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Engineering mathematics forms the backbone of accurate land surveying by providing tools to analyze, interpret, and predict data behavior. In land survey projects across Kenya, understanding statistical measures and probability is essential for managing measurement errors, analyzing terrain data, and making informed decisions. This chapter focuses on applying statistics and probability to land survey data, starting with measures of central tendency and variability that summarize and describe survey observations effectively.
In land surveying, summarizing data sets such as elevation points, distances, or angles is crucial for making sense of measurements. Measures of central tendency provide representative values that describe the typical or central point of a dataset. These measures help surveyors identify average conditions, detect outliers, and guide adjustments during data processing.
The mean or arithmetic average is the sum of all observations divided by the number of observations. It is widely used in land surveying to estimate average elevations, distances, or angles from multiple measurements.
$$\text{Mean} = \bar{x} = \frac{\sum_{i=1}^n x_i}{n}$$
where \(x_i\) are the individual observations and \(n\) is the total number of observations.
Example 1: A surveyor measures the elevation of a point five times: 102.4 m, 102.7 m, 102.5 m, 102.6 m, 102.3 m. Find the mean elevation.
Given: \(x_1 = 102.4\) m, \(x_2 = 102.7\) m, \(x_3 = 102.5\) m, \(x_4 = 102.6\) m, \(x_5 = 102.3\) m, \(n=5\)
$$\bar{x} = \frac{102.4 + 102.7 + 102.5 + 102.6 + 102.3}{5}$$
$$= \frac{512.5}{5}$$
$$= 102.5 \text{ m}$$
Answer: 102.5 m
Example 2: A land surveyor records distances between benchmarks as: 150.2 m, 149.8 m, 150.0 m, 150.1 m, 149.9 m, 150.3 m. Calculate the mean distance.
Given: \(x_1=150.2\) m, \(x_2=149.8\) m, \(x_3=150.0\) m, \(x_4=150.1\) m, \(x_5=149.9\) m, \(x_6=150.3\) m, \(n=6\)
$$\bar{x} = \frac{150.2 + 149.8 + 150.0 + 150.1 + 149.9 + 150.3}{6}$$
$$= \frac{900.3}{6}$$
$$= 150.05 \text{ m}$$
Answer: 150.05 m
Example 3: In a leveling survey, the following reduced levels (RL) were recorded: 98.75 m, 98.82 m, 98.78 m, 98.80 m, 98.79 m, 98.76 m, 98.77 m. Find the mean RL.
Given: \(x_1=98.75\) m, \(x_2=98.82\) m, \(x_3=98.78\) m, \(x_4=98.80\) m, \(x_5=98.79\) m, \(x_6=98.76\) m, \(x_7=98.77\) m, \(n=7\)
$$\bar{x} = \frac{98.75 + 98.82 + 98.78 + 98.80 + 98.79 + 98.76 + 98.77}{7}$$
$$= \frac{691.47}{7}$$
$$= 98.7814 \text{ m}$$
Answer: 98.78 m
Example 4: A surveyor records horizontal angles in degrees as: 45.2°, 45.5°, 44.9°, 45.1°, 45.3°. Compute the mean horizontal angle.
Given: \(x_1=45.2^\circ\), \(x_2=45.5^\circ\), \(x_3=44.9^\circ\), \(x_4=45.1^\circ\), \(x_5=45.3^\circ\), \(n=5\)
$$\bar{x} = \frac{45.2 + 45.5 + 44.9 + 45.1 + 45.3}{5}$$
$$= \frac{225.9}{5}$$
$$= 45.18^\circ$$
Answer: 45.18°
Example 5: A surveyor measures the thickness of a soil layer at eight points: 1.25 m, 1.27 m, 1.23 m, 1.26 m, 1.24 m, 1.28 m, 1.25 m, 1.26 m. Find the mean thickness.
Given: \(x_1=1.25\) m, \(x_2=1.27\) m, \(x_3=1.23\) m, \(x_4=1.26\) m, \(x_5=1.24\) m, \(x_6=1.28\) m, \(x_7=1.25\) m, \(x_8=1.26\) m, \(n=8\)
$$\bar{x} = \frac{1.25 + 1.27 + 1.23 + 1.26 + 1.24 + 1.28 + 1.25 + 1.26}{8}$$
$$= \frac{10.04}{8}$$
$$= 1.255 \text{ m}$$
Answer: 1.255 m
The median is the middle value of a data set when arranged in ascending or descending order. It is useful in land surveying when the data contains outliers or skewed values, providing a better central value than the mean.
Example 1: Find the median of the elevations (m): 102.4, 102.7, 102.5, 102.6, 102.3.
Given: \(n=5\), data sorted: 102.3, 102.4, 102.5, 102.6, 102.7
Since \(n\) is odd, median is the middle value:
$$\text{Median} = x_{\frac{n+1}{2}} = x_3 = 102.5 \text{ m}$$
Answer: 102.5 m
Example 2: Determine the median distance from measurements (m): 150.2, 149.8, 150.0, 150.1, 149.9, 150.3.
