5.1 Measures of Central Tendency

Measures of central tendency summarize data by identifying a central point around which data values cluster. These measures include the mean, median, and mode, each with different applications depending on data type and distribution. Understanding these measures enables professionals to analyze data effectively, whether evaluating patient wait times in a county hospital, sales volumes in a retail business, or crop yields in a cooperative farm.

5.1.1 Arithmetic Mean

The arithmetic mean is the sum of all data values divided by the number of values. It provides an average that balances all data points, widely used in performance analysis and financial reporting.

$$ \text{Mean} = \frac{\sum x_i}{n} $$

where \( x_i \) are data points and \( n \) is the number of data points.

Worked Examples

Example 1: A SACCO records the monthly loan amounts (in Ksh 1,000) disbursed to five clients as: 120, 150, 130, 140, 160. Find the average loan amount.

Given: \( x_1 = 120, x_2 = 150, x_3 = 130, x_4 = 140, x_5 = 160 \), \( n = 5 \)$$ \text{Mean} = \frac{120 + 150 + 130 + 140 + 160}{5} $$$$ = \frac{700}{5} $$$$ = 140 $$ Answer: 140,000 Ksh

Example 2: A county referral hospital measures patient waiting times (minutes) as 30, 45, 50, 35, 40, 60. Calculate the mean waiting time.

Given: \( x_1=30, x_2=45, x_3=50, x_4=35, x_5=40, x_6=60 \), \( n=6 \)$$ \text{Mean} = \frac{30 + 45 + 50 + 35 + 40 + 60}{6} $$$$ = \frac{260}{6} $$$$ = 43.33 $$ Answer: 43.33 minutes

Example 3: A hotel tracks daily guest arrivals over 7 days: 80, 95, 100, 90, 85, 110, 120. Find the mean daily arrivals.

Given: \( x_1=80, x_2=95, x_3=100, x_4=90, x_5=85, x_6=110, x_7=120 \), \( n=7 \)$$ \text{Mean} = \frac{80 + 95 + 100 + 90 + 85 + 110 + 120}{7} $$$$ = \frac{680}{7} $$$$ = 97.14 $$ Answer: 97.14 guests

Example 4: A retail store records weekly sales (Ksh 1,000) over 12 weeks: 200, 180, 210, 195, 220, 205, 190, 215, 230, 225, 205, 210. Find the average weekly sales.

Given: \( n=12 \) and sales as above$$ \text{Mean} = \frac{200 + 180 + 210 + 195 + 220 + 205 + 190 + 215 + 230 + 225 + 205 + 210}{12} $$$$ = \frac{2485}{12} $$$$ = 207.08 $$ Answer: 207,083 Ksh

Example 5: A cooperative farm measures daily milk production (litres) for 10 days: 150, 160, 155, 165, 170, 175, 160, 155, 150, 165. Calculate the mean daily production.

Given: \( n=10 \) and data as above$$ \text{Mean} = \frac{150 + 160 + 155 + 165 + 170 + 175 + 160 + 155 + 150 + 165}{10} $$$$ = \frac{1605}{10} $$$$ = 160.5 $$ Answer: 160.5 litres

5.1.2 Median

The median is the middle value in an ordered dataset, dividing the data into two equal halves. It is especially useful when data contain outliers or are skewed, such as income levels or patient recovery times.

To find the median:
- Arrange data in ascending order
- If \( n \) is odd, median is the middle value
- If \( n \) is even, median is the average of the two middle values

Worked Examples

Example 1: A county government office records the number of daily visitors over 7 days: 50, 45, 55, 60, 40, 65, 70. Find the median.

Given: \( n=7 \)
Ordered data: 40, 45, 50, 55, 60, 65, 70
Middle value is the 4th value (since \( \frac{7+1}{2} = 4 \))
Answer: 55 visitors

Example 2: A TVET college records test scores for 8 students: 70, 65, 80, 75, 85, 60, 90, 55. Find the median score.

Given: \( n=8 \)
Ordered data: 55, 60, 65, 70, 75, 80, 85, 90
Median is average of 4th and 5th values: \( \frac{70 + 75}{2} = 72.5 \)
Answer: 72.5

Example 3: A bank measures the number of customers served per hour in 9 hours: 15, 18, 20, 17, 16, 21, 19, 22, 23. Find the median.

Given: \( n=9 \)
Ordered data: 15, 16, 17, 18, 19, 20, 21, 22, 23
Median is the 5th value: 19
Answer: 19 customers

Example 4: A farm records daily rainfall (mm) over 10 days: 12, 15, 14, 10, 20, 18, 16, 13, 11, 17. Find the median rainfall.

Given: \( n=10 \)
Ordered data: 10, 11, 12, 13, 14, 15, 16, 17, 18, 20
Median is average of 5th and 6th values: \( \frac{14 + 15}{2} = 14.5 \)
Answer: 14.5 mm

Example 5: A hotel tracks the number of bookings over 11 days: 40, 42, 45, 38, 50, 48, 46, 44, 47, 49, 43. Find the median bookings.

Given: \( n=11 \)
Ordered data: 38, 40, 42, 43, 44, 45, 46, 47, 48, 49, 50
Median is the 6th value: 45
Answer: 45 bookings

5.1.3 Mode

The mode is the most frequently occurring value in a dataset. It is useful in categorical data analysis and situations where the most common value is of interest, such as the most popular product sold or common disease diagnosed.

Worked Examples

Example 1: A retail business records daily sales items sold: 10, 12, 10, 15, 12, 10, 18. Find the mode.

