Measures of Central Tendency
Key Concepts
Solved Examples: 50
Practice Questions: 5
Solved Examples
= 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.
1, 2, 3, 4, 5
Mean
= (1 + 2 + 3 + 4 + 5) ÷ 5
= 15 ÷ 5
= 3
45, 52, 48, 60, 55, 40
Find the average number of customers per day.
= 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.
Total sum of five numbers
= 24 × 5
= 120
Sum of known numbers
= 18 + 22 + 25 + 30
= 95
Missing number
= 120 − 95
= 25
(a) Group A: 15, 20, 25, 30, 35
(b) Group B: 10, 25, 30, 35, 40
= (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.
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.
= 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.
= 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.
| Number of Books (x) | 2 | 4 | 6 | 8 |
|---|---|---|---|---|
| Frequency (f) | 3 | 5 | x | 2 |
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.
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.
Original total age = 6 × 24 = 144 years
New total age = 7 × 25 = 175 years
Captain’s age = 175 − 144 = 31 years
Question 12: The average of 8 numbers is 25. If a new number 41 is added to the data set, find the new average.
= 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.
= 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?
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.
= 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
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
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.
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.
Original median = 24
Added value = 5
New median = 24 + 5 = 29
∴ The new median is 29.
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
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.
12, 15, 18, 12, 20, 15, 12, 25, 18
| 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.
| 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
| 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.
Find the value of x if x is equal to the mode.
| 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.
Which measure of central tendency is most suitable to represent the typical income of employees?
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
= (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
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.
| Observation | 10 | 15 | 20 | 25 |
|---|---|---|---|---|
| Frequency | 2 | f | 4 | 2 |
Σfᵢ = 2 + f + 4 + 2
= f + 8
The sum of fᵢxᵢ is
Σfᵢxᵢ = 10(2) + 15f + 20(4) + 25(2)
= 20 + 15f + 80 + 50
= 15f + 150
Using Mean = Σfᵢxᵢ ÷ Σfᵢ
18 = (15f + 150) ÷ (f + 8)
or, 18(f + 8) = 15f + 150
or, 18f + 144 = 15f + 150
or, 3f = 6
or, f = 2
∴ The missing frequency is 2.
| Observation | 15 | 20 | 25 | 30 |
|---|---|---|---|---|
| Frequency | 3 | 5 | 2 | 4 |
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.
| 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?
= (8 × 6) + (10 × 10) + (12 × 8) + (15 × 4) + (20 × 2)
= 48 + 100 + 96 + 60 + 40
= 344
| Goals (x) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Frequency | 2 | 5 | 6 | 4 | 3 |
= 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
| Hours Studied | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|
| Frequency | 3 | 6 | 9 | 4 | 2 |
= 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.
| Observation | 10 | 20 | 30 | 40 |
|---|---|---|---|---|
| Frequency | 5 | x | 8 | 6 |
The highest existing frequency is 8.
Therefore,
x > 8
The smallest possible value is
x = 9
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.
Mean = 24
Median = 20
Mode = 18
Only statement C is false.
Mode is not the largest measure.
Find the missing observation x.
| Observation | 10 | x | 20 |
|---|---|---|---|
| Frequency | 2 | 3 | 5 |
= 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
One more observation, 100, is added.
Which measure(s) will definitely change?
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.
Mode ≈ 3 × Median − 2 × Mean
Given:
Mean = 28
Median = 30
Substituting
Mode ≈ 3 × 30 − 2 × 28
= 90 − 56
= 34
∴ The estimated mode is 34.
Mode ≈ 3 × Median − 2 × Mean
Given
Median = 45
Mode = 51
Substituting
51 ≈ 3 × 45 − 2 × Mean
51 ≈ 135 − 2 × Mean
2 × Mean ≈ 135 − 51
2 × Mean ≈ 84
Mean ≈ 42
∴ The estimated mean is 42.
Which measure of central tendency gives the best idea of a typical employee’s salary?
The mean will be greatly increased.
The median is not affected significantly by extreme values.
Therefore, the median is the most appropriate measure.
The median represents the middle selling price and is less affected by unusually expensive houses.
Therefore, the median should be used.
Question 41: In a moderately skewed distribution, the median is twice the arithmetic mean. If the mode of the distribution is 32, find the arithmetic mean using the empirical relationship between mean, median and mode.
Mode ≈ 3 × Median − 2 × Mean
Let the arithmetic mean be x.
