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MEASURES OF CENTRAL TENDENCY

MEANING OF MEASURES OF CENTRAL TENDENCY

Definition:
Measures of central tendency are the values that describe the center or average of a set of numbers.
It helps us to find one single value that represents a whole group of data.

In simple words, it helps us to know what is normal or typical in a given set of numbers.

For example:
If we have the numbers 2, 4, 6, 8, and 10, the measure of central tendency will tell us what number stands at the middle or best represents these numbers.

THE THREE MAIN MEASURES OF CENTRAL TENDENCY

  1. Mean

  2. Median

  3. Mode

1. MEAN (ARITHMETIC MEAN)

Definition:
The mean is the sum of all the numbers in a set divided by the total number of items.
It helps us to find the average of a group of numbers.

Formula for Mean:

Mean = (Sum of all the data) ÷ (Number of data items)

Example 1:

Find the mean of 4, 6, 8, 10, and 12.

Solution:
Mean = (4 + 6 + 8 + 10 + 12) ÷ 5 = 40 ÷ 5 = 8
Mean = 8

Example 2:

Find the mean of 10, 15, 20, and 25.

Solution:
Mean = (10 + 15 + 20 + 25) ÷ 4 = 70 ÷ 4 = 17.5
Mean = 17.5

Example 3 (When Data is in a Table):

Score (x)Frequency (f)fx
236
3412
428
515
Total1031

Mean = Σfx ÷ Σf = 31 ÷ 10 = 3.1
Mean = 3.1

Importance of Mean

  1. It helps us to find the general average of data.

  2. It helps us to compare the performance of two or more groups.

  3. It helps us to analyze test scores, salaries, or results.

2. MEDIAN

Definition:
The median is the middle number when the data is arranged in order (either from smallest to largest or largest to smallest).

If the number of data items is:
- Odd: The median is the middle number.
- Even: The median is the average of the two middle numbers.

Example 1 (Odd Number of Data):

Find the median of 3, 5, 7, 9, 11.
Arrange in order: 3, 5, 7, 9, 11
Median = 7

Example 2 (Even Number of Data):

Find the median of 2, 4, 6, 8.
Arrange in order: 2, 4, 6, 8
Median = (4 + 6) ÷ 2 = 5
Median = 5

Example 3 (Repeated Data):

Find the median of 2, 2, 3, 3, 4, 4, 5, 5.
Number of items = 8 (even)
Middle two numbers = 3 and 4
Median = (3 + 4) ÷ 2 = 3.5
Median = 3.5

Steps to Find Median

  1. Arrange data in ascending order.

  2. Count how many items are in the data.

  3. If the number is odd, take the middle number.

  4. If the number is even, find the average of the two middle numbers.

Importance of Median

  1. It helps us to know the center of data when some numbers are very large or very small.

  2. It helps us to find the middle performance in a test or survey.

  3. It helps us to avoid being affected by extreme values.

3. MODE

Definition:
The mode is the number that appears most often in a data set.
It helps us to know which number occurs most frequently.

Example 1:

Find the mode of 2, 3, 3, 4, 5, 5, 5, 6.
Mode = 5

Example 2:

Find the mode of 8, 9, 9, 10, 11, 11.
Data is bimodal (two modes: 9 and 11)

Example 3:

Find the mode of 1, 2, 3, 4, 5, 6.
No mode (all appear once)

Importance of Mode

  1. It helps us to identify the most common value in a data set.

  2. It helps us to find the most popular choice or most sold item.

  3. It helps us to analyze survey or market data (for example: most preferred color or food).

COMPARISON BETWEEN MEAN, MEDIAN, AND MODE

MeasureMeaningWhen to UseExample
MeanArithmetic averageWhen data values are close togetherAverage test score
MedianMiddle valueWhen data has extreme valuesIncome distribution
ModeMost frequent valueWhen finding most common valueMost common shoe size

IMPORTANCE OF MEASURES OF CENTRAL TENDENCY

  1. It helps us to summarize a large group of data into one simple value.

  2. It helps us to compare two or more groups of data easily.

  3. It helps us to understand the general trend of data.

  4. It helps us to make better decisions in daily life, business, and research.

  5. It helps us to analyze examination results, prices, heights, weights, and survey information.

SUMMARY TABLE

MeasureFormula or MethodExampleResult
MeanSum ÷ Number of items(4+6+8+10+12) ÷ 58
MedianMiddle number when arranged3, 5, 7, 9, 11 → middle = 77
ModeNumber that appears most2, 3, 3, 4, 5 → most = 33

