By the end of this chapter, you will be able to:
Mastering these skills will help you solve real-world engineering problems involving forces, fields, and flows with precision and confidence.
Vectors are fundamental in describing quantities that possess both magnitude and direction, such as displacement, velocity, and force. In professional fields across Kenya, from hospital logistics to agricultural planning, understanding vectors enables accurate modelling of real-world phenomena involving direction and size. This chapter covers the application of vector theory, starting with the measures of central tendency and dispersion, which are essential for analyzing vector data and other numerical datasets encountered in diverse sectors like county government offices and retail businesses.
Measures of central tendency summarize data by identifying a central value that represents the dataset. These measures help professionals in all sectors, such as SACCOs and universities, to analyze and interpret data efficiently for decision making. The main measures include the mean, mode, and median, each providing a different perspective on the dataset’s center.
The mean or average is the sum of all data values divided by the number of values. It is widely used in finance, agriculture, and health sectors to find typical values such as average income, crop yield, or patient wait times.
$$ \text{Mean} = \frac{\sum_{i=1}^n x_i}{n} $$
Example 1: A retail business records daily sales (Ksh) for 5 days as 12,000; 15,000; 14,000; 13,000; 16,000. Find the mean daily sales.
Given: \( x_1=12000, x_2=15000, x_3=14000, x_4=13000, x_5=16000, n=5 \)
$$ \text{Mean} = \frac{12000 + 15000 + 14000 + 13000 + 16000}{5} $$
$$ = \frac{70000}{5} $$
$$ = 14000 \text{ Ksh} $$
Answer: 14000 Ksh
Example 2: A county hospital measures patient waiting times (minutes) over 6 days: 25, 30, 35, 40, 45, 50. Find the mean waiting time.
Given: \( x_1=25, x_2=30, x_3=35, x_4=40, x_5=45, x_6=50, n=6 \)
$$ \text{Mean} = \frac{25 + 30 + 35 + 40 + 45 + 50}{6} $$
$$ = \frac{225}{6} $$
$$ = 37.5 \text{ minutes} $$
Answer: 37.5 minutes
Example 3: A SACCO records loan amounts (Ksh) given to 4 clients: 50,000; 75,000; 65,000; 85,000. Calculate the mean loan amount.
Given: \( x_1=50000, x_2=75000, x_3=65000, x_4=85000, n=4 \)
$$ \text{Mean} = \frac{50000 + 75000 + 65000 + 85000}{4} $$
$$ = \frac{275000}{4} $$
$$ = 68750 \text{ Ksh} $$
Answer: 68750 Ksh
Example 4: A farm records daily milk production (litres) for 7 days: 120, 110, 130, 115, 125, 140, 135. Find the mean production.
Given: \( x_1=120, x_2=110, x_3=130, x_4=115, x_5=125, x_6=140, x_7=135, n=7 \)
$$ \text{Mean} = \frac{120 + 110 + 130 + 115 + 125 + 140 + 135}{7} $$
$$ = \frac{875}{7} $$
$$ \approx 125 \text{ litres} $$
Answer: 125 litres
Example 5: A university records scores of 8 students in a test: 60, 70, 75, 80, 85, 90, 95, 100. Calculate the mean score.
Given: \( x_1=60, x_2=70, x_3=75, x_4=80, x_5=85, x_6=90, x_7=95, x_8=100, n=8 \)
$$ \text{Mean} = \frac{60 + 70 + 75 + 80 + 85 + 90 + 95 + 100}{8} $$
$$ = \frac{655}{8} $$
$$ = 81.875 $$
Answer: 81.875
The mode is the data value that appears most frequently in a dataset. It is useful in retail and banking sectors to identify the most common product sold or the most frequent transaction amount.
Mode is the value \(x\) such that frequency of \(x\) is maximum.
Example 1: A hotel records the number of guests per day over a week: 20, 22, 20, 25, 20, 23, 22. Find the mode.
Given data: 20, 22, 20, 25, 20, 23, 22
Frequency: 20 (3 times), 22 (2 times), 25 (1 time), 23 (1 time)
Answer: 20 guests
Example 2: A SACCO notes monthly loan applications: 5, 7, 5, 8, 7, 7, 6. Find the mode.
Given data: 5, 7, 5, 8, 7, 7, 6
Frequency: 5 (2 times), 7 (3 times), 8 (1 time), 6 (1 time)
Answer: 7 applications
Example 3: A retail shop tracks sales of product units: 10, 12, 15, 12, 10, 12, 14. Find the mode.
