the only possible value is 0.
Therefore, 18 − (15 + Y) = 0
3 − Y = 0
∴ Y = 3
Question 12: The average of 8 numbers is 25. If a new number 41 is added to the data set, find the new average.
Original sum of observations:
= Mean × Number of observations
= 25 × 8
= 200
New sum:
= 200 + 41
= 241
New number of observations:
= 8 + 1
= 9
New Mean:
= 241 ÷ 9
= 26.78
Therefore, the new average is 26.78.
= Mean × Number of observations
= 25 × 8
= 200
New sum:
= 200 + 41
= 241
New number of observations:
= 8 + 1
= 9
New Mean:
= 241 ÷ 9
= 26.78
Therefore, the new average is 26.78.
Question 13: The average weight of 10 students is 52 kg. If a student weighing 43 kg leaves the group, find the new average weight.
Total weight of 10 students:
= 52 × 10
= 520 kg
Weight after removing the student:
= 520 − 43
= 477 kg
Number of students remaining:
= 10 − 1
= 9
New average:
= 477 ÷ 9
= 53 kg
Therefore, the new average weight is 53 kg.
= 52 × 10
= 520 kg
Weight after removing the student:
= 520 − 43
= 477 kg
Number of students remaining:
= 10 − 1
= 9
New average:
= 477 ÷ 9
= 53 kg
Therefore, the new average weight is 53 kg.
Question 14: The average of 15 observations is 32. If each observation is increased by 5, what will be the new average?
When the same value is added to every observation:
New Mean = Old Mean + Added Value
New Mean:
= 32 + 5
= 37
Therefore, the new average is 37.
New Mean = Old Mean + Added Value
New Mean:
= 32 + 5
= 37
Therefore, the new average is 37.
Question 15: The average salary of 6 employees is ¤42,000. The salary of one employee is incorrectly recorded as ¤96,000 instead of ¤36,000. Find the correct average salary.
Recorded total salary:
= 42,000 × 6
= ¤252,000
Incorrect salary exceeds actual salary by:
= ¤96,000 − ¤36,000
= ¤60,000
Correct total salary:
= ¤252,000 − ¤60,000
= ¤192,000
Correct average salary:
= ¤192,000 ÷ 6
= ¤32,000
Therefore, the correct average salary is ¤32,000.
= 42,000 × 6
= ¤252,000
Incorrect salary exceeds actual salary by:
= ¤96,000 − ¤36,000
= ¤60,000
Correct total salary:
= ¤252,000 − ¤60,000
= ¤192,000
Correct average salary:
= ¤192,000 ÷ 6
= ¤32,000
Therefore, the correct average salary is ¤32,000.
Question 16: Find the median of the following data 18, 25, 12, 30, 22, 15, 28
First arrange the data in ascending order:
12, 15, 18, 22, 25, 28, 30
Number of observations: n = 7
Since n is odd,
Median = Value of (n + 1) ÷ 2 th observation
= Value of (7 + 1) ÷ 2 th observation
= Value of 4th observation
The 4th observation is 22.
Therefore, the median is 22.
12, 15, 18, 22, 25, 28, 30
Number of observations: n = 7
Since n is odd,
Median = Value of (n + 1) ÷ 2 th observation
= Value of (7 + 1) ÷ 2 th observation
= Value of 4th observation
The 4th observation is 22.
Therefore, the median is 22.
Question 17: Find the median of the following data 14, 8, 21, 16, 25, 10, 18, 12
Arrange the data in ascending order:
8, 10, 12, 14, 16, 18, 21, 25
Number of observations: n = 8
Since n is even,
Median = Average of n ÷ 2 th and (n ÷ 2 + 1) th observations
= Average of 4th and 5th observations
= (14 + 16) ÷ 2
= 30 ÷ 2
= 15
Therefore, the median is 15.
8, 10, 12, 14, 16, 18, 21, 25
Number of observations: n = 8
Since n is even,
Median = Average of n ÷ 2 th and (n ÷ 2 + 1) th observations
= Average of 4th and 5th observations
= (14 + 16) ÷ 2
= 30 ÷ 2
= 15
Therefore, the median is 15.
Question 18: 24 is the median for the following 7 observations 12, 18, 24, 30, 35, 40, 45. If 50 is added to the data set, find the new median.