Then Median = 2x
Given Mode = 32
Substituting
32 ≈ 3 × 2x − 2x
or, 32 ≈ 6x − 2x
or, 32 ≈ 4x
or, x ≈ 8
∴ The estimated arithmetic mean is 8.
Question 42: In a moderately skewed 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?
Mode = 3 × Median − 2 × Mean
Given Mean : Median = 3 : 4
Let:
Mean = 3k
Median = 4k
Also, Mode = 30.
So, 30 = 3(4k) − 2(3k)
or, 30 = 12k − 6k
or, 30 = 6k
or, k = 5
Median = 4k = 4 × 5 = 20
∴ The median is 20.
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?
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?
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
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.
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 arithmetic mean exceeds the median by 3. Using the empirical relationship between mean, median and mode, estimate the difference between the arithmetic mean and the mode.
Mode ≈ 3 × Median − 2 × Mean
Let
Mean = M
Median = Md
Mode = Mo
Given
M − Md = 3
∴ Md = M − 3
Using the empirical relationship
Mo ≈ 3 × Md − 2 × M
Substituting Md = M − 3:
Mo ≈ 3(M − 3) − 2M
or, Mo ≈ 3M − 9 − 2M
or, Mo ≈ M − 9
∴ M − Mo ≈ 9
∴ The estimated difference between the arithmetic mean and the mode is 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.
= 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
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.
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
Question 51: The mean of the following distribution is 50. If the sum of the frequencies is 120, find the values of f₁ and f₂.
| Observation | 10 | 30 | 50 | 70 | 90 |
|---|---|---|---|---|---|
| Frequency | 17 | f₁ | 32 | f₂ | 19 |
17 + f₁ + 32 + f₂ + 19 = 120
or, f₁ + f₂ + 68 = 120
or, f₁ + f₂ = 52 … ①
The arithmetic mean is given by Mean = Σfᵢxᵢ / Σfᵢ
Since the mean is 50 and the sum of the frequencies is 120
50 = Σfᵢxᵢ / 120
∴ Σfᵢxᵢ = 6000
Now,
10(17) + 30f₁ + 50(32) + 70f₂ + 90(19) = 6000
or, 170 + 30f₁ + 1600 + 70f₂ + 1710 = 6000
or, 30f₁ + 70f₂ = 2520
or, Dividing by 10:
3f₁ + 7f₂ = 252 … ②
From equation ①:
or, f₁ = 52 − f₂
Substituting this into equation ②:
3(52 − f₂) + 7f₂ = 252
or, 156 − 3f₂ + 7f₂ = 252
or, 4f₂ = 96
or, f₂ = 24
Using equation ①:
f₁ + 24 = 52
or, f₁ = 28
∴ f₁ = 28 and f₂ = 24.
Question 52: The following table shows the marks obtained by a group of students. Find the mean marks.
| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Frequency | 4 | 6 | 10 | 7 | 3 |
Class mark = (Lower limit + Upper limit) ÷ 2
Thus
0–10 → (0 + 10) ÷ 2 = 5
10–20 → (10 + 20) ÷ 2 = 15
20–30 → (20 + 30) ÷ 2 = 25
30–40 → (30 + 40) ÷ 2 = 35
40–50 → (40 + 50) ÷ 2 = 45
The grouped-data mean is
Mean = Σfᵢxᵢ ÷ Σfᵢ where xᵢ is the class mark.
Now calculate fᵢxᵢ
4 × 5 = 20
6 × 15 = 90
10 × 25 = 250
7 × 35 = 245
3 × 45 = 135
∴ Σfᵢxᵢ = 20 + 90 + 250 + 245 + 135
= 740
Total frequency
Σfᵢ = 4 + 6 + 10 + 7 + 3
= 30
Therefore Mean = 740 ÷ 30
= 24.67
∴ The mean marks are approximately 24.67.
| Marks | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 |
|---|---|---|---|---|---|
| Frequency | 5 | 8 | 12 | 10 | 5 |
Class mark = (Lower limit + Upper limit) ÷ 2
Thus
10–20 → 15
20–30 → 25
30–40 → 35
40–50 → 45
50–60 → 55
Now calculate fᵢxᵢ
5 × 15 = 75
8 × 25 = 200
12 × 35 = 420
10 × 45 = 450
5 × 55 = 275
∴ Σfᵢxᵢ = 75 + 200 + 420 + 450 + 275
= 1420
Total frequency Σfᵢ = 5 + 8 + 12 + 10 + 5 = 40
Mean = Σfᵢxᵢ ÷ Σfᵢ
= 1420 ÷ 40
= 35.5
∴ The mean marks are 35.5.
