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Compiled by: Elijah Ochieng
Applied statistics and probability are essential tools for agricultural engineers in Kenya, enabling them to analyze data related to crop yields, soil properties, machinery performance, and environmental factors. Accurate data presentation helps in making informed decisions, optimizing resource use, and improving agricultural productivity. This chapter focuses on the mathematical treatment of different types of data encountered in agricultural engineering, emphasizing clear and effective presentation methods.
Data presentation involves organizing and displaying collected data in a form suitable for analysis and interpretation. Agricultural engineers often deal with various data types such as measurements from soil tests, crop counts, or machinery operating parameters. Understanding how to present continuous, discrete, grouped, and ungrouped data mathematically is critical for effective data analysis and reporting.
Continuous variables represent measurements that can take any value within a range, such as soil moisture content, temperature, or crop height. These variables are fundamental in agricultural engineering for monitoring and controlling processes.
The mathematical representation of continuous data often involves frequency distributions and histograms. The frequency density \( f_d \) for a class interval is given by:
$$ f_d = \frac{f}{c} $$
where \( f \) is the frequency of the class and \( c \) is the class width.
Example 1: A soil moisture content test recorded the following frequencies in percentage intervals: 10–15% (5 samples), 15–20% (8 samples), 20–25% (12 samples). Calculate the frequency density for each interval.
Given:
Class intervals and frequencies as above.
Formula:$$f_d = \frac{f}{c}$$
For 10–15%:$$c = 15 - 10 = 5$$$$f_d = \frac{5}{5} = 1$$
For 15–20%:$$c = 20 - 15 = 5$$$$f_d = \frac{8}{5} = 1.6$$
For 20–25%:$$c = 25 - 20 = 5$$$$f_d = \frac{12}{5} = 2.4$$
Answer: Frequency densities are 1, 1.6, and 2.4 respectively.
Example 2: The heights of maize plants were recorded in the intervals 100–110 cm (7 plants), 110–120 cm (15 plants), 120–130 cm (18 plants), and 130–140 cm (10 plants). Find the frequency densities.
Given: Frequencies and class widths.
Formula:$$f_d = \frac{f}{c}$$
Calculate class width:$$c = 10 \text{ cm}$$ for all classes.
Calculate frequency densities:
For 100–110 cm:$$f_d = \frac{7}{10} = 0.7$$
For 110–120 cm:$$f_d = \frac{15}{10} = 1.5$$
For 120–130 cm:$$f_d = \frac{18}{10} = 1.8$$
For 130–140 cm:$$f_d = \frac{10}{10} = 1.0$$
Answer: Frequency densities are 0.7, 1.5, 1.8, and 1.0 respectively.
Example 3: A temperature sensor in a greenhouse recorded temperatures in intervals: 22–24°C (3 readings), 24–27°C (5 readings), 27–30°C (7 readings), and 30–33°C (5 readings). Calculate the frequency densities.
Given: Frequencies and class widths.
Calculate class widths:
Calculate frequency densities:
For 22–24°C:$$f_d = \frac{3}{2} = 1.5$$
For 24–27°C:$$f_d = \frac{5}{3} \approx 1.67$$
For 27–30°C:$$f_d = \frac{7}{3} \approx 2.33$$
For 30–33°C:$$f_d = \frac{5}{3} \approx 1.67$$
Answer: Frequency densities are approximately 1.5, 1.67, 2.33, and 1.67 respectively.
Discrete variables take specific countable values, such as the number of tractors in a farm yard or the number of pest occurrences in a plot. These variables are crucial in agricultural engineering for inventory, quality control, and monitoring.
The probability mass function (pmf) \( P(X = x_i) \) for discrete variables satisfies:
$$ \sum_i P(X = x_i) = 1 $$
and the expected value \( E(X) \) is:
$$ E(X) = \sum_i x_i P(X = x_i) $$
Example 1: A farm has 0, 1, 2, or 3 irrigation pumps with probabilities 0.1, 0.3, 0.4, and 0.2 respectively. Find the expected number of pumps.
Given:
\(x_i = 0, 1, 2, 3\)
\(P(X = x_i) = 0.1, 0.3, 0.4, 0.2\)
Formula:$$E(X) = \sum_i x_i P(X = x_i)$$
Calculate:$$E(X) = 0 \times 0.1 + 1 \times 0.3 + 2 \times 0.4 + 3 \times 0.2$$$$E(X) = 0 + 0.3 + 0.8 + 0.6 = 1.7$$
Answer: Expected number of pumps is 1.7.