Original data:
12, 18, 24, 30, 35, 40, 45
After adding 50:
12, 18, 24, 30, 35, 40, 45, 50
Number of observations:
n = 8
Since n is even,
Median = Average of 4th and 5th observations
= (30 + 35) ÷ 2
= 32.5
Therefore, the new median is 32.5.
12, 18, 24, 30, 35, 40, 45
After adding 50:
12, 18, 24, 30, 35, 40, 45, 50
Number of observations:
n = 8
Since n is even,
Median = Average of 4th and 5th observations
= (30 + 35) ÷ 2
= 32.5
Therefore, the new median is 32.5.
Question 19: The median of the following observations is 25-
12, 18, 25, x, 35, 40, 45
Find the value of x if x is greater than 25 but less than 35.
12, 18, 25, x, 35, 40, 45
Find the value of x if x is greater than 25 but less than 35.
The data is already arranged in ascending order:
12, 18, 25, x, 35, 40, 45
Number of observations:
n = 7
Since n is odd,
Median = Value of 4th observation
The median is given as 25.
Therefore,
x cannot be the median value.
For the 4th observation to remain 25, the arrangement must be:
12, 18, x, 25, 35, 40, 45
Therefore,
x must be greater than 18 and less than 25.
The given condition is not possible.
Hence, no such value of x exists.
12, 18, 25, x, 35, 40, 45
Number of observations:
n = 7
Since n is odd,
Median = Value of 4th observation
The median is given as 25.
Therefore,
x cannot be the median value.
For the 4th observation to remain 25, the arrangement must be:
12, 18, x, 25, 35, 40, 45
Therefore,
x must be greater than 18 and less than 25.
The given condition is not possible.
Hence, no such value of x exists.
Question 20: The following are the daily incomes (in ¤) of 5 workers
500, 600, 700, 800, 900
Find the median. If the highest income changes from ¤900 to ¤9000, find the new median.
500, 600, 700, 800, 900
Find the median. If the highest income changes from ¤900 to ¤9000, find the new median.
Original data:
500, 600, 700, 800, 900
Number of observations:
n = 5
Median = Value of 3rd observation
Median = ¤700
New data:
500, 600, 700, 800, 9000
Median = Value of 3rd observation
Median = ¤700
Therefore, the median remains unchanged even after the extreme value increases.
500, 600, 700, 800, 900
Number of observations:
n = 5
Median = Value of 3rd observation
Median = ¤700
New data:
500, 600, 700, 800, 9000
Median = Value of 3rd observation
Median = ¤700
Therefore, the median remains unchanged even after the extreme value increases.
Question 21: Find the mode of the following data:
12, 15, 18, 12, 20, 15, 12, 25, 18
12, 15, 18, 12, 20, 15, 12, 25, 18
The frequency of each observation is calculated as follows:
| Observation | 12 | 15 | 18 | 20 | 25 |
|---|---|---|---|---|---|
| Frequency | 3 | 2 | 2 | 1 | 1 |
The observation with the highest frequency is 12.
Therefore, the mode is 12.
Question 22: Find the mode of the following data- 8, 12, 15, 8, 20, 12, 25, 15, 8, 12
The frequency table is:
| Observation | 8 | 12 | 15 | 20 | 25 |
|---|---|---|---|---|---|
| Frequency | 3 | 3 | 2 | 1 | 1 |
The highest frequency is 3.
Both 8 and 12 occur 3 times.
Therefore, the data has two modes:
Mode = 8 and 12
Question 23: Find the mode of the following data- 14, 18, 22, 25, 30, 35
The frequency table is:
| Observation | 14 | 18 | 22 | 25 | 30 | 35 |
|---|---|---|---|---|---|---|
| Frequency | 1 | 1 | 1 | 1 | 1 | 1 |
Every observation occurs only once.
Therefore, the data has no mode.
Question 24: The mode of the following data is 25- 18, 22, 25, 30, 25, x, 40, 25
Find the value of x if x is equal to the mode.
Find the value of x if x is equal to the mode.
The frequency table is:
| Observation | 18 | 22 | 25 | 30 | 40 |
|---|---|---|---|---|---|
| Frequency | 1 | 1 | 3 | 1 | 1 |
The highest frequency is of 25.
For x to remain equal to the mode: x = 25
Therefore, the missing value is 25.
Question 25: The monthly income (in ¤) of five employees in a company is- 25,000; 27,000; 30,000; 32,000; 1,50,000
Which measure of central tendency is most suitable to represent the typical income of employees?