Given: \(n=6\), sorted data: 149.8, 149.9, 150.0, 150.1, 150.2, 150.3
Median is average of 3rd and 4th values:
$$\text{Median} = \frac{150.0 + 150.1}{2} = \frac{300.1}{2} = 150.05 \text{ m}$$
Answer: 150.05 m
Example 3: Find the median RL from: 98.75, 98.82, 98.78, 98.80, 98.79, 98.76, 98.77.
Sorted data: 98.75, 98.76, 98.77, 98.78, 98.79, 98.80, 98.82
\(n=7\), median is the 4th value:
$$\text{Median} = x_4 = 98.78 \text{ m}$$
Answer: 98.78 m
Example 4: Find the median horizontal angle (degrees): 45.2, 45.5, 44.9, 45.1, 45.3.
Sorted: 44.9, 45.1, 45.2, 45.3, 45.5
Median is 3rd value:
$$\text{Median} = 45.2^\circ$$
Answer: 45.2°
Example 5: A soil thickness dataset (m): 1.25, 1.27, 1.23, 1.26, 1.24, 1.28, 1.25, 1.26.
Sorted: 1.23, 1.24, 1.25, 1.25, 1.26, 1.26, 1.27, 1.28
\(n=8\), median is average of 4th and 5th values:
$$\text{Median} = \frac{1.25 + 1.26}{2} = \frac{2.51}{2} = 1.255 \text{ m}$$
Answer: 1.255 m
The mode is the value that appears most frequently in a dataset. It is useful in land surveying to identify repeated measurements or common categories, such as the most frequent soil type or common elevation.
Example 1: Find the mode of elevations (m): 102.4, 102.7, 102.5, 102.6, 102.3, 102.5.
Values and frequencies:
Mode is 102.5 m (appears twice).
Answer: 102.5 m
Example 2: Determine the mode of distances (m): 150.2, 149.8, 150.0, 150.1, 149.9, 150.3, 150.0.
Frequencies:
Mode is 150.0 m.
Answer: 150.0 m
Example 3: Find the mode RL (m): 98.75, 98.82, 98.78, 98.80, 98.79, 98.76, 98.77, 98.78.
Frequencies:
Mode is 98.78 m.
Answer: 98.78 m
Example 4: Find the mode horizontal angle (degrees): 45.2, 45.5, 44.9, 45.1, 45.3, 45.1.
Frequencies:
Mode is 45.1°.
Answer: 45.1°
Example 5: Find the mode soil thickness (m): 1.25, 1.27, 1.23, 1.26, 1.24, 1.28, 1.25, 1.26, 1.25.
Frequencies:
Mode is 1.25 m.
Answer: 1.25 m
The standard deviation measures the spread or dispersion of data around the mean. It quantifies variability, which is critical in land surveying to assess measurement precision and reliability.
$$s = \sqrt{\frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n-1}}$$
where \(s\) is the sample standard deviation, \(x_i\) are observations, \(\bar{x}\) is the mean, and \(n\) is the number of observations.
Example 1: Calculate the standard deviation of elevations (m): 102.4, 102.7, 102.5, 102.6, 102.3.
Given: \(n=5\), \(\bar{x} = 102.5\) m (from previous example)
Calculate squared deviations:
\((102.4 - 102.5)^2 = (-0.1)^2 = 0.01\)
\((102.7 - 102.5)^2 = (0.2)^2 = 0.04\)
\((102.5 - 102.5)^2 = 0^2 = 0\)
\((102.6 - 102.5)^2 = (0.1)^2 = 0.01\)
\((102.3 - 102.5)^2 = (-0.2)^2 = 0.04\)
Sum of squared deviations:
$$0.01 + 0.04 + 0 + 0.01 + 0.04 = 0.10$$
Standard deviation:
$$s = \sqrt{\frac{0.10}{5-1}} = \sqrt{\frac{0.10}{4}} = \sqrt{0.025} = 0.1581 \text{ m}$$
Answer: 0.158 m
Example 2: Compute standard deviation of distances (m): 150.2, 149.8, 150.0, 150.1, 149.9, 150.3.
Given: \(n=6\), \(\bar{x} = 150.05\) m
Squared deviations:
\((150.2 - 150.05)^2 = (0.15)^2 = 0.0225\)
\((149.8 - 150.05)^2 = (-0.25)^2 = 0.0625\)
\((150.0 - 150.05)^2 = (-0.05)^2 = 0.0025\)
\((150.1 - 150.05)^2 = (0.05)^2 = 0.0025\)
\((149.9 - 150.05)^2 = (-0.15)^2 = 0.0225\)
\((150.3 - 150.05)^2 = (0.25)^2 = 0.0625\)
Sum:
$$0.0225 + 0.0625 + 0.0025 + 0.0025 + 0.0225 + 0.0625 = 0.175$$
Standard deviation:
$$s = \sqrt{\frac{0.175}{6-1}} = \sqrt{\frac{0.175}{5}} = \sqrt{0.035} = 0.1871 \text{ m}$$
Answer: 0.187 m
Example 3: Calculate standard deviation of RLs (m): 98.75, 98.82, 98.78, 98.80, 98.79, 98.76, 98.77.
