Statistics and probability are essential tools in agricultural engineering for making informed decisions based on data collected from farms, soil tests, crop yields, and machinery performance. This chapter focuses on methods to organize, summarize, and interpret data relevant to agricultural engineering projects in Kenya. Understanding how to present data effectively enables clearer communication of findings to stakeholders such as farmers, extension officers, and agricultural equipment manufacturers. Accurate data presentation also forms the basis for further statistical analysis and probability calculations.

5.1 Data Presentation

Data presentation is the first step in analyzing agricultural engineering data. Proper organization of data collected from field measurements, equipment tests, or survey responses ensures clarity and facilitates interpretation. Different types of data require different presentation methods depending on whether the data are continuous measurements like soil moisture or discrete counts like number of tractors.

5.1.1 Continuous Variables

Continuous variables are measurable quantities that can take any value within a range. In agricultural engineering, examples include soil pH, moisture content, temperature, and crop height. Such variables are recorded with decimal precision and often require grouping for meaningful analysis.

The primary formula for continuous data grouping is the class interval width:

$$ \text{Class interval width} = \frac{\text{Range}}{\text{Number of classes}} $$

where Range = Maximum value - Minimum value.

Worked Examples

Example 1: Soil moisture content was measured in 10 soil samples with values (in %): 12.5, 15.2, 13.8, 14.6, 16.0, 12.9, 15.5, 14.3, 13.0, 15.8. Find the range and class interval width if 4 classes are desired.

Given:
Soil moisture values = 12.5, 15.2, 13.8, 14.6, 16.0, 12.9, 15.5, 14.3, 13.0, 15.8 (%)
Number of classes = 4

Calculate Range:$$ \text{Range} = 16.0 - 12.5 = 3.5 $$

Calculate class interval width:$$ \text{Class interval width} = \frac{3.5}{4} = 0.875 $$

Answer: Range = 3.5%, Class interval width = 0.875%

Example 2: The height of maize plants in a field varies from 80 cm to 140 cm. If data are grouped into 6 classes, find the class interval width.

Given:
Minimum height = 80 cm
Maximum height = 140 cm
Number of classes = 6

Calculate Range:$$ 140 - 80 = 60 $$

Calculate class interval width:$$ \frac{60}{6} = 10 $$

Answer: Class interval width = 10 cm

Example 3: A tractor fuel efficiency test recorded values between 3.2 and 7.8 liters per hour. If 5 class intervals are to be used, calculate the class width.

Given:
Minimum = 3.2 L/hr
Maximum = 7.8 L/hr
Classes = 5

Range:$$ 7.8 - 3.2 = 4.6 $$

Class interval width:$$ \frac{4.6}{5} = 0.92 $$

Answer: Class interval width = 0.92 L/hr

Example 4: Temperature readings from a greenhouse range from 18.4°C to 28.9°C. For 8 classes, find the interval width.

Given:
Min = 18.4°C
Max = 28.9°C
Classes = 8

Range:$$ 28.9 - 18.4 = 10.5 $$

Class width:$$ \frac{10.5}{8} = 1.3125 $$

Answer: Class interval width = 1.3125°C

Example 5: Soil bulk density varies from 1.1 to 1.6 g/cm³. Group into 5 classes and find the class interval width.

Given:
Min = 1.1 g/cm³
Max = 1.6 g/cm³
Classes = 5

Range:$$ 1.6 - 1.1 = 0.5 $$

Class width:$$ \frac{0.5}{5} = 0.1 $$

Answer: Class interval width = 0.1 g/cm³

5.1.2 Discrete Variables

Discrete variables represent countable items that take only integer values. In agricultural engineering, examples include number of irrigation pumps, number of tractors, or defects in machinery. Data presentation for discrete variables often involves frequency tabulation.

The frequency of an observation is the number of times it occurs in the dataset.

Worked Examples

Example 1: Count the frequency of tractors in 5 farms: 2, 3, 1, 3, 2. Construct a frequency distribution table.