| Class Interval | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Frequency | 1 | x | 1 | 1 | 1 |
0–10 → 5
10–20 → 15
20–30 → 25
30–40 → 35
40–50 → 45
The total frequency is Σfᵢ = 1 + x + 1 + 1 + 1
= x + 4
The sum of fᵢxᵢ is
Σfᵢxᵢ = 1(5) + x(15) + 1(25) + 1(35) + 1(45)
= 5 + 15x + 25 + 35 + 45
= 15x + 110
Using Mean = Σfᵢxᵢ ÷ Σfᵢ
20 = (15x + 110) ÷ (x + 4)
or, 20(x + 4) = 15x + 110
or, 20x + 80 = 15x + 110
or, 5x = 30
or, x = 6
∴ The missing frequency is 6.
| Age (years) | 20–30 | 30–40 | 40–50 | 50–60 | 60–70 |
|---|---|---|---|---|---|
| Frequency | 4 | 7 | 10 | 6 | 3 |
The class mark is
xᵢ = (Lower limit + Upper limit) ÷ 2
Let dᵢ = xᵢ − A
where A = 45.
The calculation table is
| Class Interval | Class Mark (xᵢ) | Frequency (fᵢ) | dᵢ = xᵢ − 45 | fᵢdᵢ |
|---|---|---|---|---|
| 20–30 | 25 | 4 | −20 | −80 |
| 30–40 | 35 | 7 | −10 | −70 |
| 40–50 | 45 | 10 | 0 | 0 |
| 50–60 | 55 | 6 | 10 | 60 |
| 60–70 | 65 | 3 | 20 | 60 |
| Σ | — | 30 | — | −30 |
Thus, Σfᵢ = 30 and Σfᵢdᵢ = −30
Using the assumed mean formula
Mean = A + Σfᵢdᵢ ÷ Σfᵢ
= 45 + (−30 ÷ 30)
= 45 − 1
= 44
∴ The mean age is 44 years.
| Hours Studied | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 |
|---|---|---|---|---|---|
| Frequency | 3 | 7 | 12 | 6 | 2 |
The class mark is
xᵢ = (Lower limit + Upper limit) ÷ 2
The class width is 10.
Let uᵢ = (xᵢ − A) ÷ h
where A = 35 and h = 10
The calculation table is:
| Class Interval | Class Mark (xᵢ) | Frequency (fᵢ) | uᵢ | fᵢuᵢ |
|---|---|---|---|---|
| 10–20 | 15 | 3 | −2 | −6 |
| 20–30 | 25 | 7 | −1 | −7 |
| 30–40 | 35 | 12 | 0 | 0 |
| 40–50 | 45 | 6 | 1 | 6 |
| 50–60 | 55 | 2 | 2 | 4 |
| Σ | — | 30 | — | −3 |
Thus, Σfᵢ = 30 and Σfᵢuᵢ = −3
Using the step-deviation formula
Mean = A + h × (Σfᵢuᵢ ÷ Σfᵢ)
= 35 + 10 × (−3 ÷ 30)
= 35 − 1
= 34
∴ The mean number of hours studied is 34 hours.
| Class Interval | 20–30 | 30–40 | 40–50 | 50–60 | 60–70 | 70–80 |
|---|---|---|---|---|---|---|
| Frequency | 10 | 6 | 8 | 12 | 5 | 9 |
20–30 → 25
30–40 → 35
40–50 → 45
50–60 → 55
60–70 → 65
70–80 → 75
Take 55 as the assumed mean. The class width is 10.
Let uᵢ = (yᵢ − a) ÷ c
where a = 55 and c = 10
The calculation table is:
| Class Interval | Class Mark (yᵢ) | uᵢ = (yᵢ − a) ÷ c | Frequency (fᵢ) | fᵢuᵢ |
|---|---|---|---|---|
| 20–30 | 25 | −3 | 10 | −30 |
| 30–40 | 35 | −2 | 6 | −12 |
| 40–50 | 45 | −1 | 8 | −8 |
| 50–60 | 55 | 0 | 12 | 0 |
| 60–70 | 65 | 1 | 5 | 5 |
| 70–80 | 75 | 2 | 9 | 18 |
| Σ | — | — | 50 | −27 |
Thus, Σfᵢ = 50 and Σfᵢuᵢ = −27
Using the step-deviation formula:
Mean = a + c × (Σfᵢuᵢ ÷ Σfᵢ)
= 55 + 10 × (−27 ÷ 50)
= 55 − 5.4
= 49.6
∴ The mean is 49.6.
| Donation (¤) | 20–60 | 60–100 | 100–150 | 150–250 | 250–350 | 350–450 |
|---|---|---|---|---|---|---|
| Number of Individuals | 7 | 5 | 16 | 12 | 2 | 3 |
20–60 → 40
60–100 → 80
100–150 → 125
150–250 → 200
250–350 → 300
350–450 → 400
Take 200 as the assumed mean.
