Descriptive statistics are fundamental in banking and finance for summarizing and interpreting large volumes of financial data. Measures of central tendency such as mean, mode, and median provide concise summaries that inform decision-making, risk assessment, and customer analysis. Mastery of these measures enables professionals to understand typical values in datasets such as loan amounts, interest rates, or customer deposits. This chapter focuses on calculating and interpreting these key statistics within a Kenyan banking context.

5.1 Measures of Central Tendency Descriptive

5.1.1 Mean Statistics

The mean is the arithmetic average representing the central value of a dataset. In banking, the mean helps summarize data such as average loan size or average monthly deposit, offering insights into typical financial behavior. The mean is calculated by summing all data points and dividing by the number of observations.

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

Worked Examples

Example 1: A SACCO recorded loan amounts (in Ksh) for 5 members as 20,000; 25,000; 30,000; 22,000; and 28,000. Calculate the mean loan amount.

Given: \(x_1=20000\), \(x_2=25000\), \(x_3=30000\), \(x_4=22000\), \(x_5=28000\), \(n=5\)

$$ \text{Mean} = \frac{20000 + 25000 + 30000 + 22000 + 28000}{5} $$

$$ = \frac{125000}{5} $$

$$ = 25000 \text{ Ksh} $$

Answer: 25000 Ksh

Example 2: A bank's interest rates (in %) on fixed deposits over 6 months are 7.2, 7.5, 7.8, 7.4, 7.6, and 7.3. Find the mean interest rate.

Given: \(x_1=7.2\), \(x_2=7.5\), \(x_3=7.8\), \(x_4=7.4\), \(x_5=7.6\), \(x_6=7.3\), \(n=6\)

$$ \text{Mean} = \frac{7.2 + 7.5 + 7.8 + 7.4 + 7.6 + 7.3}{6} $$

$$ = \frac{44.8}{6} $$

$$ = 7.4667\% $$

Answer: 7.47%

Example 3: A retail bank collected monthly deposits from 8 customers (in Ksh): 15,000; 12,000; 18,000; 20,000; 16,000; 14,000; 19,000; 17,000. Calculate the mean deposit.

Given: \(x_1=15000\), \(x_2=12000\), \(x_3=18000\), \(x_4=20000\), \(x_5=16000\), \(x_6=14000\), \(x_7=19000\), \(x_8=17000\), \(n=8\)

$$ \text{Mean} = \frac{15000 + 12000 + 18000 + 20000 + 16000 + 14000 + 19000 + 17000}{8} $$

$$ = \frac{131000}{8} $$

$$ = 16375 \text{ Ksh} $$

Answer: 16375 Ksh

Example 4: A county government office records daily cash inflows (in Ksh) over 4 days: 45,000; 50,000; 42,000; 48,000. Find the average daily cash inflow.

Given: \(x_1=45000\), \(x_2=50000\), \(x_3=42000\), \(x_4=48000\), \(n=4\)

$$ \text{Mean} = \frac{45000 + 50000 + 42000 + 48000}{4} $$

$$ = \frac{185000}{4} $$

$$ = 46250 \text{ Ksh} $$

Answer: 46250 Ksh

Example 5: A cooperative has monthly savings (in Ksh) from 10 members: 1000, 1200, 1100, 1300, 1250, 1400, 1150, 1350, 1450, 1500. Calculate the mean savings per member.

Given: \(n=10\), sum = \(1000 + 1200 + 1100 + 1300 + 1250 + 1400 + 1150 + 1350 + 1450 + 1500\)

$$ \text{Mean} = \frac{12700}{10} $$

$$ = 1270 \text{ Ksh} $$

Answer: 1270 Ksh

5.1.2 Mode

The mode is the value that occurs most frequently in a dataset. In banking, mode identifies the most common transaction amount or interest rate, useful for understanding customer behavior patterns. Unlike mean, mode is less affected by extreme values.

Worked Examples

Example 1: A bank records the following deposit amounts (in Ksh): 10,000; 15,000; 10,000; 20,000; 15,000; 10,000. Find the mode.

Given: Data set = 10,000; 15,000; 10,000; 20,000; 15,000; 10,000

• 10,000 appears 3 times
• 15,000 appears 2 times
• 20,000 appears 1 time

Answer: 10,000 Ksh

Example 2: Interest rates (%) offered by a bank over a week: 7.5, 7.7, 7.5, 7.8, 7.7, 7.5, 7.9. Find the mode.