Example 2: The number of daily pest attacks on a crop follows the distribution: 0 (0.5), 1 (0.3), 2 (0.15), 3 (0.05). Calculate the expected number of attacks.
Given:
\(x_i = 0,1,2,3\)
\(P = 0.5, 0.3, 0.15, 0.05\)
Calculate:$$E(X) = 0 \times 0.5 + 1 \times 0.3 + 2 \times 0.15 + 3 \times 0.05$$$$E(X) = 0 + 0.3 + 0.3 + 0.15 = 0.75$$
Answer: Expected daily pest attacks are 0.75.
Example 3: A tractor has 1 to 4 filters in use with probabilities 0.25, 0.35, 0.25, and 0.15 respectively. Find the expected number of filters.
Given:
\(x_i = 1,2,3,4\)
\(P = 0.25, 0.35, 0.25, 0.15\)
Calculate:$$E(X) = 1 \times 0.25 + 2 \times 0.35 + 3 \times 0.25 + 4 \times 0.15$$$$E(X) = 0.25 + 0.7 + 0.75 + 0.6 = 2.3$$
Answer: Expected number of filters is 2.3.
For Kenyan context, consider the following:
Example 4: The number of daily pest attacks on maize crops at Kenya Agricultural and Livestock Research Organization (KALRO) follows the distribution: 0 (0.5), 1 (0.3), 2 (0.15), 3 (0.05). Calculate the expected number of attacks.
Given: \(x_i = 0,1,2,3\) \(P = 0.5, 0.3, 0.15, 0.05\)
Calculate:$$E(X) = 0 \times 0.5 + 1 \times 0.3 + 2 \times 0.15 + 3 \times 0.05$$$$E(X) = 0 + 0.3 + 0.3 + 0.15 = 0.75$$
Answer: Expected daily pest attacks at KALRO are 0.75.
Grouped data arise when raw data are organized into intervals or classes, common when measuring continuous variables like rainfall amounts or grain sizes. Grouping simplifies large datasets and aids in identifying data patterns.
The midpoint \( x_m \) of each class is:
$$ x_m = \frac{\text{Lower Class Limit} + \text{Upper Class Limit}}{2} $$
The mean \( \bar{x} \) for grouped data is:
$$ \bar{x} = \frac{\sum f_i x_{m_i}}{\sum f_i} $$
where \( f_i \) is the frequency of class \( i \).
Example 1: Rainfall (mm) data grouped as follows: 0–10 (5 days), 10–20 (8 days), 20–30 (12 days), 30–40 (5 days). Calculate the mean rainfall.
Given:
Classes and frequencies as above.
Calculate midpoints:$$x_m = \frac{0+10}{2} = 5$$$$x_m = \frac{10+20}{2} = 15$$$$x_m = \frac{20+30}{2} = 25$$$$x_m = \frac{30+40}{2} = 35$$
Calculate numerator:$$\sum f_i x_{m_i} = 5 \times 5 + 8 \times 15 + 12 \times 25 + 5 \times 35$$$$= 25 + 120 + 300 + 175 = 620$$
Calculate denominator:$$\sum f_i = 5 + 8 + 12 + 5 = 30$$
Calculate mean:$$\bar{x} = \frac{620}{30} \approx 20.67 \text{ mm}$$
Answer: Mean rainfall is approximately 20.67 mm.
Example 2: Soil particle sizes (micrometers) grouped: 0–50 (10 samples), 50–100 (15 samples), 100–150 (5 samples). Find the mean particle size.
Given: Frequencies and classes.
Calculate midpoints:$$x_m = 25, 75, 125$$
Calculate numerator:$$10 \times 25 + 15 \times 75 + 5 \times 125 = 250 + 1125 + 625 = 2000$$
Calculate denominator:$$10 + 15 + 5 = 30$$
Calculate mean:$$\bar{x} = \frac{2000}{30} \approx 66.67 \text{ micrometers}$$
Answer: Mean particle size is approximately 66.67 micrometers.
Example 3: Grain yield (kg/ha) grouped: 1000–1500 (6 farms), 1500–2000 (10 farms), 2000–2500 (4 farms). Calculate the mean yield.
Calculate midpoints:$$x_m = 1250, 1750, 2250$$
Calculate numerator:$$6 \times 1250 + 10 \times 1750 + 4 \times 2250 = 7500 + 17500 + 9000 = 34000$$
Calculate denominator:$$6 + 10 + 4 = 20$$
Calculate mean:$$\bar{x} = \frac{34000}{20} = 1700 \text{ kg/ha}$$
Answer: Mean grain yield is 1700 kg/ha.