Which measure of central tendency is most suitable to represent the typical income of employees?
Calculate the mean:
Mean = (25,000 + 27,000 + 30,000 + 32,000 + 1,50,000) ÷ 5
= 2,64,000 ÷ 5
= ¤52,800
The mean is much higher because the income of ¤1,50,000 is an extreme value.
Arrange the data:
25,000, 27,000, 30,000, 32,000, 1,50,000
Median = Middle value
= ¤30,000
Since the data contains an extremely large value, median gives a better representation.
Therefore, Median is the most suitable measure.
Mean = (25,000 + 27,000 + 30,000 + 32,000 + 1,50,000) ÷ 5
= 2,64,000 ÷ 5
= ¤52,800
The mean is much higher because the income of ¤1,50,000 is an extreme value.
Arrange the data:
25,000, 27,000, 30,000, 32,000, 1,50,000
Median = Middle value
= ¤30,000
Since the data contains an extremely large value, median gives a better representation.
Therefore, Median is the most suitable measure.
Question 26: Find the mean and median of the following data- 12, 15, 18, 20, 22, 25, 28
Mean:
= (12 + 15 + 18 + 20 + 22 + 25 + 28) ÷ 7
= 140 ÷ 7
= 20
Since the number of observations is 7:
Median position
= (n + 1) ÷ 2
= (7 + 1) ÷ 2
= 4th observation
Median = 20
Comparison:
Mean = 20
Median = 20
Both values are equal.
Therefore, the data is approximately symmetrical.
= (12 + 15 + 18 + 20 + 22 + 25 + 28) ÷ 7
= 140 ÷ 7
= 20
Since the number of observations is 7:
Median position
= (n + 1) ÷ 2
= (7 + 1) ÷ 2
= 4th observation
Median = 20
Comparison:
Mean = 20
Median = 20
Both values are equal.
Therefore, the data is approximately symmetrical.
Question 27: Find the mean, median and mode of the following data- 5, 8, 8, 10, 12, 8, 15
Arrange the data: 5, 8, 8, 8, 10, 12, 15
Mean:
= (5 + 8 + 8 + 10 + 12 + 8 + 15) ÷ 7
= 66 ÷ 7
= 9.43
Median:
Number of observations = 7
Median position:
= (7 + 1) ÷ 2
= 4th observation
Median = 8
Mode:
8 occurs three times, which is the highest frequency.
Mode = 8
Comparison:
Mean = 9.43
Median = 8
Mode = 8
Therefore, when a data set contains repeated values, mode may be equal to median but different from mean.
Mean:
= (5 + 8 + 8 + 10 + 12 + 8 + 15) ÷ 7
= 66 ÷ 7
= 9.43
Median:
Number of observations = 7
Median position:
= (7 + 1) ÷ 2
= 4th observation
Median = 8
Mode:
8 occurs three times, which is the highest frequency.
Mode = 8
Comparison:
Mean = 9.43
Median = 8
Mode = 8
Therefore, when a data set contains repeated values, mode may be equal to median but different from mean.
Question 28: The mean of the following frequency distribution is 18
Find the missing frequency x.
Find the missing frequency x.
| Observation | 10 | x | 20 | 25 |
|---|---|---|---|---|
| Frequency | 2 | 5 | 3 | 2 |
Σf
= 2 + 5 + 3 + 2
= 12
Since Mean = 18,
Σfx = 18 × 12
= 216
Now,
(10 × 2) + (x × 5) + (20 × 3) + (25 × 2) = 216
20 + 5x + 60 + 50 = 216
5x + 130 = 216
5x = 86
x = 17.2
= 2 + 5 + 3 + 2
= 12
Since Mean = 18,
Σfx = 18 × 12
= 216
Now,
(10 × 2) + (x × 5) + (20 × 3) + (25 × 2) = 216
20 + 5x + 60 + 50 = 216
5x + 130 = 216
5x = 86
x = 17.2
Question 29: The frequency of observation 25 is increased by 4. Find the new mean.
| Observation | 15 | 20 | 25 | 30 |
|---|---|---|---|---|
| Frequency | 3 | 5 | 2 | 4 |
Original:
Σfx
= 45 + 100 + 50 + 120
= 315
Σf
= 14
The frequency of 25 increases by 4.
New Σfx
= 315 + (25 × 4)
= 415
New Σf
= 18
Mean
= 415 ÷ 18
= 23.06
Therefore, the new mean is 23.06.
