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This chapter introduces key statistical and probability concepts essential for land survey professionals in Kenya. Accurate interpretation of survey data relies heavily on understanding measures of central tendency and variability, which help summarize and analyze the distribution of measurements such as distances, angles, and elevations. Mastery of these statistical tools enables surveyors to assess data quality, detect anomalies, and make informed decisions in fieldwork and office computations.
Measures of central tendency summarize a data set by identifying a central or typical value. In land surveying, these measures help condense numerous observations into a representative value, aiding in error analysis and data adjustment. The main measures are mean, median, and mode, each capturing centrality differently depending on data distribution characteristics.
The mean or average is the sum of all data values divided by the number of values. It is widely used in surveying to find the average of repeated measurements, such as multiple readings of a baseline length or elevation point.
$$\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$$
Example 1: A surveyor records the following distances (in meters) between two points: 100.2, 99.8, 100.5, 100.1, 99.9. Find the mean distance.
Given: \(x_1=100.2\), \(x_2=99.8\), \(x_3=100.5\), \(x_4=100.1\), \(x_5=99.9\), \(n=5\)
$$\bar{x} = \frac{100.2 + 99.8 + 100.5 + 100.1 + 99.9}{5}$$
$$\bar{x} = \frac{500.5}{5}$$
$$\bar{x} = 100.1 \text{ m}$$
Answer: 100.1 m
Example 2: A survey team measures the height of a benchmark five times: 150.3 m, 150.1 m, 150.4 m, 150.2 m, 150.5 m. Calculate the mean height.
Given: \(x_1=150.3\), \(x_2=150.1\), \(x_3=150.4\), \(x_4=150.2\), \(x_5=150.5\), \(n=5\)
$$\bar{x} = \frac{150.3 + 150.1 + 150.4 + 150.2 + 150.5}{5}$$
$$\bar{x} = \frac{751.5}{5}$$
$$\bar{x} = 150.3 \text{ m}$$
Answer: 150.3 m
Example 3: A surveyor measures the angle between two roads three times: 45° 12', 45° 15', 45° 10'. Convert to decimal degrees and find the mean angle.
Given: \(x_1=45 + \frac{12}{60} = 45.2^\circ\), \(x_2=45 + \frac{15}{60} = 45.25^\circ\), \(x_3=45 + \frac{10}{60} = 45.1667^\circ\), \(n=3\)
$$\bar{x} = \frac{45.2 + 45.25 + 45.1667}{3}$$
$$\bar{x} = \frac{135.6167}{3}$$
$$\bar{x} = 45.2056^\circ$$
Convert back to degrees and minutes:
$$0.2056 \times 60 = 12.34'$$
Answer: 45° 12.3'
Example 4: A leveling survey records the following reduced levels (RLs) at a point: 102.45 m, 102.50 m, 102.48 m, 102.46 m, 102.49 m, 102.47 m. Find the mean RL.
Given: \(n=6\), data as above.
$$\bar{x} = \frac{102.45 + 102.50 + 102.48 + 102.46 + 102.49 + 102.47}{6}$$
$$\bar{x} = \frac{614.85}{6}$$
$$\bar{x} = 102.475 \text{ m}$$
Answer: 102.475 m
Example 5: A surveyor records the time taken (in seconds) to set up equipment five times: 120, 125, 118, 122, 121. Calculate the mean setup time.
Given: \(n=5\), data as above.
$$\bar{x} = \frac{120 + 125 + 118 + 122 + 121}{5}$$
$$\bar{x} = \frac{606}{5}$$
$$\bar{x} = 121.2 \text{ s}$$
Answer: 121.2 s
The median is the middle value in an ordered data set and is useful when data contain outliers or are skewed, which often happens with survey errors or unusual terrain measurements.
Example 1: Find the median of the distances (m): 99.8, 100.2, 100.5, 100.1, 99.9.
Ordered: 99.8, 99.9, 100.1, 100.2, 100.5
Median is the 3rd value (middle):
Answer: 100.1 m
Example 2: Find the median of the RLs (m): 102.45, 102.50, 102.48, 102.46, 102.49, 102.47.