EXERCISE (For Practice)

  1. Find the mean of 5, 10, 15, 20, and 25.

  2. Find the median of 8, 12, 10, 15, and 20.

  3. Find the mode of 6, 7, 7, 8, 9, 9, 9.

  4. Find the mean and median of 10, 20, 30, 40, 50, 60.

  5. Explain one real-life situation where you can apply the mode.






APPLICATION OF MEASURES OF CENTRAL TENDENCY TO ANALYZE ANY GIVEN INFORMATION

Meaning:
The application of measures of central tendency means using mean, median, and mode to study, interpret, and make sense of data or information. It helps us to find out the general behavior or trend of a group of numbers, people, or things.

1. APPLICATION OF MEAN

Definition:
The mean is the average of a set of numbers. It is found by adding up all the values and dividing by the number of values.

Application:

  1. It helps us to find the average score of students in a test.

  2. It helps us to know the average daily temperature in a week or month.

  3. It helps businesses to find the average sales or profit made over a period.

  4. It helps government and researchers to find the average income of people in a community.

Example:

Five students scored 10, 12, 8, 14, and 16 in a test.
To find the mean score:
Mean = (10 + 12 + 8 + 14 + 16) ÷ 5 = 60 ÷ 5 = 12.

The average score of the students is 12.

Interpretation:
It means most students performed around a score of 12 marks. The mean helps us to understand the overall class performance.

2. APPLICATION OF MEDIAN

Definition:
The median is the middle value when data is arranged in order of size (either from smallest to largest or largest to smallest).

Application:

  1. It helps us to find the central value in a list of incomes, ages, or heights.

  2. It helps when data has extreme values (very high or very low numbers) that may affect the mean.

  3. It is useful for understanding the typical or middle position in a set of values.

Example:

The ages of five pupils are 8, 10, 9, 11, and 12.
First, arrange them in order: 8, 9, 10, 11, 12.
The middle value is 10.

The median age is 10.

Interpretation:
This means half of the pupils are younger than 10 years, and half are older than 10 years. The median helps us to see the central point of the data.

3. APPLICATION OF MODE

Definition:
The mode is the number that appears most often in a set of data.

Application:

  1. It helps traders or manufacturers to know which product sells the most.

  2. It helps teachers to see which score is most common in a test.

  3. It helps fashion designers or sellers to know which size or color of dress is most popular.

  4. It helps government to know which age group or area has the highest population.

Example:

The shoe sizes of 8 pupils are 37, 38, 37, 39, 38, 37, 38, 40.
The mode is the most frequent number.
37 appears 3 times, and 38 appears 3 times.

The data is bimodal (two modes): 37 and 38.

Interpretation:
This means most pupils wear shoe sizes 37 and 38. This helps a seller know which sizes to stock more.

4. GENERAL APPLICATIONS OF MEASURES OF CENTRAL TENDENCY

In general, mean, median, and mode can be used to analyze and understand any given information, such as:

  1. Finding the average result of students in different subjects.

  2. Comparing performances of two schools or two groups of people.

  3. Understanding the pattern of prices, wages, heights, or weights in a population.

  4. Planning for production, budgeting, and market supply based on common values.

  5. Making fair judgments or decisions based on summarized data.

SUMMARY TABLE

MeasureDefinitionApplication
MeanSum of values ÷ Number of valuesUsed for finding average score, income, temperature, etc.
MedianMiddle value when data is arranged in orderUsed for understanding the central value of data, especially when data has extreme values.
ModeMost frequent valueUsed for identifying the most common or popular item, product, or number.

CONCLUSION

It helps us to analyze information by showing the central or most typical value in a set of data. This makes it easier to compare, describe, and understand the behavior of different data sets in real life.





PRACTICE QUESTIONS

Instructions: Show all working steps. Use mean, median, and mode where required and give interpretation of results.

  1. Question 1
    Find the mean, median and mode of the data set: 6, 8, 10, 12, 14, 16, 18.

    Solution (step-by-step):
    1. Count the number of items: n = 7.