Σfx
= 45 + 100 + 50 + 120
= 315
Σf
= 14
The frequency of 25 increases by 4.
New Σfx
= 315 + (25 × 4)
= 415
New Σf
= 18
Mean
= 415 ÷ 18
= 23.06
Therefore, the new mean is 23.06.
Question 30: A shopkeeper records the number of pens sold per day.
| Pens Sold | 8 | 10 | 12 | 15 | 20 |
|---|---|---|---|---|---|
| Number of Days | 6 | 10 | 8 | 4 | 2 |
How many pens were sold in all during the recorded period?
Σfx
= (8 × 6) + (10 × 10) + (12 × 8) + (15 × 4) + (20 × 2)
= 48 + 100 + 96 + 60 + 40
= 344
= (8 × 6) + (10 × 10) + (12 × 8) + (15 × 4) + (20 × 2)
= 48 + 100 + 96 + 60 + 40
= 344
Question 31: The following table shows the number of goals scored by players in a football tournament.
| Goals (x) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Frequency | 2 | 5 | 6 | 4 | 3 |
Total frequency
= 2 + 5 + 6 + 4 + 3
= 20
Since n = 20,
Median is the average of the 10th and 11th observations.
Cumulative frequencies:
0 → 2
1 → 7
2 → 13
3 → 17
4 → 20
Both the 10th and 11th observations lie in the value 2.
Median = 2
= 2 + 5 + 6 + 4 + 3
= 20
Since n = 20,
Median is the average of the 10th and 11th observations.
Cumulative frequencies:
0 → 2
1 → 7
2 → 13
3 → 17
4 → 20
Both the 10th and 11th observations lie in the value 2.
Median = 2
Question 32: The table below shows the number of hours students studied during a week.
| Hours Studied | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|
| Frequency | 3 | 6 | 9 | 4 | 2 |
Total frequency
= 3 + 6 + 9 + 4 + 2
= 24
Median:
The 12th and 13th observations are required.
Cumulative frequencies:
2 → 3
3 → 9
4 → 18
5 → 22
6 → 24
The 12th and 13th observations both correspond to 4.
Median = 4
Mode:
Highest frequency = 9
Mode = 4
Therefore, Median = 4 and Mode = 4.
= 3 + 6 + 9 + 4 + 2
= 24
Median:
The 12th and 13th observations are required.
Cumulative frequencies:
2 → 3
3 → 9
4 → 18
5 → 22
6 → 24
The 12th and 13th observations both correspond to 4.
Median = 4
Mode:
Highest frequency = 9
Mode = 4
Therefore, Median = 4 and Mode = 4.
Question 33: The mode of the following frequency table is 20. Find the smallest possible value of x.
| Observation | 10 | 20 | 30 | 40 |
|---|---|---|---|---|
| Frequency | 5 | x | 8 | 6 |
For 20 to be the mode, its frequency must be greater than every other frequency.
The highest existing frequency is 8.
Therefore,
x > 8
The smallest possible value is
x = 9
The highest existing frequency is 8.
Therefore,
x > 8
The smallest possible value is
x = 9
Question 34: Identifying the Incorrect Statement
A data set has
Mean = 24
Median = 20
Mode = 18
Which of the following statements is incorrect?
A. Mean is greater than the median.
B. Median is greater than the mode.
C. Mode is the largest measure.
D. Mean is affected by extreme values.
A data set has
Mean = 24
Median = 20
Mode = 18
Which of the following statements is incorrect?
A. Mean is greater than the median.
B. Median is greater than the mode.
C. Mode is the largest measure.
D. Mean is affected by extreme values.
Given,
Mean = 24
Median = 20
Mode = 18
Only statement C is false.
Mode is not the largest measure.
Mean = 24
Median = 20
Mode = 18
Only statement C is false.
Mode is not the largest measure.
Question 35: The following frequency table has a mean of 16.
Find the missing observation x.