Given: Data set = 7.5, 7.7, 7.5, 7.8, 7.7, 7.5, 7.9

• 7.5 appears 3 times
• 7.7 appears 2 times
• 7.8 appears 1 time
• 7.9 appears 1 time

Answer: 7.5%

Example 3: A cooperative records monthly loan requests (in Ksh): 12,000; 15,000; 12,000; 18,000; 12,000; 15,000; 18,000; 15,000. Determine the mode.

Given:

• 12,000 appears 3 times
• 15,000 appears 3 times
• 18,000 appears 2 times

Since 12,000 and 15,000 both appear 3 times, the dataset is bimodal.

Answer: 12,000 Ksh and 15,000 Ksh (bimodal)

Example 4: Daily customer visits to a bank branch over 6 days: 120, 130, 125, 130, 120, 130. Find the mode.

Given:

• 120 appears 2 times
• 130 appears 3 times
• 125 appears 1 time

Answer: 130 customers

Example 5: A retail bank records the number of cheques processed daily: 50, 55, 50, 60, 55, 50, 55. Find the mode.

Given:

• 50 appears 3 times
• 55 appears 3 times
• 60 appears 1 time

The dataset is bimodal.

Answer: 50 and 55 cheques

5.1.3 Median

The median is the middle value when data is arranged in ascending or descending order. It divides the dataset into two equal halves and is especially useful in banking when data is skewed by extreme values such as very large loans or deposits.

To find the median:

• If \(n\) (number of observations) is odd, median is the middle value.
• If \(n\) is even, median is the average of the two middle values.

Worked Examples

Example 1: Find the median loan amount (Ksh) for 5 customers: 15,000; 22,000; 18,000; 25,000; 20,000.

Given: \(n=5\)

Step 1: Arrange in ascending order: 15,000; 18,000; 20,000; 22,000; 25,000

Step 2: \(n\) is odd, median is middle value (3rd value)

Answer: 20,000 Ksh

Example 2: Find the median interest rate (%) from 6 values: 7.2, 7.5, 7.8, 7.4, 7.6, 7.3.

Given: \(n=6\)

Step 1: Arrange ascending: 7.2; 7.3; 7.4; 7.5; 7.6; 7.8

Step 2: \(n\) is even, median is average of 3rd and 4th values:

$$ \text{Median} = \frac{7.4 + 7.5}{2} $$

$$ = \frac{14.9}{2} $$

$$ = 7.45\% $$

Answer: 7.45%

Example 3: Find the median daily deposits (Ksh) for 7 days: 10,000; 12,000; 9,000; 15,000; 11,000; 13,000; 14,000.

Given: \(n=7\)

Step 1: Arrange ascending: 9,000; 10,000; 11,000; 12,000; 13,000; 14,000; 15,000

Step 2: \(n\) odd, median is 4th value

Answer: 12,000 Ksh

Example 4: Find the median number of transactions per day over 8 days: 150; 160; 155; 165; 170; 150; 160; 175.

Given: \(n=8\)

Step 1: Arrange ascending: 150; 150; 155; 160; 160; 165; 170; 175

Step 2: \(n\) even, median is average of 4th and 5th values:

$$ \text{Median} = \frac{160 + 160}{2} $$

$$ = \frac{320}{2} $$

$$ = 160 $$

Answer: 160 transactions

Example 5: Find the median monthly savings (Ksh) for 9 members: 1,200; 1,100; 1,500; 1,300; 1,400; 1,250; 1,350; 1,450; 1,300.

Given: \(n=9\)

Step 1: Arrange ascending: 1,100; 1,200; 1,250; 1,300; 1,300; 1,350; 1,400; 1,450; 1,500

Step 2: \(n\) odd, median is 5th value

Answer: 1,300 Ksh

Chapter Summary

This chapter introduced measures of central tendency, focusing on the mean, mode, and median as key statistics that summarize data sets by identifying their central points. It then examined measures of dispersion, specifically variance and standard deviation, which quantify the spread or variability of data around the mean. The chapter also covered measures of skewness, which describe the asymmetry of data distributions, indicating whether values tend to cluster more on one side of the average. Finally, it explored measures of kurtosis, explaining how they assess the peakedness or flatness of a distribution relative to a normal curve. Together, these descriptive statistics provide a comprehensive toolkit for summarizing, analyzing, and interpreting data in business contexts. Mastery of these concepts enables better decision-making through clearer understanding of data patterns and variability.