    2. Mean: Add all numbers: 6+8+10+12+14+16+18 = 84. Mean = 84 ÷ 7 = 12.

    3. Median: Data are already ordered and n is odd, middle item is the 4th item → 12. Median = 12.

    4. Mode: All numbers occur once, so there is no mode.

    5. Interpretation: The typical or central value is 12. Mean and median agree, so the data are symmetric.


  2. Question 2
    The marks of ten students are: 45, 52, 60, 60, 72, 80, 80, 80, 90, 95. Find the mean, median and mode. Explain which measure best describes the class performance.

    Solution (step-by-step):
    1. n = 10.

    2. Mean: Sum = 45+52+60+60+72+80+80+80+90+95 = 714. Mean = 714 ÷ 10 = 71.4.

    3. Median: n is even so median is average of 5th and 6th items. 5th = 72, 6th = 80. Median = (72 + 80) ÷ 2 = 76.

    4. Mode: 80 appears three times while others appear less. Mode = 80.

    5. Interpretation: The mean is 71.4, median 76 and mode 80. Mode and median are higher than the mean because lower marks (for example 45 and 52) pull the mean down. For typical student performance, median or mode may be better since they resist extremes.


  3. Question 3
    A teacher records the number of books read by 12 pupils: 0, 1, 1, 2, 2, 2, 3, 4, 4, 5, 6, 12. Find the mean, median and mode. Comment on the effect of the value 12.

    Solution (step-by-step):
    1. n = 12.

    2. Mean: Sum = 0+1+1+2+2+2+3+4+4+5+6+12 = 42. Mean = 42 ÷ 12 = 3.5.

    3. Median: n even so median is average of 6th and 7th items. 6th = 2, 7th = 3. Median = (2 + 3) ÷ 2 = 2.5.

    4. Mode: 2 appears three times, more than any other. Mode = 2.

    5. Comment: The single large value 12 raises the mean (3.5) above the median (2.5) and mode (2). This shows the mean is sensitive to extreme values while median and mode are more robust.


  4. Question 4
    Use the grouped frequency table below to find the estimated mean (use class midpoints) and the modal class. Interpret the results.
    ClassFrequency (f)
    0 – 94
    10 – 196
    20 – 2910
    30 – 395
    40 – 493

    Solution (step-by-step):
    1. Compute class midpoints: 0–9 → 4.5; 10–19 → 14.5; 20–29 → 24.5; 30–39 → 34.5; 40–49 → 44.5.

    2. Compute f × midpoint: 4×4.5 = 18; 6×14.5 = 87; 10×24.5 = 245; 5×34.5 = 172.5; 3×44.5 = 133.5.

    3. Sum of frequencies Σf = 4+6+10+5+3 = 28. Sum of f×midpoint Σfx = 18+87+245+172.5+133.5 = 656.

    4. Estimated mean = Σfx ÷ Σf = 656 ÷ 28 ≈ 23.43.

    5. Modal class = class with highest frequency = 20 – 29 (frequency 10). Interpretation: Most observations lie in 20–29 and the average is about 23.4; data are concentrated in that region.


  5. Question 5
    A shop sold the following numbers of shirts per week for 8 weeks: 7, 8, 7, 10, 9, 8, 7, 20. Find mean, median, and mode and explain what the shopkeeper will learn from each measure.

    Solution (step-by-step):
    1. n = 8. Ordered data: 7, 7, 7, 8, 8, 9, 10, 20.

    2. Mean: Sum = 7+7+7+8+8+9+10+20 = 76. Mean = 76 ÷ 8 = 9.5.

    3. Median: average of 4th and 5th items: (8 + 8) ÷ 2 = 8.

    4. Mode: 7 appears three times. Mode = 7.

    5. Explanation: Mean = 9.5 suggests average weekly sales are 9.5 shirts, but this is raised by the one week of 20 sales. Median = 8 is better for typical week. Mode = 7 indicates the most common weekly sale count. The shopkeeper will get that most weeks sell about 7 to 8 shirts; the large week is unusual.


  6. Question 6
    The incomes (in thousands of naira) of 9 households are: 20, 25, 18, 30, 22, 21, 19, 28, 27. Find mean and median and state which measure would be preferred if one household earned 200 instead of 30.