Given data: 10, 12, 15, 12, 10, 12, 14
Frequency: 10 (2 times), 12 (3 times), 15 (1 time), 14 (1 time)
Answer: 12 units
Example 4: A county office records daily visitors: 30, 35, 30, 40, 35, 35, 30. Find the mode.
Given data: 30, 35, 30, 40, 35, 35, 30
Frequency: 30 (3 times), 35 (3 times), 40 (1 time)
Answer: 30 and 35 (bimodal)
Example 5: A farm records number of animals sold weekly: 8, 9, 8, 10, 8, 9, 9. Find the mode.
Given data: 8, 9, 8, 10, 8, 9, 9
Frequency: 8 (3 times), 9 (3 times), 10 (1 time)
Answer: 8 and 9 (bimodal)
The median is the middle value in an ordered dataset, separating the higher half from the lower half. It is a robust measure of central tendency especially useful when data contains outliers, such as income surveys at county government offices.
To find median:
Example 1: Find the median of daily sales (Ksh) for 5 days: 15000, 12000, 17000, 13000, 16000.
Given data: 15000, 12000, 17000, 13000, 16000
Sort: 12000, 13000, 15000, 16000, 17000
Number of data points \(n=5\) (odd)
Median position = \( \frac{5+1}{2} = 3 \)
Median = 3rd value = 15000 Ksh
Answer: 15000 Ksh
Example 2: Find the median patient stay (days) for 6 patients: 3, 5, 2, 4, 6, 7.
Given data: 3, 5, 2, 4, 6, 7
Sort: 2, 3, 4, 5, 6, 7
Number \(n=6\) (even)
Median position = average of 3rd and 4th values
Median = \( \frac{4 + 5}{2} = 4.5 \) days
Answer: 4.5 days
Example 3: Find the median loan amount (Ksh) of 7 clients: 40000, 60000, 50000, 70000, 55000, 65000, 45000.
Given data: 40000, 60000, 50000, 70000, 55000, 65000, 45000
Sort: 40000, 45000, 50000, 55000, 60000, 65000, 70000
\(n=7\) (odd)
Median position = \( \frac{7+1}{2} = 4 \)
Median = 4th value = 55000 Ksh
Answer: 55000 Ksh
Example 4: Find the median number of books borrowed per month by 8 students: 5, 7, 6, 8, 7, 9, 10, 6.
Given data: 5, 7, 6, 8, 7, 9, 10, 6
Sort: 5, 6, 6, 7, 7, 8, 9, 10
\(n=8\) (even)
Median position = average of 4th and 5th values
Median = \( \frac{7 + 7}{2} = 7 \)
Answer: 7 books
Example 5: Find the median daily energy consumption (kWh) for 9 days: 12, 15, 14, 13, 16, 18, 17, 19, 20.
Given data: 12, 15, 14, 13, 16, 18, 17, 19, 20
Sort: 12, 13, 14, 15, 16, 17, 18, 19, 20
\(n=9\) (odd)
Median position = \( \frac{9+1}{2} = 5 \)
Median = 5th value = 16 kWh
Answer: 16 kWh
Create a free account to open more of this chapter.
Free: practical guides, quick cards, workplace scenarios and more.
Create a free accountThis chapter covered fundamental statistical concepts starting with measures of central tendency, including mean, mode, and median, which summarize data sets by indicating central values. It then explored measures of dispersion that describe the spread or variability of data around the central tendency. Variance and standard deviation were introduced as key numerical indicators of this variability. The chapter defined probability as a measure of the likelihood of events and explained the basic laws of probability governing event combinations. It proceeded to discuss expectation, variance, and standard deviation in the context of random variables, providing tools for analyzing uncertain outcomes. Calculations involving both discrete and continuous random variables demonstrated practical applications of these concepts. The chapter examined important types of probability distributions such as binomial, Poisson, and normal distributions, detailing their characteristics. Finally, it addressed how to determine the mean, variance, and standard deviation of these distributions and highlighted their applications in engineering problem solving.
A county hospital recorded the daily number of patients visiting its outpatient department over 5 days as follows: 40, 50, 45, 55, and 60. Calculate the mean number of patients per day. (2 marks)
A retail business collected data on the number of products sold per hour over 8 hours: 15, 18, 20, 15, 22, 20, 18, 20. Determine the mode of the products sold per hour. (2 marks)
At the start of this chapter we promised you would be able to:
Tick each one you can genuinely do.
So, are you there yet?
You're competent when you can confidently do 50% or more of what this chapter promised.
Sign in to record how you're doing.