Find the missing observation x.
| Observation | 10 | x | 20 |
|---|---|---|---|
| Frequency | 2 | 3 | 5 |
Σf
= 2 + 3 + 5
= 10
Mean
= Σfx ÷ Σf
16 = Σfx ÷ 10
Σfx = 160
Now,
(10 × 2) + (3x) + (20 × 5)
= 160
20 + 3x + 100
= 160
3x = 40
x = 40⁄3
= 13⅓
Missing observation = 13⅓
= 2 + 3 + 5
= 10
Mean
= Σfx ÷ Σf
16 = Σfx ÷ 10
Σfx = 160
Now,
(10 × 2) + (3x) + (20 × 5)
= 160
20 + 3x + 100
= 160
3x = 40
x = 40⁄3
= 13⅓
Missing observation = 13⅓
Question 36: The observations are 8, 9, 10, 10, 11
One more observation, 100, is added.
Which measure(s) will definitely change?
One more observation, 100, is added.
Which measure(s) will definitely change?
Original mode = 10
After adding 100,
Mode remains 10.
Original median = 10
New observations:
8, 9, 10, 10, 11, 100
Median
= (10 + 10) ÷ 2
= 10
Median remains unchanged.
Mean increases because 100 is much larger than the other observations.
After adding 100,
Mode remains 10.
Original median = 10
New observations:
8, 9, 10, 10, 11, 100
Median
= (10 + 10) ÷ 2
= 10
Median remains unchanged.
Mean increases because 100 is much larger than the other observations.
Only the mean changes.
Question 37: The mean of a data set is 28 and the median is 30. Find the mode.
Mode ≈ 3 × Median − 2 × Mean
= (3 × 30) − (2 × 28)
= 90 − 56
= 34
Mode = 34
= (3 × 30) − (2 × 28)
= 90 − 56
= 34
Mode = 34
Question 38: The median of a data set is 45 and the mode is 51. Find the mean.
Mode = 3 × Median − 2 × Mean
51 = (3 × 45) − 2Mean
51 = 135 − 2Mean
2Mean = 84
Mean = 42
Mean = 42
51 = (3 × 45) − 2Mean
51 = 135 − 2Mean
2Mean = 84
Mean = 42
Mean = 42
Question 39: A company has 50 employees. Most employees earn between $35,000 and $45,000 per year, but the Managing Director earns $900,000.
Which measure of central tendency gives the best idea of a typical employee’s salary?
Which measure of central tendency gives the best idea of a typical employee’s salary?
The Managing Director’s salary is an extreme value.
The mean will be greatly increased.
The median is not affected significantly by extreme values.
Therefore, the median is the most appropriate measure.
The mean will be greatly increased.
The median is not affected significantly by extreme values.
Therefore, the median is the most appropriate measure.
Question 40: A real-estate agent wants to report the typical selling price of houses in an area where a few luxury houses cost several million dollars while most houses are moderately priced.
Luxury houses are extreme values. The mean will be affected considerably.
The median represents the middle selling price and is less affected by unusually expensive houses.
Therefore, the median should be used.
The median represents the middle selling price and is less affected by unusually expensive houses.
Therefore, the median should be used.
Question 41: In an asymmetrical frequency distribution, the value of the median is twice the value of the mean. If the mode of this distribution is 32, what is the value of its arithmetic mean?
Given:
Median = 2 × Mean
Let Mean = x
Therefore:
Median = 2x
Mode = 32
Substituting:
32 = 3 × 2x − 2x
32 = 6x − 2x
32 = 4x
x = 8
Therefore:
Arithmetic Mean = 8
Median = 2 × Mean
Let Mean = x
Therefore:
Median = 2x
Mode = 32
Substituting:
32 = 3 × 2x − 2x
32 = 6x − 2x
32 = 4x
x = 8
Therefore:
Arithmetic Mean = 8
Question 42: In a moderately asymmetrical distribution of student marks, the ratio of the arithmetic mean to the median is 3:4. If the mode of this distribution is 30, what is the value of its median?
Given:
Median = 2 × Mean
Let Mean = x
Therefore:
Median = 2x
Mode = 32
Substituting:
32 = 3 × 2x − 2x
32 = 6x − 2x
32 = 4x
x = 8
Therefore:
Arithmetic Mean = 8
Median = 2 × Mean
Let Mean = x
Therefore:
Median = 2x
Mode = 32
Substituting:
32 = 3 × 2x − 2x
32 = 6x − 2x
32 = 4x
x = 8
Therefore:
Arithmetic Mean = 8
Question 43: A teacher calculates the mean and median marks of 25 students as 48 and 45, respectively. Later, she decides to scale up the performance by doubling each student’s score and then subtracting 5 marks. What is the difference between the new mean and the new median?
