By the end of this chapter, you will be able to:
Mastering these skills will help you analyze data confidently, making you a valuable problem solver in any business environment.
Descriptive statistics provide essential tools for summarizing and interpreting data in professional settings across Kenya. Whether managing patient records at a county referral hospital, analysing customer transactions in a bank, or evaluating student performance in a TVET college, understanding measures of central tendency helps in making informed decisions. This chapter focuses on calculating and applying key measures such as mean, median, and mode to real-world data sets relevant to various sectors.
Measures of central tendency describe the center point or typical value of a data set, providing a summary statistic that represents the entire distribution. These measures are fundamental in business mathematics and statistics because they simplify complex data into understandable values for decision-making. The three main measures are the mean, median, and mode, each suited for different types of data and contexts.
The mean is the sum of all data values divided by the number of values. It is widely used in financial reports, performance metrics, and quality control in sectors such as retail businesses and county government offices. The formula for the mean of \(n\) observations \(x_1, x_2, \ldots, x_n\) is
$$ \bar{x} = \frac{\sum_{i=1}^n x_i}{n} $$
Example 1: A retail shop records daily sales (in Ksh) for 5 days as follows: 12,000; 15,000; 13,500; 14,000; 16,500. Find the mean daily sales.
Given: \(n=5\), \(x_1=12,000\), \(x_2=15,000\), \(x_3=13,500\), \(x_4=14,000\), \(x_5=16,500\)
$$ \bar{x} = \frac{12,000 + 15,000 + 13,500 + 14,000 + 16,500}{5} $$
$$ = \frac{71,000}{5} $$
$$ = 14,200 \text{ Ksh} $$
Answer: Mean daily sales = 14,200 Ksh
Example 2: A SACCO tracks loan amounts (in Ksh) disbursed to 6 members: 50,000; 65,000; 55,000; 70,000; 60,000; 80,000. Calculate the mean loan amount.
Given: \(n=6\), \(x_1=50,000\), \(x_2=65,000\), \(x_3=55,000\), \(x_4=70,000\), \(x_5=60,000\), \(x_6=80,000\)
$$ \bar{x} = \frac{50,000 + 65,000 + 55,000 + 70,000 + 60,000 + 80,000}{6} $$
$$ = \frac{380,000}{6} $$
$$ = 63,333.33 \text{ Ksh} $$
Answer: Mean loan amount = 63,333.33 Ksh
Example 3: A county hospital measures the waiting times (in minutes) of 7 patients: 20, 25, 30, 22, 28, 24, 26. Find the mean waiting time.
Given: \(n=7\), \(x_1=20\), \(x_2=25\), \(x_3=30\), \(x_4=22\), \(x_5=28\), \(x_6=24\), \(x_7=26\)
$$ \bar{x} = \frac{20 + 25 + 30 + 22 + 28 + 24 + 26}{7} $$
$$ = \frac{175}{7} $$
$$ = 25 \text{ minutes} $$
Answer: Mean waiting time = 25 minutes
Example 4: A cooperative records the monthly milk production (litres) of 8 farmers: 120, 135, 140, 115, 130, 125, 145, 110. Compute the mean production.
Given: \(n=8\), values as above.
$$ \bar{x} = \frac{120 + 135 + 140 + 115 + 130 + 125 + 145 + 110}{8} $$
$$ = \frac{1020}{8} $$
$$ = 127.5 \text{ litres} $$
Answer: Mean milk production = 127.5 litres
Example 5: A university department records marks of 10 students out of 100: 65, 70, 58, 82, 77, 69, 74, 80, 60, 68. Find the mean mark.
Given: \(n=10\), values as above.
$$ \bar{x} = \frac{65 + 70 + 58 + 82 + 77 + 69 + 74 + 80 + 60 + 68}{10} $$
$$ = \frac{703}{10} $$
$$ = 70.3 $$
Answer: Mean mark = 70.3
The median is the middle value in an ordered data set. It is less affected by extreme values and is useful in income distribution analysis, patient stay durations, or any skewed data in sectors like banks or hospitals. To find the median:
Example 1: Find the median of daily customer visits at a hotel recorded over 7 days: 45, 50, 48, 55, 60, 52, 49.
Given: \(n=7\), values unordered.