Given:
Tractors per farm = 2, 3, 1, 3, 2

Values: 1, 2, 3

Frequency:

• 1 tractor: count how many times 1 appears = 1
• 2 tractors: count = 2
• 3 tractors: count = 2

Answer: Frequency table as above.

Example 2: Number of irrigation pumps in 8 farms: 4, 5, 4, 3, 4, 5, 3, 5. Prepare a frequency distribution.

Given:
Pumps = 4, 5, 4, 3, 4, 5, 3, 5

Values: 3, 4, 5

Frequency:

• 3 pumps: 2
• 4 pumps: 3
• 5 pumps: 3

Answer: Frequency table as above.

Example 3: Count the number of defective parts in 10 machinery units: 0, 1, 2, 0, 1, 3, 2, 1, 0, 2.

Given:
Defects = 0, 1, 2, 0, 1, 3, 2, 1, 0, 2

Values: 0, 1, 2, 3

Frequency:

• 0 defects: 3
• 1 defect: 3
• 2 defects: 3
• 3 defects: 1

Answer: Frequency table as above.

Example 4: Number of water tanks in 6 farms: 1, 2, 2, 1, 3, 2. Prepare frequency distribution.

Given:
Tanks = 1, 2, 2, 1, 3, 2

Values: 1, 2, 3

Frequency:

• 1 tank: 2
• 2 tanks: 3
• 3 tanks: 1

Answer: Frequency table as above.

Example 5: Count number of failures in 7 irrigation systems: 0, 1, 1, 2, 0, 1, 0.

Given:
Failures = 0, 1, 1, 2, 0, 1, 0

Values: 0, 1, 2

Frequency:

• 0 failures: 3
• 1 failure: 3
• 2 failures: 1

Answer: Frequency table as above.

5.1.3 Grouped Data

Grouped data refers to continuous or discrete data organized into classes or intervals. Grouping simplifies large datasets and facilitates calculation of statistics like mean and standard deviation. Grouped frequency tables are common in agricultural experiments measuring crop yields or soil properties.

The midpoint of a class interval is calculated as:

$$ \text{Midpoint} = \frac{\text{Lower class limit} + \text{Upper class limit}}{2} $$

Worked Examples

Example 1: Group crop yield data (kg/plot) into classes and find midpoints:

Data: 12, 15, 14, 13, 16, 15, 14, 13
Classes: 12–13, 14–15, 16–17

Calculate midpoints:

• For 12–13:$$ \frac{12 + 13}{2} = 12.5 $$

• For 14–15:$$ \frac{14 + 15}{2} = 14.5 $$

• For 16–17:$$ \frac{16 + 17}{2} = 16.5 $$

Answer: Midpoints are 12.5, 14.5, 16.5 kg/plot.

Example 2: Soil pH readings grouped as 5.5–6.0, 6.1–6.5, 6.6–7.0. Find midpoints.

Given:
Classes = 5.5–6.0, 6.1–6.5, 6.6–7.0

Midpoints:

• 5.5–6.0:$$ \frac{5.5 + 6.0}{2} = 5.75 $$

• 6.1–6.5:$$ \frac{6.1 + 6.5}{2} = 6.3 $$

• 6.6–7.0:$$ \frac{6.6 + 7.0}{2} = 6.8 $$

Answer: Midpoints are 5.75, 6.3, 6.8 pH.

Example 3: Group tractor speeds (km/h) into 3 classes: 10–20, 21–30, 31–40. Calculate midpoints.

Given:
Classes = 10–20, 21–30, 31–40

Midpoints:

• 10–20:$$ \frac{10 + 20}{2} = 15 $$

• 21–30:$$ \frac{21 + 30}{2} = 25.5 $$

• 31–40:$$ \frac{31 + 40}{2} = 35.5 $$

Answer: Midpoints 15, 25.5, 35.5 km/h.

Example 4: Group water flow rates (L/s) into intervals 5–10, 11–15, 16–20. Find midpoints.