If every observation is changed as:
New value = 2 × Original value − 5
then:
New Mean = 2 × 48 − 5 = 91
New Median = 2 × 45 − 5 = 85
Difference = 91 − 85 = 6
New value = 2 × Original value − 5
then:
New Mean = 2 × 48 − 5 = 91
New Median = 2 × 45 − 5 = 85
Difference = 91 − 85 = 6
Question 44: The mean of a set of 11 distinct observations is 35. If the largest observation is increased by 22 and the smallest observation is decreased by 11, what will be the new mean of the dataset?
Original sum:
11 × 35 = 385
Change in total:
Largest increased by 22 → +22
Smallest decreased by 11 → −11
Net increase:
22 − 11 = 11
New sum:
385 + 11 = 396
New mean:
396 ÷ 11 = 36
11 × 35 = 385
Change in total:
Largest increased by 22 → +22
Smallest decreased by 11 → −11
Net increase:
22 − 11 = 11
New sum:
385 + 11 = 396
New mean:
396 ÷ 11 = 36
Question 45: Five distinct positive integers a, b, c, d, and e are arranged in ascending order. The median of these numbers is 12. If the mean of the first three numbers is 8 and the mean of the last three numbers is 18, find the mean of all five numbers.
The numbers are arranged in ascending order:
a < b < c < d < e Since there are five observations, the median is the middle value. Therefore: c = 12 Mean of the first three numbers = 8 (a + b + c) ÷ 3 = 8 Substituting c = 12: (a + b + 12) ÷ 3 = 8 a + b + 12 = 24 a + b = 12 Mean of the last three numbers = 18 (c + d + e) ÷ 3 = 18 Substituting c = 12: (12 + d + e) ÷ 3 = 18 12 + d + e = 54 d + e = 42 Sum of all five numbers: a + b + c + d + e = (a + b) + c + (d + e) = 12 + 12 + 42 = 66 Mean of all five numbers: Mean = Sum of observations ÷ Number of observations = 66 ÷ 5 = 13.2[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row admin_label="SE_Question" _builder_version="4.27.6" _module_preset="default" custom_margin="-50px||||false|false" global_colors_info="{}"][et_pb_column type="4_4" _builder_version="4.27.6" _module_preset="default" global_colors_info="{}"][et_pb_text _builder_version="4.27.6" _module_preset="default" text_font="|300|||||||" text_font_size="18px" custom_padding="0.5%||0.5%|1%|false|false" text_font_size_tablet="17px" text_font_size_phone="16px" text_font_size_last_edited="on|phone" border_radii="on|3px|3px|3px|3px" border_width_all="1px" border_color_all="#0d9488" border_width_all_tablet="1px" border_width_all_phone="2px" border_width_all_last_edited="on|phone" global_colors_info="{}"]Question 46: Set A consists of 7 consecutive even integers starting with 2. Set B consists of 5 consecutive odd integers starting with 1. If Set A and Set B are combined into a single data set, find the median of the combined data set.
a < b < c < d < e Since there are five observations, the median is the middle value. Therefore: c = 12 Mean of the first three numbers = 8 (a + b + c) ÷ 3 = 8 Substituting c = 12: (a + b + 12) ÷ 3 = 8 a + b + 12 = 24 a + b = 12 Mean of the last three numbers = 18 (c + d + e) ÷ 3 = 18 Substituting c = 12: (12 + d + e) ÷ 3 = 18 12 + d + e = 54 d + e = 42 Sum of all five numbers: a + b + c + d + e = (a + b) + c + (d + e) = 12 + 12 + 42 = 66 Mean of all five numbers: Mean = Sum of observations ÷ Number of observations = 66 ÷ 5 = 13.2[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row admin_label="SE_Question" _builder_version="4.27.6" _module_preset="default" custom_margin="-50px||||false|false" global_colors_info="{}"][et_pb_column type="4_4" _builder_version="4.27.6" _module_preset="default" global_colors_info="{}"][et_pb_text _builder_version="4.27.6" _module_preset="default" text_font="|300|||||||" text_font_size="18px" custom_padding="0.5%||0.5%|1%|false|false" text_font_size_tablet="17px" text_font_size_phone="16px" text_font_size_last_edited="on|phone" border_radii="on|3px|3px|3px|3px" border_width_all="1px" border_color_all="#0d9488" border_width_all_tablet="1px" border_width_all_phone="2px" border_width_all_last_edited="on|phone" global_colors_info="{}"]Question 46: Set A consists of 7 consecutive even integers starting with 2. Set B consists of 5 consecutive odd integers starting with 1. If Set A and Set B are combined into a single data set, find the median of the combined data set.