Ordered: 102.45, 102.46, 102.47, 102.48, 102.49, 102.50
Median is average of 3rd and 4th values:
$$\frac{102.47 + 102.48}{2} = \frac{204.95}{2} = 102.475 \text{ m}$$
Answer: 102.475 m
Example 3: A surveyor measures angles (degrees): 45.2, 45.25, 44.8, 45.15, 44.9.
Ordered: 44.8, 44.9, 45.15, 45.2, 45.25
Median is 3rd value:
Answer: 45.15°
Example 4: Find the median of the setup times (seconds): 120, 125, 118, 122.
Ordered: 118, 120, 122, 125
Median is average of 2nd and 3rd:
$$\frac{120 + 122}{2} = 121 \text{ s}$$
Answer: 121 s
Example 5: Find the median of the following distances (m): 100.5, 100.7, 100.2, 100.8, 100.3, 100.1, 100.4.
Ordered: 100.1, 100.2, 100.3, 100.4, 100.5, 100.7, 100.8
Median is the 4th value:
Answer: 100.4 m
The mode is the most frequently occurring value in a data set. It is useful in surveys when repeated measurements or common values appear, such as typical soil types or frequently measured angles.
Example 1: Find the mode of these distances (m): 100.1, 100.2, 100.1, 100.3, 100.1.
The value 100.1 appears 3 times, others once.
Answer: 100.1 m
Example 2: Mode of RLs (m): 102.45, 102.50, 102.48, 102.50, 102.49, 102.50.
102.50 appears 3 times.
Answer: 102.50 m
Example 3: Angles (degrees): 45.25, 45.30, 45.25, 45.20, 45.25.
45.25 appears 3 times.
Answer: 45.25°
Example 4: Setup times (seconds): 120, 122, 121, 120, 123, 120.
120 appears 3 times.
Answer: 120 s
Example 5: Soil types coded as numbers: 1, 2, 2, 3, 2, 1, 2.
2 appears 4 times.
Answer: 2 (soil type code)
Standard deviation measures the spread of data around the mean, indicating precision and consistency of survey measurements. Smaller standard deviation means higher precision.
Formula for sample standard deviation:
$$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}$$
Example 1: Find standard deviation of distances (m): 100.2, 99.8, 100.5, 100.1, 99.9.
Given: \(n=5\)
Step 1: Find mean \(\bar{x}\):
$$\bar{x} = \frac{100.2 + 99.8 + 100.5 + 100.1 + 99.9}{5} = 100.1 \text{ m}$$
Step 2: Calculate squared deviations:
\((100.2 - 100.1)^2 = 0.01^2 = 0.01\)
\((99.8 - 100.1)^2 = (-0.3)^2 = 0.09\)
\((100.5 - 100.1)^2 = 0.4^2 = 0.16\)
\((100.1 - 100.1)^2 = 0^2 = 0\)
\((99.9 - 100.1)^2 = (-0.2)^2 = 0.04\)
Sum:
$$0.01 + 0.09 + 0.16 + 0 + 0.04 = 0.3$$
Step 3: Divide by \(n-1 = 4\):
$$\frac{0.3}{4} = 0.075$$
Step 4: Square root:
$$s = \sqrt{0.075} = 0.274 \text{ m}$$
Answer: 0.274 m
Example 2: RLs (m): 102.45, 102.50, 102.48, 102.46, 102.49, 102.47.
\(n=6\), mean \(\bar{x}=102.475\)
Squared deviations:
\((102.45 - 102.475)^2 = (-0.025)^2 = 0.000625\)
\((102.50 - 102.475)^2 = 0.025^2 = 0.000625\)
\((102.48 - 102.475)^2 = 0.005^2 = 0.000025\)
\((102.46 - 102.475)^2 = (-0.015)^2 = 0.000225\)
\((102.49 - 102.475)^2 = 0.015^2 = 0.000225\)
\((102.47 - 102.475)^2 = (-0.005)^2 = 0.000025\)
Sum:
$$0.000625 + 0.000625 + 0.000025 + 0.000225 + 0.000225 + 0.000025 = 0.00175$$
Divide by \(n-1=5\):
$$\frac{0.00175}{5} = 0.00035$$
Square root:
$$s = \sqrt{0.00035} = 0.0187 \text{ m}$$
Answer: 0.0187 m
Example 3: Angles (degrees): 45.2, 45.25, 45.15, 45.3, 45.1.