For the step-deviation method, take c = 50
Let uᵢ = (yᵢ − a) ÷ c
where a = 200 and c = 50
The calculation table is:
| Class Interval | Class Mark (yᵢ) | uᵢ = (yᵢ − a) ÷ c | Frequency (fᵢ) | fᵢuᵢ |
|---|---|---|---|---|
| 20–60 | 40 | −3.2 | 7 | −22.4 |
| 60–100 | 80 | −2.4 | 5 | −12 |
| 100–150 | 125 | −1.5 | 16 | −24 |
| 150–250 | 200 | 0 | 12 | 0 |
| 250–350 | 300 | 2 | 2 | 4 |
| 350–450 | 400 | 4 | 3 | 12 |
| Σ | — | — | 45 | −42.4 |
Thus, Σfᵢ = 45 and Σfᵢuᵢ = −42.4
Using the step-deviation formula:
Mean = a + c × (Σfᵢuᵢ ÷ Σfᵢ)
= 200 + 50 × (−42.4 ÷ 45)
= 200 − 47.111…
= 152.888…
∴ The mean donation is approximately ¤152.89.
| Consumption below (in units) | 30 | 40 | 50 | 60 | 70 | 80 |
|---|---|---|---|---|---|---|
| Number of households | 100 | 220 | 350 | 750 | 950 | 1000 |
The first class begins at 20 units.
For the first class f₁ = 100
For the subsequent classes
f₂ = 220 − 100 = 120
f₃ = 350 − 220 = 130
f₄ = 750 − 350 = 400
f₅ = 950 − 750 = 200
f₆ = 1000 − 950 = 50
The class marks are calculated using
Class Mark = (Lower limit + Upper limit) ÷ 2
| Class Interval | Class Mark (xᵢ) | Cumulative Frequency | Frequency (fᵢ) | fᵢxᵢ |
|---|---|---|---|---|
| 20–30 | 25 | 100 | 100 | 2500 |
| 30–40 | 35 | 220 | 120 | 4200 |
| 40–50 | 45 | 350 | 130 | 5850 |
| 50–60 | 55 | 750 | 400 | 22000 |
| 60–70 | 65 | 950 | 200 | 13000 |
| 70–80 | 75 | 1000 | 50 | 3750 |
| Σ | — | 1000 | 1000 | 51300 |
Thus, Σfᵢ = 1000 and Σfᵢxᵢ = 51300
Using the mean formula
Mean = Σfᵢxᵢ ÷ Σfᵢ
= 51300 ÷ 1000
= 51.3
∴ The mean monthly electricity consumption is 51.3 units.
| Class Interval | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 | 60–70 | 70–80 |
|---|---|---|---|---|---|---|---|
| Frequency | 5 | 3 | f | 7 | 2 | 6 | 13 |
10–20 → 15
20–30 → 25
30–40 → 35
40–50 → 45
50–60 → 55
60–70 → 65
70–80 → 75
Now construct the required values
| Class Interval | Class Mark (xᵢ) | Frequency (fᵢ) | fᵢxᵢ |
|---|---|---|---|
| 10–20 | 15 | 5 | 75 |
| 20–30 | 25 | 3 | 75 |
| 30–40 | 35 | f | 35f |
| 40–50 | 45 | 7 | 315 |
| 50–60 | 55 | 2 | 110 |
| 60–70 | 65 | 6 | 390 |
| 70–80 | 75 | 13 | 975 |
| Σ | — | 36 + f | 1940 + 35f |
Using the mean formula Mean = Σfᵢxᵢ ÷ Σfᵢ
Given mean = 52,
52 = (1940 + 35f) ÷ (36 + f)
or, 52(36 + f) = 1940 + 35f
or, 1872 + 52f = 1940 + 35f
or, 52f − 35f = 1940 − 1872
or, 17f = 68
∴ f = 4
The missing frequency is 4.