Given: \(n=7\), \(\bar{x} = 98.7814\) m
Squared deviations:
\((98.75 - 98.7814)^2 = (-0.0314)^2 = 0.00098596\)
\((98.82 - 98.7814)^2 = (0.0386)^2 = 0.00148996\)
\((98.78 - 98.7814)^2 = (-0.0014)^2 = 0.00000196\)
\((98.80 - 98.7814)^2 = (0.0186)^2 = 0.00034596\)
\((98.79 - 98.7814)^2 = (0.0086)^2 = 0.00007396\)
\((98.76 - 98.7814)^2 = (-0.0214)^2 = 0.00045796\)
\((98.77 - 98.7814)^2 = (-0.0114)^2 = 0.00012996\)
Sum:
$$0.00098596 + 0.00148996 + 0.00000196 + 0.00034596 + 0.00007396 + 0.00045796 + 0.00012996 = 0.00348568$$
Standard deviation:
$$s = \sqrt{\frac{0.00348568}{7-1}} = \sqrt{\frac{0.00348568}{6}} = \sqrt{0.00058095} = 0.0241 \text{ m}$$
Answer: 0.024 m
Example 4: Calculate standard deviation of horizontal angles (degrees): 45.2, 45.5, 44.9, 45.1, 45.3.
Given: \(n=5\), \(\bar{x} = 45.18^\circ\)
Squared deviations:
\((45.2 - 45.18)^2 = (0.02)^2 = 0.0004\)
\((45.5 - 45.18)^2 = (0.32)^2 = 0.1024\)
\((44.9 - 45.18)^2 = (-0.28)^2 = 0.0784\)
\((45.1 - 45.18)^2 = (-0.08)^2 = 0.0064\)
\((45.3 - 45.18)^2 = (0.12)^2 = 0.0144\)
Sum:
$$0.0004 + 0.1024 + 0.0784 + 0.0064 + 0.0144 = 0.202$$
Standard deviation:
$$s = \sqrt{\frac{0.202}{5-1}} = \sqrt{\frac{0.202}{4}} = \sqrt{0.0505} = 0.2247^\circ$$
Answer: 0.225°
Example 5: Calculate standard deviation of soil thickness (m): 1.25, 1.27, 1.23, 1.26, 1.24, 1.28, 1.25, 1.26.
Given: \(n=8\), \(\bar{x} = 1.255\) m
Squared deviations:
\((1.25 - 1.255)^2 = (-0.005)^2 = 0.000025\)
\((1.27 - 1.255)^2 = (0.015)^2 = 0.000225\)
\((1.23 - 1.255)^2 = (-0.025)^2 = 0.000625\)
\((1.26 - 1.255)^2 = (0.005)^2 = 0.000025\)
\((1.24 - 1.255)^2 = (-0.015)^2 = 0.000225\)
\((1.28 - 1.255)^2 = (0.025)^2 = 0.000625\)
\((1.25 - 1.255)^2 = (-0.005)^2 = 0.000025\)
\((1.26 - 1.255)^2 = (0.005)^2 = 0.000025\)
Sum:
$$0.000025 + 0.000225 + 0.000625 + 0.000025 + 0.000225 + 0.000625 + 0.000025 + 0.000025 = 0.0018$$
Standard deviation:
$$s = \sqrt{\frac{0.0018}{8-1}} = \sqrt{\frac{0.0018}{7}} = \sqrt{0.000257} = 0.0160 \text{ m}$$
Answer: 0.016 m
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Create a free accountThis chapter covered key statistical tools used in engineering analysis, beginning with measures of central tendency including mean, median, and mode, which describe the central point of data sets. It also introduced standard deviation as a measure of data dispersion around the mean, essential for understanding variability. The distinction between grouped and ungrouped data was explained, focusing on how to organize and analyze written test scores in both formats. Probability distributions were explored to model the likelihood of different outcomes in engineering contexts. Finally, the chapter examined confidence intervals, providing methods to estimate population parameters with a specified level of certainty. Together, these concepts equip students to analyze data rigorously and make informed decisions based on statistical evidence.
A land surveyor recorded the following distances (in metres) between survey markers: 12, 15, 14, 13, 16. Calculate the mean distance. (2 marks)
The following elevation readings (in metres) were recorded along a survey line: 102, 105, 101, 107, 106, 104. Find the median elevation. (2 marks)
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