Set A:
The consecutive even integers starting with 2 are:
2, 4, 6, 8, 10, 12, 14
Set B:
The consecutive odd integers starting with 1 are:
1, 3, 5, 7, 9
Combining both sets:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14
Number of observations: n = 12
Since the number of observations is even, the median is the average of the two middle observations.
The two middle positions are:
n ÷ 2 = 12 ÷ 2 = 6th position
and
(n ÷ 2) + 1 = 6 + 1 = 7th position
The 6th and 7th observations are: 6 and 7
Median: = (6 + 7) ÷ 2
= 13 ÷ 2
= 6.5
The consecutive even integers starting with 2 are:
2, 4, 6, 8, 10, 12, 14
Set B:
The consecutive odd integers starting with 1 are:
1, 3, 5, 7, 9
Combining both sets:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14
Number of observations: n = 12
Since the number of observations is even, the median is the average of the two middle observations.
The two middle positions are:
n ÷ 2 = 12 ÷ 2 = 6th position
and
(n ÷ 2) + 1 = 6 + 1 = 7th position
The 6th and 7th observations are: 6 and 7
Median: = (6 + 7) ÷ 2
= 13 ÷ 2
= 6.5
Question 47: For a moderately skewed data set, the difference between the arithmetic mean and the median is exactly 3. What is the difference between the arithmetic mean and the mode of this data set?
For a moderately skewed distribution, the empirical relationship between mean, median and mode is:
Mode ≈ 3 × Median − 2 × Mean
Let:
Mean = M
Median = Md
Mode = Mo
Given: Mean − Median = 3
Therefore: M − Md = 3
Rearranging: Md = M − 3
Now using the empirical relationship:
Mo = 3 × Md − 2 × M
Substitute Md = M − 3:
Mo = 3 × (M − 3) − 2M
Mo = 3M − 9 − 2M
Mo = M − 9
Therefore: Mean − Mode
= M − (M − 9)
= 9
Mode ≈ 3 × Median − 2 × Mean
Let:
Mean = M
Median = Md
Mode = Mo
Given: Mean − Median = 3
Therefore: M − Md = 3
Rearranging: Md = M − 3
Now using the empirical relationship:
Mo = 3 × Md − 2 × M
Substitute Md = M − 3:
Mo = 3 × (M − 3) − 2M
Mo = 3M − 9 − 2M
Mo = M − 9
Therefore: Mean − Mode
= M − (M − 9)
= 9
Question 48: The mean of 15 observations is 20. If two observations, originally 10 and 15, are replaced by 20 and x respectively, the new mean becomes 21. Find the value of x.
Original sum: = 15 × 20
= 300
New sum: = 15 × 21
= 315
Increase in total: = 315 − 300
= 15
Change in values:
= (20 + x) − (10 + 15)
= x − 5
Therefore:
x − 5 = 15
x = 20
= 300
New sum: = 15 × 21
= 315
Increase in total: = 315 − 300
= 15
Change in values:
= (20 + x) − (10 + 15)
= x − 5
Therefore:
x − 5 = 15
x = 20
Question 49: For a dataset of n observations, the arithmetic mean is x̄. If a new observation with value k is added to the dataset, the new mean becomes x̄ − 2. Find the value of k in terms of n and x̄.
Original sum of observations: = n × x̄
After adding the new observation:
Total observations = n + 1
New mean: = x̄ − 2
New sum: = (n + 1)(x̄ − 2)
The new sum is also:
= Original sum + k
Therefore:
n × x̄ + k = (n + 1)(x̄ − 2)
k = (n + 1)(x̄ − 2) − n × x̄
Expanding:
k = n x̄ + x̄ − 2n − 2 − n x̄
k = x̄ − 2n − 2
After adding the new observation:
Total observations = n + 1
New mean: = x̄ − 2
New sum: = (n + 1)(x̄ − 2)
The new sum is also:
= Original sum + k
Therefore:
n × x̄ + k = (n + 1)(x̄ − 2)
k = (n + 1)(x̄ − 2) − n × x̄
Expanding:
k = n x̄ + x̄ − 2n − 2 − n x̄
k = x̄ − 2n − 2
Question 50: Let Set A be the set of all positive integers less than 20 that are multiples of 3. Let Set B be the set of all positive integers less than 20 that are multiples of 4. If Set A and Set B are combined into a single data set, find the median of the combined data set.