| Class Interval | 0–20 | 20–40 | 40–60 | 60–80 | 80–100 | 100–120 |
|---|---|---|---|---|---|---|
| Frequency | 7 | p | 12 | q | 8 | 5 |
0–20 → 10
20–40 → 30
40–60 → 50
60–80 → 70
80–100 → 90
100–120 → 110
Now construct the required values
| Class Interval | Class Mark (xᵢ) | Frequency (fᵢ) | fᵢxᵢ |
|---|---|---|---|
| 0–20 | 10 | 7 | 70 |
| 20–40 | 30 | p | 30p |
| 40–60 | 50 | 12 | 600 |
| 60–80 | 70 | q | 70q |
| 80–100 | 90 | 8 | 720 |
| 100–120 | 110 | 5 | 550 |
| Σ | — | 32 + p + q | 1940 + 30p + 70q |
Since the sum of the frequencies is 50
p + q + 32 = 50
or, p + q = 18 …(1)
Using the mean formula Mean = Σfᵢxᵢ ÷ Σfᵢ
Given mean = 57.6,
57.6 = (1940 + 30p + 70q) ÷ 50
or, 1940 + 30p + 70q = 2880
or, 30p + 70q = 940
or, 3p + 7q = 94 …(2)
or, From (1) p = 18 − q
Substituting in (2)
3(18 − q) + 7q = 94
or, 54 − 3q + 7q = 94
or, 4q = 40
or, q = 10
∴ p = 18 − 10
or, p = 8
The values are p = 8 and q = 10.
| Class Interval | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Frequency | 5 | 8 | 12 | 10 | 5 |
| Class Interval | Frequency (fᵢ) | Cumulative Frequency |
|---|---|---|
| 0–10 | 5 | 5 |
| 10–20 | 8 | 13 |
| 20–30 | 12 | 25 |
| 30–40 | 10 | 35 |
| 40–50 | 5 | 40 |
Total frequency N = 40
Therefore, N ÷ 2 = 40 ÷ 2 = 20
The cumulative frequency first exceeding 20 is 25, corresponding to the class 20–30.
Thus, the median class is 20–30.
Using the median formula
Median = l + [(N/2 − cf) ÷ f] × h
Here, l = 20, N = 40, cf = 13, f = 12, h = 10
So, Median = 20 + [(20 − 13) ÷ 12] × 10
= 20 + (7 ÷ 12) × 10
= 20 + 70/12
= 20 + 5.83
∴ Median ≈ 25.83
| Age (years) | 20–25 | 25–30 | 30–35 | 35–40 | 40–45 |
|---|---|---|---|---|---|
| Frequency | 6 | 9 | 15 | 12 | 8 |
| Age (years) | Frequency | Cumulative Frequency |
|---|---|---|
| 20–25 | 6 | 6 |
| 25–30 | 9 | 15 |
| 30–35 | 15 | 30 |
| 35–40 | 12 | 42 |
| 40–45 | 8 | 50 |
Total frequency N = 50
So N ÷ 2 = 25
The cumulative frequency first exceeding 25 is 30.
Therefore, the median class is 30–35.
Here, l = 30, cf = 15, f = 15, h = 5
Median = 30 + [(25 − 15) ÷ 15] × 5
= 30 + (10 ÷ 15) × 5
= 30 + 10/3
= 30 + 3.33
∴ Median ≈ 33.33 years
| Score Interval | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|
| Frequency | 6 | 11 | 9 | 4 |
| Score Interval | Frequency (fᵢ) | Cumulative Frequency (cf) |
|---|---|---|
| 10–20 | 6 | 6 |
| 20–30 | 11 | 17 |
| 30–40 | 9 | 26 |
| 40–50 | 4 | 30 |
Total frequency N = 30
Median position N ÷ 2 = 30 ÷ 2 = 15
The cumulative frequency first exceeding 15 is 17.
The median class is 20–30.
For the median class
l = 20
cf = 6
f = 11
h = 10
Using the median formula
Median = l + [(N/2 − cf) ÷ f] × h
= 20 + [(15 − 6) ÷ 11] × 10
= 20 + (9 ÷ 11) × 10
= 20 + 90/11
= 20 + 8.18
= 28.18
∴ The median score is approximately 28.18.
| Class Interval | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 |
|---|---|---|---|---|---|---|
| Frequency | 10 | x | 25 | 30 | y | 5 |
Total frequency = 100
10 + x + 25 + 30 + y + 5 = 100
or, 70 + x + y = 100
∴ x + y = 30 …(1)
Next, construct the cumulative frequencies
| Class Interval | Frequency (fᵢ) | Cumulative Frequency (cf) |
|---|---|---|
| 0–10 | 10 | 10 |
| 10–20 | x | 10 + x |
| 20–30 | 25 | 35 + x |
| 30–40 | 30 | 65 + x |
| 40–50 | y | 65 + x + y |
| 50–60 | 5 | 70 + x + y |
Since the median is 32, the median class is 30–40.