    Solution (step-by-step):
    1. Original data ordered: 18, 19, 20, 21, 22, 25, 27, 28, 30. n = 9.

    2. Mean: Sum = 18+19+20+21+22+25+27+28+30 = 210. Mean = 210 ÷ 9 ≈ 23.33.

    3. Median: middle item is 5th = 22. Median = 22.

    4. If one household earned 200 instead of 30, recompute mean: new sum = 210 − 30 + 200 = 380. New mean = 380 ÷ 9 ≈ 42.22. Median remains 22. The median would be preferred because it is not affected by the extreme value 200 while the mean is heavily affected.


  7. Question 7
    A small factory records daily production for 7 days as follows: 120, 125, 118, 130, 127, 122, 121. Compute the mean and explain whether the mean is appropriate to report the typical daily production.

    Solution (step-by-step):
    1. n = 7. Sum = 120+125+118+130+127+122+121 = 863.

    2. Mean = 863 ÷ 7 = 123.285714... ≈ 123.29 (rounded to two decimal places).

    3. All values are close; no extreme outliers. Therefore mean is appropriate and it helps us to understand the typical daily output which is about 123 units.


  8. Question 8
    Given the frequency distribution below, find the median class and estimate the median using linear interpolation.
    ClassFrequency (f)
    0 – 95
    10 – 198
    20 – 2912
    30 – 395

    Solution (step-by-step):
    1. Total frequency N = 5+8+12+5 = 30. Median position = (N+1) ÷ 2 = 31 ÷ 2 = 15.5.

    2. Cumulative frequencies: up to 0–9 = 5; up to 10–19 = 13; up to 20–29 = 25.

    3. Median class is 20–29 because cumulative frequency just reaches 25 which contains position 15.5.

    4. Use linear interpolation: L = lower limit of median class = 20. c.f. before median class = 13. f = frequency of median class = 12. class width = 10.

    5. Median ≈ L + [(N/2 − c.f. before) ÷ f] × class width = 20 + [(15 − 13) ÷ 12] × 10. (Use N/2 = 15 for grouped median formula.)

    6. Compute: (15 − 13) = 2. 2 ÷ 12 = 1/6 ≈ 0.1666667. Multiply by 10 → ≈ 1.6667. Median ≈ 20 + 1.6667 = 21.6667.

    7. Estimated median ≈ 21.67.

    8. Interpretation: The middle observation lies in 20–29 and is about 21.7.


  9. Question 9
    The number of goals scored by a football team over 11 matches are: 0, 1, 2, 1, 3, 2, 4, 2, 1, 0, 5. Find the mode and explain how this helps the coach.

    Solution (step-by-step):
    1. Order the data or count frequencies: 0 appears 2 times; 1 appears 3 times; 2 appears 3 times; 3 appears 1; 4 appears 1; 5 appears 1.

    2. 1 and 2 both appear 3 times → bimodal with modes 1 and 2.

    3. Interpretation: The most common match results are scoring 1 or 2 goals. The coach will get that in most matches the team scores 1 or 2 goals and can plan tactics accordingly.


  10. Question 10
    A test has 5 questions. Scores (out of 5) by 15 pupils are: 5, 4, 3, 5, 2, 3, 4, 4, 5, 1, 0, 3, 2, 5, 4. Find the mean, median and mode and say which measure best represents the class performance.

    Solution (step-by-step):
    1. Order data: 0,1,2,2,3,3,3,4,4,4,4,5,5,5,5. n = 15.

    2. Mean: sum = 0+1+2+2+3+3+3+4+4+4+4+5+5+5+5 = 46. Mean = 46 ÷ 15 ≈ 3.0667.

    3. Median: middle item is 8th item = 4. Median = 4.

    4. Mode: 4 appears four times and 5 appears four times. Bimodal (4 and 5). Mode = 4 and 5.

    5. Conclusion: Median 4 is a strong representative since many pupils scored 4 or 5 and the mean is slightly above 3. The median or mode better represents typical performance because they show most pupils scored 4 or 5.


  11. Question 11
    The weekly expenses (in naira) of 7 families are: 3,200; 3,500; 3,450; 3,300; 7,000; 3,400; 3,350. Find the mean and median and recommend which measure to use for comparison of typical expense when one family reports 70,000 instead of 7,000.