Set A:
Multiples of 3 less than 20:
3, 6, 9, 12, 15, 18
Set B:
Multiples of 4 less than 20:
4, 8, 12, 16
Combining both sets:
3, 4, 6, 8, 9, 12, 12, 15, 16, 18
Number of observations:
n = 10
Since the number of observations is even:
Median = Average of the 5th and 6th observations
5th observation = 9
6th observation = 12
Median:
= (9 + 12) ÷ 2
= 21 ÷ 2
= 10.5
Multiples of 3 less than 20:
3, 6, 9, 12, 15, 18
Set B:
Multiples of 4 less than 20:
4, 8, 12, 16
Combining both sets:
3, 4, 6, 8, 9, 12, 12, 15, 16, 18
Number of observations:
n = 10
Since the number of observations is even:
Median = Average of the 5th and 6th observations
5th observation = 9
6th observation = 12
Median:
= (9 + 12) ÷ 2
= 21 ÷ 2
= 10.5
Common Mistakes
- Confusing Mean with Median or Mode. Students often use the arithmetic mean when the question asks for the median or mode. Read the question carefully and identify which measure of central tendency is required.
- Ignoring the Formula for Mean. While finding the mean, students sometimes divide by the wrong number. The mean is always calculated by dividing the sum of all observations by the total number of observations.
- Arranging Data Incorrectly Before Finding the Median. Students often calculate the median without arranging the observations in ascending or descending order. The data must always be arranged first.
- Using the Wrong Median Rule. Students sometimes use the odd-number formula for an even number of observations or vice versa. For an odd number of observations, the median is the middle value. For an even number of observations, the median is the average of the two middle values.
- Confusing Mode with the Largest Observation. Students often assume that the highest value is the mode. The mode is the observation that occurs most frequently, regardless of its magnitude.
- Ignoring Repeated Observations While Finding the Mode. Students sometimes overlook repeated values and identify the wrong mode. Count the frequency of every observation carefully before selecting the mode.
- Using Frequencies Incorrectly While Finding the Mean of a Frequency Distribution. Students often add only the observations instead of multiplying each observation by its corresponding frequency. Always calculate Σfx before dividing by Σf.
- Ignoring the Total Frequency. While working with frequency distributions, students sometimes divide by the number of distinct observations instead of the total frequency. The denominator should always be Σf.
- Applying the Empirical Relationship Incorrectly. Students often use the wrong formula while relating the mean, median and mode. The correct empirical relationship is: Mode ≈ 3 × Median − 2 × Mean.
- Assuming the Empirical Relationship is Always Exact. The empirical relationship is only an approximate relationship and generally holds for moderately skewed distributions. It should not be treated as an exact formula for every data set.
- Choosing the Wrong Measure of Central Tendency. Students often use the mean even when the data contains extremely large or small values. In such situations, the median usually provides a better representation of the data, while the mode is most suitable for identifying the most frequently occurring value.
- Failing to Verify the Final Answer. After calculating the mean, median or mode, students often forget to check whether the answer is reasonable and lies within the range of the given data wherever applicable.
Practice Questions
Question 1: The average marks obtained by boys in an examination are 71, while the average marks obtained by girls are 73. If the average marks of all the students together are 71.8, find the ratio of the number of boys to the number of girls who appeared for the examination.
Question 2: The median of the following observations arranged in ascending order is 27.
9, 13, 18, 22, (2x + 3), 31, 36, 40, 45
9, 13, 18, 22, (2x + 3), 31, 36, 40, 45
Question 3: The median of the following observations arranged in ascending order is 41.
16, 21, 28, 35, (4x + 1), 44, 49, 55, 60
16, 21, 28, 35, (4x + 1), 44, 49, 55, 60
Question 4: The following table shows the number of customers visiting a shop on different days.
| Number of Customers | 20 | 30 | 40 | 50 |
|---|---|---|---|---|
| Frequency | 3 | 5 | 4 | 2 |
Find the mean number of customers visiting the shop.
Question 5: IA company wants to report the typical salary of its employees. Most employees earn between ¤30,000 and ¤50,000 per month, but a few senior executives earn more than ¤10,00,000 per month.
Which measure of central tendency should the company use and why?
Related Topics
 |
|
|
|
|
|