For the median class
l = 30
N ÷ 2 = 100 ÷ 2 = 50
cf = 35 + x
f = 30
h = 10
Using the median formula
Median = l + [((N/2) − cf) ÷ f] × h
32 = 30 + [((50 − (35 + x)) ÷ 30)] × 10
or, 32 − 30 = [(15 − x) ÷ 30] × 10
or, 2 = (15 − x) ÷ 3
or, 6 = 15 − x
or, x = 9
Substitute x = 9 into Equation (1):
or, 9 + y = 30
or, y = 21
∴ The missing frequencies are x = 9 and y = 21.
| Marks | 0–10 | 10–20 | 20–30 | 30–x | x–50 |
|---|---|---|---|---|---|
| Number of Students | 4 | 6 | 8 | 15 | 17 |
Total number of observations
N = 4 + 6 + 8 + 15 + 17
N = 50
Median position N ÷ 2 = 50 ÷ 2 = 25
Construct the cumulative frequency table
| Marks | Frequency (fᵢ) | Cumulative Frequency (cf) |
|---|---|---|
| 0–10 | 4 | 4 |
| 10–20 | 6 | 10 |
| 20–30 | 8 | 18 |
| 30–x | 15 | 33 |
| x–50 | 17 | 50 |
Since 18 < 25 ≤ 33, the median class is 30–x.
For the median class
l = 30
cf = 18
f = 15
h = x − 30
Using the median formula:
Median = l + [((N/2) − cf) ÷ f] × h
37 = 30 + [((25 − 18) ÷ 15)] × (x − 30)
37 = 30 + (7 ÷ 15)(x − 30)
7 = (7 ÷ 15)(x − 30)
x − 30 = 15
∴ x = 45
The class intervals therefore become:
0–10, 10–20, 20–30, 30–45, 45–50
These form a continuous, non-overlapping sequence.
∴ The missing class boundary is x = 45.
| Group A | 12 | 15 | 18 | 21 | 24 | 29 | 31 | 35 | 40 | ||
|---|---|---|---|---|---|---|---|---|---|---|---|
| Group B | 10 | 14 | 19 | 22 | 25 | x | 30 | 33 | 37 | 42 | 45 |
Since N is even, the median is the average of the 10th and 11th observations in the combined ordered data.
Given median = 27
(10th observation + 11th observation) ÷ 2 = 27
∴ 10th observation + 11th observation = 54
Now arrange the relevant observations from both groups in ascending order:
10, 12, 14, 15, 18, 19, 21, 22, 24, 25, x, 29, 30, 31, 33, 35, 37, 40, 42, 45
For x to be in its stated position, 25 ≤ x ≤ 29.
Thus, the 10th observation is 25 and the 11th observation is x.
∴ 25 + x = 54
or, x = 29
| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 |
|---|---|---|---|---|---|---|
| Frequency | 7 | 11 | 16 | 18 | 15 | 13 |
The highest frequency is 15.
Therefore, the modal class is 30–40.
For the modal class
l = 30
f₁ = 15
f₀ = 12
f₂ = 8
h = 10
Using the mode formula
Mode = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h
= 30 + [(15 − 12) ÷ (2 × 15 − 12 − 8)] × 10
= 30 + [3 ÷ (30 − 20)] × 10
= 30 + (3 ÷ 10) × 10
= 30 + 3
∴ The mode is 33.
| Study Time (hours) | 0–2 | 2–4 | 4–6 | 6–8 | 8–10 | 10–12 |
|---|---|---|---|---|---|---|
| Frequency | 5 | 9 | 14 | 18 | 10 | 4 |
Therefore, the modal class is 6–8 hours.
For the modal class
l = 6
f₁ = 18
f₀ = 14
f₂ = 10
h = 2
Using the mode formula
Mode = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h
= 6 + [(18 − 14) ÷ (2 × 18 − 14 − 10)] × 2
= 6 + [4 ÷ (36 − 24)] × 2
= 6 + (4 ÷ 12) × 2
= 6 + 2/3
= 6.67
∴ The mode is approximately 6.67 hours.
| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 |
|---|---|---|---|---|---|---|
| Frequency | 4 | 7 | 12 | 15 | 8 | 4 |
The highest frequency is 15.