Data: 10 (3 times), 12 (2 times), 15 (1), 18 (1)
Answer: 10 items

Example 2: A hospital records daily patient diagnoses counts: 5, 7, 5, 9, 7, 5, 8, 7. Find the mode.

Data: 5 (3 times), 7 (3 times), 8 (1), 9 (1)
Two modes: 5 and 7 (bimodal)
Answer: 5 and 7 diagnoses

Example 3: A cooperative records crop yield categories (tons): 4, 5, 6, 5, 7, 6, 5, 6, 6. Find the mode.

Data: 5 (3 times), 6 (4 times), 4 (1), 7 (1)
Answer: 6 tons

Example 4: A county office records the number of vehicles arriving per hour: 20, 25, 20, 30, 25, 20, 35. Find the mode.

Data: 20 (3 times), 25 (2 times), 30 (1), 35 (1)
Answer: 20 vehicles

Example 5: A school records students' preferred lunch options: Rice (10), Ugali (15), Rice (10), Chapati (15), Ugali (15), Rice (10). Find the mode.

Count: Rice (3 times), Ugali (3 times), Chapati (1)
Two modes: Rice and Ugali
Answer: Rice and Ugali

5.1.4 Weighted Mean

The weighted mean accounts for differing importance or frequency of data points, common in financial and operational analyses such as average cost calculations or quality scores.

The formula is:

$$ \text{Weighted Mean} = \frac{\sum w_i x_i}{\sum w_i} $$

where \( w_i \) are weights and \( x_i \) are data values.

Worked Examples

Example 1: A bank calculates the average interest rate on loans: 8% on Ksh 500,000, 10% on Ksh 300,000, and 9% on Ksh 200,000. Find the weighted average interest rate.

Given: \( w_1=500,000, x_1=8\% \), \( w_2=300,000, x_2=10\% \), \( w_3=200,000, x_3=9\% \)$$ \text{Weighted Mean} = \frac{(500,000 \times 8) + (300,000 \times 10) + (200,000 \times 9)}{500,000 + 300,000 + 200,000} $$$$ = \frac{4,000,000 + 3,000,000 + 1,800,000}{1,000,000} $$$$ = \frac{8,800,000}{1,000,000} $$$$ = 8.8\% $$ Answer: 8.8% interest rate

Example 2: A school reports exam scores weighted by number of students: 70 (20 students), 80 (25 students), 90 (15 students). Find the weighted mean score.

Given: \( w_1=20, x_1=70 \), \( w_2=25, x_2=80 \), \( w_3=15, x_3=90 \)$$ \text{Weighted Mean} = \frac{(20 \times 70) + (25 \times 80) + (15 \times 90)}{20 + 25 + 15} $$$$ = \frac{1400 + 2000 + 1350}{60} $$$$ = \frac{4750}{60} $$$$ = 79.17 $$ Answer: 79.17 marks

Example 3: A cooperative sells three types of produce with prices and quantities: Maize 150 bags at Ksh 2,000 each, Beans 100 bags at Ksh 3,000, and Coffee 50 bags at Ksh 4,000. Find the weighted average price per bag.

Given: \( w_1=150, x_1=2000 \), \( w_2=100, x_2=3000 \), \( w_3=50, x_3=4000 \)$$ \text{Weighted Mean} = \frac{(150 \times 2000) + (100 \times 3000) + (50 \times 4000)}{150 + 100 + 50} $$$$ = \frac{300,000 + 300,000 + 200,000}{300} $$$$ = \frac{800,000}{300} $$$$ = 2666.67 $$ Answer: 2,666.67 Ksh per bag

Example 4: A hotel rates service scores with weights: Cleanliness 9 (weight 3), Food quality 8 (weight 4), Staff behaviour 7 (weight 2). Find the weighted average service score.

Given: \( w_1=3, x_1=9 \), \( w_2=4, x_2=8 \), \( w_3=2, x_3=7 \)$$ \text{Weighted Mean} = \frac{(3 \times 9) + (4 \times 8) + (2 \times 7)}{3 + 4 + 2} $$$$ = \frac{27 + 32 + 14}{9} $$$$ = \frac{73}{9} $$$$ = 8.11 $$ Answer: 8.11 score

Example 5: A county government evaluates project costs weighted by importance: Road repair Ksh 5 million (weight 5), Water supply Ksh 3 million (weight 3), Health center Ksh 4 million (weight 4). Find the weighted average cost.

Given: \( w_1=5, x_1=5,000,000 \), \( w_2=3, x_2=3,000,000 \), \( w_3=4, x_3=4,000,000 \)$$ \text{Weighted Mean} = \frac{(5 \times 5,000,000) + (3 \times 3,000,000) + (4 \times 4,000,000)}{5 + 3 + 4} $$$$ = \frac{25,000,000 + 9,000,000 + 16,000,000}{12} $$$$ = \frac{50,000,000}{12} $$$$ = 4,166,666.67 $$ Answer: Ksh 4,166,666.67

Chapter Summary

This chapter provided a comprehensive overview of descriptive statistics, beginning with measures of central tendency that summarize data sets by identifying typical values. It explored different types of means including the arithmetic mean, weighted arithmetic mean, geometric mean, and harmonic mean, each suited for specific data scenarios. The chapter then examined the mode as the most frequently occurring value and the median as the middle value that divides data into two equal halves. It further introduced measures of dispersion that describe the spread or variability within data, covering the range, quartiles, deciles, percentiles, mean deviation, standard deviation, and coefficient of variation. The chapter concluded with a discussion on measures of skewness and kurtosis, which describe the shape and distribution characteristics of data without focusing on their computational formulas. Together, these concepts equip students to effectively summarize, analyze, and interpret quantitative business data.