Order the data:
$$ 45, 48, 49, 50, 52, 55, 60 $$
Since \(n=7\) (odd), the median is the 4th value:
$$ \text{Median} = 50 $$
Answer: Median daily visits = 50
Example 2: A county government office records the number of documents processed daily for 8 days: 120, 115, 125, 130, 110, 135, 140, 118. Find the median.
Given: \(n=8\), values unordered.
Order the data:
$$ 110, 115, 118, 120, 125, 130, 135, 140 $$
Since \(n=8\) (even), median is average of 4th and 5th values:
$$ \text{Median} = \frac{120 + 125}{2} = \frac{245}{2} = 122.5 $$
Answer: Median documents processed = 122.5
Example 3: A SACCO records the loan repayment durations (months) of 5 members: 12, 15, 10, 20, 18. Find the median repayment duration.
Given: \(n=5\), unordered.
Order:
$$ 10, 12, 15, 18, 20 $$
Median is the 3rd value:
$$ \text{Median} = 15 $$
Answer: Median repayment duration = 15 months
Example 4: A TVET college records test scores of 6 students: 78, 85, 90, 70, 88, 82. Find the median score.
Given: \(n=6\), unordered.
Order:
$$ 70, 78, 82, 85, 88, 90 $$
Median is average of 3rd and 4th values:
$$ \text{Median} = \frac{82 + 85}{2} = \frac{167}{2} = 83.5 $$
Answer: Median test score = 83.5
Example 5: A farm records the weights (kg) of 9 cattle: 250, 270, 260, 255, 275, 265, 280, 290, 240. Find the median weight.
Given: \(n=9\), unordered.
Order:
$$ 240, 250, 255, 260, 265, 270, 275, 280, 290 $$
Median is the 5th value:
$$ \text{Median} = 265 $$
Answer: Median cattle weight = 265 kg
The mode is the value or values that appear most frequently in a data set. It is useful in identifying popular products, common patient diagnoses, or frequent transaction amounts in banks or retail. A data set can be unimodal (one mode), bimodal (two modes), or multimodal (more than two modes).
Example 1: Find the mode of daily sales (in Ksh) over 10 days: 12,000; 13,000; 12,000; 14,000; 15,000; 12,000; 14,000; 13,000; 15,000; 12,000.
Given: \(n=10\), values as above.
Frequency count:
Mode is 12,000.
Answer: Mode daily sales = 12,000 Ksh
Example 2: A bank records the number of transactions per customer in a day: 3, 4, 5, 3, 5, 6, 3, 4, 5, 6, 4. Find the mode.
Frequency count:
Modes are 3, 4, and 5 (multimodal).
Answer: Modes = 3, 4, 5 transactions
Example 3: A hotel records room bookings per day over 8 days: 10, 12, 11, 12, 10, 11, 12, 10. Find the mode.
Frequency count:
Modes are 10 and 12 (bimodal).
Answer: Modes = 10, 12 bookings
Example 4: Student attendance (days) in a course over 9 weeks: 5, 5, 5, 4, 4, 3, 5, 3, 4. Find the mode.
Frequency count:
Mode is 5.
Answer: Mode attendance = 5 days
Example 5: Milk production (litres) per day for 7 cows: 20, 22, 20, 23, 22, 24, 20. Find the mode.
Frequency count:
Mode is 20.
Answer: Mode milk production = 20 litres
Selecting the correct measure depends on the data type and distribution. The mean is sensitive to outliers, the median is robust for skewed data, and the mode is useful for categorical or discrete data. Professionals in various Kenyan sectors must understand these differences to analyze data accurately.
Example 1: A county referral hospital records daily patient waiting times (minutes): 10, 12, 15, 14, 100. Calculate mean and median and interpret which is better for central tendency.
Given: \(n=5\), values as above.
Mean:
$$ \bar{x} = \frac{10 + 12 + 15 + 14 + 100}{5} = \frac{151}{5} = 30.2 \text{ minutes} $$
Order data:
$$ 10, 12, 14, 15, 100 $$
Median is 3rd value:
$$ \text{Median} = 14 \text{ minutes} $$
Interpretation: The mean is inflated by the outlier (100 minutes), so the median better represents typical waiting time.
Answer: Mean = 30.2 minutes; Median = 14 minutes (better measure)
Example 2: In a retail store, the number of items sold per day over 6 days is: 5, 5, 5, 10, 10, 15. Find mean, median, and mode.