Therefore, the modal class is 30–40.
For the modal class:
l = 30
f₁ = 15
f₀ = 12
f₂ = 8
h = 10
Using the mode formula:
Mode = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h
= 30 + [(15 − 12) ÷ (2 × 15 − 12 − 8)] × 10
= 30 + [3 ÷ (30 − 20)] × 10
= 30 + (3 ÷ 10) × 10
= 30 + 3
∴ The mode is 33.
| Travel Time (minutes) | 0–20 | 20–40 | 40–60 | 60–80 | 80–100 | 100–120 |
|---|---|---|---|---|---|---|
| Frequency | 6 | 11 | 19 | 16 | 12 | 6 |
Therefore, the modal class is 40–60 minutes.
For the modal class
l = 40
f₁ = 19
f₀ = 11
f₂ = 16
h = 20
Using the mode formula
Mode = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h
= 40 + [(19 − 11) ÷ (2 × 19 − 11 − 16)] × 20
= 40 + [8 ÷ (38 − 27)] × 20
= 40 + (8 ÷ 11) × 20
= 40 + 160/11
= 40 + 14.55
= 54.55
∴ The modal travel time is approximately 54.55 minutes.
| Height (cm) | 151 | 152 | 153 | 154 | 155 | 156 | 157 |
|---|---|---|---|---|---|---|---|
| Number of Players | 6 | 4 | 11 | 9 | 16 | 12 | 2 |
N = 6 + 4 + 11 + 9 + 16 + 12 + 2
or, N = 60
Construct the cumulative frequency table:
| Height (cm) | Frequency (fᵢ) | Cumulative Frequency (cf) |
|---|---|---|
| 151 | 6 | 6 |
| 152 | 4 | 10 |
| 153 | 11 | 21 |
| 154 | 9 | 30 |
| 155 | 16 | 46 |
| 156 | 12 | 58 |
| 157 | 2 | 60 |
Median
Since N = 60 is even, the median is the average of the 30th and 31st observations.
From the cumulative frequencies
30th observation = 154 cm
31st observation = 155 cm
Median = (154 + 155) ÷ 2
= 309 ÷ 2
= 154.5 cm
∴ The median = 154.5 cm.
Lower Quartile (Q₁)
Q₁ position = (N + 1) ÷ 4
= (60 + 1) ÷ 4
= 61 ÷ 4
= 15.25th observation
The 15th and 16th observations both have the value 153 cm.
∴ Q₁ = 153 cm.
Upper Quartile (Q₃)
Q₃ position = 3(N + 1) ÷ 4
= 3 × 61 ÷ 4
= 45.75th observation
The 45th and 46th observations both have the value 155 cm.
∴ Q₃ = 155 cm.
Interquartile Range
IQR = Q₃ − Q₁
= 155 − 153
= 2 cm
∴ The median = 154.5 cm, Q₁ = 153 cm, Q₃ = 155 cm, and the interquartile range = 2 cm.
| Height (cm) | 25 | 31 | 34 | 40 | 45 | 48 | 50 | 60 |
|---|---|---|---|---|---|---|---|---|
| Number of Players | 3 | 8 | 10 | 15 | 10 | 9 | 6 | 2 |
N = 3 + 8 + 10 + 15 + 10 + 9 + 6 + 2
or, N = 63
Construct the cumulative frequency table:
| Height (cm) | Frequency (fᵢ) | Cumulative Frequency (cf) |
|---|---|---|
| 25 | 3 | 3 |
| 31 | 8 | 11 |
| 34 | 10 | 21 |
| 40 | 15 | 36 |
| 45 | 10 | 46 |
| 48 | 9 | 55 |
| 50 | 6 | 61 |
| 60 | 2 | 63 |
Median
Since N = 63 is odd
Median position = (N + 1) ÷ 2
= (63 + 1) ÷ 2
= 64 ÷ 2
= 32nd observation
The cumulative frequency first reaching or exceeding 32 is 36, corresponding to the variate 40.
∴ Median = 40
Lower Quartile (Q₁)
Q₁ position = (N + 1) ÷ 4
= 64 ÷ 4
= 16th observation
The cumulative frequency first reaching or exceeding 16 is 21, corresponding to the variate 34.