    Solution (step-by-step):
    1. Order original data: 3,200; 3,300; 3,350; 3,400; 3,450; 3,500; 7,000. n = 7.

    2. Mean = sum ÷ 7 = (3,200+3,500+3,450+3,300+7,000+3,400+3,350) = 27,200 ÷ 7 ≈ 3,885.71.

    3. Median = middle item (4th) = 3,400.

    4. If one family reported 70,000 instead of 7,000, new sum = 27,200 − 7,000 + 70,000 = 90,200. New mean = 90,200 ÷ 7 ≈ 12,885.71. Median remains 3,400.

    5. Recommendation: The median should be used to compare typical expenses because the mean is badly affected by the extreme value 70,000.


  12. Question 12
    A factory manager wants to know the typical number of defective items per day. The counts for 10 days are: 2, 0, 1, 3, 2, 2, 1, 20, 2, 1. Compute mean and median and advise which is safer for decision making.

    Solution (step-by-step):
    1. Order data: 0,1,1,1,2,2,2,2,3,20. n = 10.

    2. Mean: sum = 0+1+1+1+2+2+2+2+3+20 = 34. Mean = 34 ÷ 10 = 3.4.

    3. Median: average of 5th and 6th items: (2 + 2) ÷ 2 = 2.

    4. Advice: Use median (2) for typical daily defect because single extreme day with 20 defects inflates the mean to 3.4. Median gives a safer operational target.


  13. Question 13
    The heights (in cm) of 9 pupils are: 120, 125, 130, 128, 122, 126, 129, 127, 124. Find mean and median and draw one line text diagram showing ordered heights with the median marked.

    Solution (step-by-step):
    1. Order data: 120, 122, 124, 125, 126, 127, 128, 129, 130.

    2. Mean: sum = 120+122+124+125+126+127+128+129+130 = 1,131. Mean = 1,131 ÷ 9 = 125.666666... ≈ 125.67 cm.

    3. Median: middle item 5th = 126 cm.

    4. Text diagram (ordered with median marked):
      120 — 122 — 124 — 125 — [126] — 127 — 128 — 129 — 130

    5. Interpretation: The typical height is about 126 cm and the average is about 125.67 cm. Both are close indicating little skew.


  14. Question 14
    A market vendor recorded the sizes sold in a day: S, M, M, L, M, S, XL, M, L, M. Convert S, M, L, XL to numbers S=1, M=2, L=3, XL=4 and find the mode and median of the coded data. Explain what the result says about demand.

    Solution (step-by-step):
    1. Code the data: S(1), M(2), M(2), L(3), M(2), S(1), XL(4), M(2), L(3), M(2).

    2. Coded list: 1,2,2,3,2,1,4,2,3,2.

    3. Order: 1,1,2,2,2,2,2,2,3,3,4 (count frequencies). M (2) appears 6 times, S(1) appears 2 times, L(3) appears 2 times, XL(4) appears once.

    4. Mode = 2 (M). Median: n = 10, median is average of 5th and 6th values. Both 5th and 6th are 2. Median = 2 (M).

    5. Explanation: Most customers demand size M. Both median and mode being M confirm size M is the typical demand.


  15. Question 15
    Using the following test scores: 40, 42, 45, 48, 52, 55, 58, 60, 62, 70, find the mean and median. If a teacher removes the lowest and highest score and recomputes the mean, what does the teacher learn about the effect of trimming extremes?

    Solution (step-by-step):
    1. n = 10. Sum = 40+42+45+48+52+55+58+60+62+70 = 532. Mean = 532 ÷ 10 = 53.2.

    2. Median: average of 5th and 6th values: (52 + 55) ÷ 2 = 53.5.

    3. Trim extremes: remove 40 and 70. New list: 42,45,48,52,55,58,60,62. New sum = 42+45+48+52+55+58+60+62 = 422. New n = 8. New mean = 422 ÷ 8 = 52.75.

    4. Lesson: Trimming extremes changed mean from 53.2 to 52.75, a small change. Trimming reduces the influence of outliers and may give a measure closer to the central bulk of data.





CHECK OTHER RELATED TOPICS HERE


  1. AREA OF PLANE SHAPES

  2. CONSTRUCTION


  3. MEASURES OF CENTRAL TENDENCY

  4. DATA PRESENTATION





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