\(n=5\), mean \(\bar{x} = \frac{45.2 + 45.25 + 45.15 + 45.3 + 45.1}{5} = 45.2^\circ\)
Squared deviations:
\((45.2 - 45.2)^2 = 0\)
\((45.25 - 45.2)^2 = 0.05^2 = 0.0025\)
\((45.15 - 45.2)^2 = (-0.05)^2 = 0.0025\)
\((45.3 - 45.2)^2 = 0.1^2 = 0.01\)
\((45.1 - 45.2)^2 = (-0.1)^2 = 0.01\)
Sum:
$$0 + 0.0025 + 0.0025 + 0.01 + 0.01 = 0.025$$
Divide by \(n-1=4\):
$$\frac{0.025}{4} = 0.00625$$
Square root:
$$s = \sqrt{0.00625} = 0.0791^\circ$$
Answer: 0.0791°
Example 4: Setup times (s): 120, 125, 118, 122, 121.
\(n=5\), mean \(\bar{x} = \frac{120+125+118+122+121}{5} = 121.2\)
Squared deviations:
\((120 - 121.2)^2 = (-1.2)^2=1.44\)
\((125 - 121.2)^2 = 3.8^2=14.44\)
\((118 - 121.2)^2 = (-3.2)^2=10.24\)
\((122 - 121.2)^2 = 0.8^2=0.64\)
\((121 - 121.2)^2 = (-0.2)^2=0.04\)
Sum:
$$1.44 + 14.44 + 10.24 + 0.64 + 0.04 = 26.8$$
Divide by \(n-1=4\):
$$\frac{26.8}{4} = 6.7$$
Square root:
$$s = \sqrt{6.7} = 2.588 \text{ s}$$
Answer: 2.588 s
Example 5: Distances (m): 100.5, 100.7, 100.2, 100.8, 100.3, 100.1, 100.4.
\(n=7\), mean:
$$\bar{x} = \frac{100.5 + 100.7 + 100.2 + 100.8 + 100.3 + 100.1 + 100.4}{7} = \frac{703.0}{7} = 100.4286 \text{ m}$$
Squared deviations:
\((100.5 - 100.4286)^2 = 0.0714^2 = 0.0051\)
\((100.7 - 100.4286)^2 = 0.2714^2 = 0.0737\)
\((100.2 - 100.4286)^2 = (-0.2286)^2 = 0.0523\)
\((100.8 - 100.4286)^2 = 0.3714^2 = 0.1380\)
\((100.3 - 100.4286)^2 = (-0.1286)^2 = 0.0165\)
\((100.1 - 100.4286)^2 = (-0.3286)^2 = 0.1070\)
\((100.4 - 100.4286)^2 = (-0.0286)^2 = 0.0008\)
Sum:
$$0.0051 + 0.0737 + 0.0523 + 0.1380 + 0.0165 + 0.1070 + 0.0008 = 0.3934$$
Divide by \(n-1=6\):
$$\frac{0.3934}{6} = 0.0656$$
Square root:
$$s = \sqrt{0.0656} = 0.256 \text{ m}$$
Answer: 0.256 m
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Create a free accountThis chapter covered essential statistical tools starting with measures of central tendency, including the calculation and interpretation of mean, median, mode, and standard deviation to summarize data sets. It distinguished between grouped and ungrouped data, explaining how to organize and analyze each type effectively. The discussion then advanced to probability distributions, detailing the characteristics and applications of binomial, Poisson, normal, chi-square, and t-distributions in engineering contexts. Each distribution was explored for its relevance in modeling different types of random phenomena. Finally, the chapter introduced confidence intervals as a method for estimating population parameters with a specified level of certainty, highlighting their importance in decision-making and quality control. Together, these topics provide a comprehensive foundation for applying statistics and probability in engineering analysis and problem-solving.
A land surveyor recorded the lengths (in metres) of 5 consecutive boundary lines of a plot as follows: 12, 15, 14, 17, and 16. Calculate the mean length of the boundary lines. (2 marks)
The following elevation readings (in metres) were taken at 7 survey points: 102.5, 103.0, 101.8, 102.2, 103.7, 102.9, 102.1. Find the median elevation reading. (3 marks)
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