Given:
Classes = 5–10, 11–15, 16–20

Midpoints:

• 5–10:$$ \frac{5 + 10}{2} = 7.5 $$

• 11–15:$$ \frac{11 + 15}{2} = 13 $$

• 16–20:$$ \frac{16 + 20}{2} = 18 $$

Answer: Midpoints 7.5, 13, 18 L/s.

Example 5: Group soil bulk density (g/cm³) into 1.2–1.3, 1.4–1.5, 1.6–1.7. Find midpoints.

Given:
Classes = 1.2–1.3, 1.4–1.5, 1.6–1.7

Midpoints:

• 1.2–1.3:$$ \frac{1.2 + 1.3}{2} = 1.25 $$

• 1.4–1.5:$$ \frac{1.4 + 1.5}{2} = 1.45 $$

• 1.6–1.7:$$ \frac{1.6 + 1.7}{2} = 1.65 $$

Answer: Midpoints 1.25, 1.45, 1.65 g/cm³.

5.1.4 Ungrouped Data

Ungrouped data is raw data presented as individual values without classification. This form is common in small-sample agricultural engineering studies or when data is limited. It allows direct calculation of statistics such as mean and variance.

The mean of ungrouped data is calculated as:

$$ \bar{x} = \frac{\Sigma x_i}{n} $$

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

Worked Examples

Example 1: Calculate the mean of soil moisture values: 12.5, 15.2, 13.8, 14.6, 16.0.

Given:
Values = 12.5, 15.2, 13.8, 14.6, 16.0
Number of values, \(n = 5\)

Calculate sum:$$ 12.5 + 15.2 + 13.8 + 14.6 + 16.0 = 72.1 $$

Calculate mean:$$ \frac{72.1}{5} = 14.42 $$

Answer: Mean soil moisture = 14.42%

Example 2: Find the mean height of maize plants: 80, 90, 85, 95, 100 cm.

Given:
Values = 80, 90, 85, 95, 100
\(n = 5\)

Sum:$$ 80 + 90 + 85 + 95 + 100 = 450 $$

Mean:$$ \frac{450}{5} = 90 $$

Answer: Mean height = 90 cm

Example 3: Calculate the average number of irrigation pumps in 6 farms: 3, 4, 3, 5, 4, 3.

Given:
Values = 3, 4, 3, 5, 4, 3
\(n = 6\)

Sum:$$ 3 + 4 + 3 + 5 + 4 + 3 = 22 $$

Mean:$$ \frac{22}{6} \approx 3.67 $$

Answer: Mean number of pumps = 3.67

Example 4: Determine the average soil pH from readings: 6.2, 6.5, 6.8, 6.3, 6.7.

Given:
Values = 6.2, 6.5, 6.8, 6.3, 6.7
\(n = 5\)

Sum:$$ 6.2 + 6.5 + 6.8 + 6.3 + 6.7 = 32.5 $$

Mean:$$ \frac{32.5}{5} = 6.5 $$

Answer: Mean soil pH = 6.5

Example 5: Calculate mean fuel consumption (L/hr) for tractors: 4.2, 5.0, 4.8, 5.1, 4.6.

Given:
Values = 4.2, 5.0, 4.8, 5.1, 4.6
\(n = 5\)

Sum:$$ 4.2 + 5.0 + 4.8 + 5.1 + 4.6 = 23.7 $$

Mean:$$ \frac{23.7}{5} = 4.74 $$

Answer: Mean fuel consumption = 4.74 L/hr

Chapter Summary

This chapter covered the presentation of data using appropriate graphical methods for continuous and discrete variables, including histograms, line graphs, bar graphs, and pie charts. It explained how grouped and ungrouped data can be effectively displayed through histograms, bar graphs, cumulative frequency curves, and ogives. Measures of central tendency such as mean, mode, and median were discussed with techniques for calculating them from both grouped and ungrouped data sets. The chapter also explored measures of dispersion, focusing on standard deviation and variance, and how to compute these for different types of data. Probability concepts were introduced, distinguishing between experiments conducted with and without replacement. The study of probability distribution functions included detailed treatment of the binomial and Poisson distributions. Finally, the chapter explained the characteristics and applications of the normal distribution as a fundamental model in statistics.