For Kenyan context, consider the following:
Example 4: Grain yield (kg/ha) grouped data from the National Cereals and Produce Board (NCPB): 1000–1500 (6 farms), 1500–2000 (10 farms), 2000–2500 (4 farms). Calculate the mean yield.
Calculate midpoints:$$x_m = 1250, 1750, 2250$$
Calculate numerator:$$6 \times 1250 + 10 \times 1750 + 4 \times 2250 = 7500 + 17500 + 9000 = 34000$$
Calculate denominator:$$6 + 10 + 4 = 20$$
Calculate mean:$$\bar{x} = \frac{34000}{20} = 1700 \text{ kg/ha}$$
Answer: Mean grain yield at NCPB is 1700 kg/ha.
Ungrouped data present individual observations without classification into intervals. These are typical in small sample studies such as measurements of plant heights or seed weights.
The mean \( \bar{x} \) of ungrouped data is:
$$ \bar{x} = \frac{\sum x_i}{n} $$
where \( x_i \) are the data points and \( n \) is the number of observations.
Example 1: Heights of five tomato plants (cm): 45, 50, 47, 52, 49. Find the mean height.
Given: Data points and \( n = 5 \).
Calculate sum:$$\sum x_i = 45 + 50 + 47 + 52 + 49 = 243$$
Calculate mean:$$\bar{x} = \frac{243}{5} = 48.6 \text{ cm}$$
Answer: Mean height is 48.6 cm.
Example 2: Weights of six seed samples (g): 12, 15, 14, 13, 16, 14. Calculate the mean weight.
Sum:$$12 + 15 + 14 + 13 + 16 + 14 = 84$$
Number of samples:$$n = 6$$
Mean:$$\bar{x} = \frac{84}{6} = 14 \text{ g}$$
Answer: Mean weight is 14 g.
Example 3: Daily milk yields (litres) from four cows: 8.5, 9.0, 7.8, 8.2. Find the average yield.
Sum:$$8.5 + 9.0 + 7.8 + 8.2 = 33.5$$
Number of observations:$$n = 4$$
Mean:$$\bar{x} = \frac{33.5}{4} = 8.375 \text{ litres}$$
Answer: Average milk yield is 8.375 litres.
For Kenyan context, consider the following:
Example 4: Daily milk yields (litres) from four cows at Brookside Dairy Ltd: 8.5, 9.0, 7.8, 8.2. Find the average yield.
Sum:$$8.5 + 9.0 + 7.8 + 8.2 = 33.5$$
Number of observations:$$n = 4$$
Mean:$$\bar{x} = \frac{33.5}{4} = 8.375 \text{ litres}$$
Answer: Average milk yield at Brookside Dairy Ltd is 8.375 litres.
Calculate the frequency densities for continuous data representing fertilizer application rates (kg/ha) grouped as 0–5 (4 farms), 5–10 (9 farms), 10–15 (7 farms). (5 marks)
The number of irrigation systems installed in a cooperative follows the distribution: 0 (0.2), 1 (0.5), 2 (0.2), 3 (0.1). Find the expected number of systems. (5 marks)
Grouped data on maize cob weights (g) are: 100–120 (8 cobs), 120–140 (12 cobs), 140–160 (10 cobs). Calculate the mean cob weight. (5 marks)
Given ungrouped data on pesticide usage in litres per hectare: 2.5, 3.0, 2.8, 3.2, 2.9, find the average usage. (5 marks)
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Create a free accountThis chapter explored various methods for presenting data, including the use of histograms and line graphs for continuous variables, bar and pie charts for discrete variables, and specialized techniques such as cumulative frequency and ogive graphs for grouped data. It then examined measures of central tendency, detailing how to calculate the mean, mode, and median for both grouped and ungrouped data sets. Following this, the chapter addressed measures of dispersion, focusing on standard deviation and variance, again differentiating between grouped and ungrouped data calculations. The concept of probability was introduced with examples of scenarios involving replacement and without replacement, highlighting how these affect outcomes. Probability distribution functions were discussed with emphasis on the binomial and Poisson distributions, explaining their applications and characteristics. Finally, the chapter concluded with the normal distribution, describing its properties and significance in statistical analysis. Together, these topics provide a comprehensive foundation for applying statistics and probability in practical contexts.
A farmer recorded the daily milk yield (in litres) from a cow for 7 days as follows: 10, 12, 11, 13, 12, 14, 11. Calculate the mean daily milk yield. (2 marks)
The number of maize cobs harvested from 5 plots are: 15, 18, 20, 17, 16. Find the mode of the data. (2 marks)
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