∴ Q₁ = 34
Upper Quartile (Q₃)
Q₃ position = 3(N + 1) ÷ 4
= 3 × 64 ÷ 4
= 48th observation
The cumulative frequency first reaching or exceeding 48 is 55, corresponding to the variate 48.
Q₃ = 48
∴ The median = 40, lower quartile Q₁ = 34, and upper quartile Q₃ = 48.
| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Frequency | 5 | 7 | x | 8 | 5 |
N = 5 + 7 + x + 8 + 5
or, N = 25 + x
Since the lower quartile Q₁ = 17.5, the Q₁ class is 10–20.
For the Q₁ class
l = 10
cf = 5
f = 7
h = 10
Using the lower quartile formula
Q₁ = l + [((N/4) − cf) ÷ f] × h
Substitute the known values
17.5 = 10 + [((N/4 − 5) ÷ 7)] × 10
or, 17.5 − 10 = [(N/4 − 5) ÷ 7] × 10
or, 7.5 = [(N/4 − 5) ÷ 7] × 10
or, 7.5 × 7 = 10(N/4 − 5)
or, 52.5 = 10(N/4 − 5)
or, 5.25 = N/4 − 5
or, N/4 = 10.25
or, N = 41
But N = 25 + x
Therefore 25 + x = 41
∴ x = 16
| Marks | 0–10 | 10–20 | 20–30 | 30–x | x–50 |
|---|---|---|---|---|---|
| Frequency | 6 | 9 | 12 | 10 | 8 |
N = 6 + 9 + 12 + 10 + 8
or, N = 45
Upper quartile position
3N ÷ 4 = (3 × 45) ÷ 4
= 33.75
Construct the cumulative frequency table
| Marks | Frequency (fᵢ) | Cumulative Frequency (cf) |
|---|---|---|
| 0–10 | 6 | 6 |
| 10–20 | 9 | 15 |
| 20–30 | 12 | 27 |
| 30–x | 10 | 37 |
| x–50 | 8 | 45 |
Since 27 < 33.75 ≤ 37
the Q₃ class is 30–x.
For the Q₃ class
l = 30
cf = 27
f = 10
h = x − 30
Using the upper quartile formula
Q₃ = l + [((3N/4) − cf) ÷ f] × h
or, 36.75 = 30 + [((33.75 − 27) ÷ 10)] × (x − 30)
or, 36.75 − 30 = (6.75 ÷ 10)(x − 30)
or, 6.75 = 0.675(x − 30)
or, x − 30 = 10
or, x = 40
| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 |
|---|---|---|---|---|---|---|
| Frequency | 5 | 8 | 12 | 15 | 7 | 3 |
N = 5 + 8 + 12 + 15 + 7 + 3
or, N = 50
Construct the cumulative frequency table:
| Marks | Frequency (fᵢ) | Cumulative Frequency (cf) |
|---|---|---|
| 0–10 | 5 | 5 |
| 10–20 | 8 | 13 |
| 20–30 | 12 | 25 |
| 30–40 | 15 | 40 |
| 40–50 | 7 | 47 |
| 50–60 | 3 | 50 |
Lower quartile position N ÷ 4 = 50 ÷ 4
= 12.5
Since 5 < 12.5 ≤ 13, the Q₁ class is 10–20.
Q₁ = 10 + [((12.5 − 5) ÷ 8)] × 10
= 10 + (7.5 ÷ 8) × 10
= 19.375
∴ Q₁ = 19.375
Median position
N ÷ 2 = 50 ÷ 2
= 25
Since 13 < 25 ≤ 25, the median class is 20–30.
Median = 20 + [((25 − 13) ÷ 12)] × 10
= 20 + 10
= 30
∴ Median = 30
Upper quartile position:
3N ÷ 4 = (3 × 50) ÷ 4
= 37.5
Since 25 < 37.5 ≤ 40, the Q₃ class is 30–40.
Q₃ = 30 + [((37.5 − 25) ÷ 15)] × 10
= 30 + (12.5 ÷ 15) × 10
≈ 38.33
∴ Q₃ ≈ 38.33
Comparison
Q₁ = 19.375
Median = 30
Q₃ ≈ 38.33
∴ Q₁ < Median < Q₃.
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
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)
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
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 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
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
| 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
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
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
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. |
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.
9, 13, 18, 22, (2x + 3), 31, 36, 40, 45
16, 21, 28, 35, (4x + 1), 44, 49, 55, 60
| 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?
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