Mean:
$$ \bar{x} = \frac{5 + 5 + 5 + 10 + 10 + 15}{6} = \frac{50}{6} = 8.33 $$
Order data:
$$ 5, 5, 5, 10, 10, 15 $$
Median is average of 3rd and 4th values:
$$ \text{Median} = \frac{5 + 10}{2} = 7.5 $$
Mode is 5.
Interpretation: Mode shows most frequent sales; median and mean give central tendency but differ due to skew.
Answer: Mean = 8.33; Median = 7.5; Mode = 5
Example 3: A bank analyzes loan amounts (Ksh) disbursed: 20,000; 25,000; 30,000; 35,000; 40,000. Calculate mean, median, and mode.
Mean:
$$ \bar{x} = \frac{20,000 + 25,000 + 30,000 + 35,000 + 40,000}{5} = \frac{150,000}{5} = 30,000 $$
Order data (already ordered).
Median is 3rd value:
$$ \text{Median} = 30,000 $$
Mode: No repeated value, no mode.
Interpretation: Mean and median are equal; mode does not exist.
Answer: Mean = 30,000; Median = 30,000; No mode
Example 4: A TVET college records final exam scores: 70, 75, 80, 75, 85, 90, 75. Find mean, median, and mode.
Mean:
$$ \bar{x} = \frac{70 + 75 + 80 + 75 + 85 + 90 + 75}{7} = \frac{550}{7} = 78.57 $$
Order data:
$$ 70, 75, 75, 75, 80, 85, 90 $$
Median is 4th value:
$$ \text{Median} = 75 $$
Mode is 75.
Interpretation: Mode and median show common mark; mean is slightly higher due to high scores.
Answer: Mean = 78.57; Median = 75; Mode = 75
Example 5: A farm measures daily egg production for 8 days: 100, 110, 120, 130, 200, 115, 125, 135. Find mean and median.
Mean:
$$ \bar{x} = \frac{100 + 110 + 120 + 130 + 200 + 115 + 125 + 135}{8} = \frac{1035}{8} = 129.38 $$
Order data:
$$ 100, 110, 115, 120, 125, 130, 135, 200 $$
Median is average of 4th and 5th values:
$$ \text{Median} = \frac{120 + 125}{2} = 122.5 $$
Interpretation: Mean is higher due to outlier 200; median better represents typical production.
Answer: Mean = 129.38; Median = 122.5
A retail store records weekly sales (Ksh) for 6 weeks: 25,000; 30,000; 28,000; 35,000; 32,000; 30,000. Calculate the mean sales. (4 marks)
Find the median of the following daily customer counts at a hotel: 45, 50, 40, 60, 55, 48, 52. (4 marks)
Determine the mode of the number of products sold per day over 10 days: 12, 15, 12, 14, 15, 12, 13, 14, 15, 13. (4 marks)
For the data set of loan amounts (Ksh): 100,000; 120,000; 110,000; 130,000; 500,000; 115,000, calculate both mean and median and state which is more appropriate. (6 marks)
A cooperative records milk production (litres) for 7 farmers: 200, 220, 210, 230, 240, 210, 220. Calculate mean, median, and mode. (6 marks)
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Create a free accountThis chapter covered the fundamental concepts of descriptive statistics essential for analyzing business data. It began with measures of central tendency, introducing the arithmetic mean, weighted arithmetic mean, geometric mean, and harmonic mean, each serving different purposes in data summarization. The chapter then explained the mode as the most frequently occurring value and the median as the middle value separating ordered data. It proceeded to explore measures of dispersion, including range, quartiles, deciles, percentiles, mean deviation, standard deviation, and coefficient of variation, which describe the spread and variability within data sets. Finally, the chapter discussed measures of skewness and kurtosis to understand data distribution shapes, excluding the detailed computation of their coefficients. Together, these tools provide a comprehensive approach to summarizing and interpreting business data effectively.
A SACCO recorded the monthly loan amounts (in Ksh 1,000) for five members as follows: 15, 20, 18, 22, and 25. Calculate the arithmetic mean loan amount. (3 marks)
A county hospital has patient satisfaction scores weighted by number of respondents: Scores 70 (40 patients), 80 (30 patients), 90 (30 patients). Find the weighted arithmetic mean satisfaction score. (4 marks)
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