In industrial control and installation, professionals rely on statistics to analyze data from manufacturing processes, equipment tests, and quality assurance checks. Understanding how to classify, collect, and analyze data ensures that systems remain efficient, safe, and compliant with industry standards. This chapter covers the essential statistical methods and calculations needed for real-world industrial control applications in Kenya, focusing on both grouped and ungrouped data, sampling methods, and recognizing errors.

1.1 Classification of Data

Statistical data in engineering is typically categorized as either grouped or ungrouped. Knowing how to work with each type is crucial for interpreting measurements from sensors, production lines, or maintenance logs.

1.1.1 Grouped Data

Grouped data arises when raw measurements are organized into intervals (classes), making it easier to analyze large datasets, such as readings from automated process controls or batch quality checks. The key parameters often calculated are the mean, median, and mode.

The formula for the mean of grouped data is:$$ \bar{x} = \frac{\Sigma f x}{\Sigma f} $$ where \(f\) is the frequency and \(x\) is the class midpoint.

Worked Examples

Example 1:

A maintenance engineer records the response times (in seconds) of a relay in an automated system over 50 cycles. The results are grouped as follows:

Find the mean response time.

Given: Frequency table as above.

Step 1: Find midpoints for each class.

• \(0.9–1.1\): \(x_1 = 1.0\)
• \(1.2–1.4\): \(x_2 = 1.3\)
• \(1.5–1.7\): \(x_3 = 1.6\)
• \(1.8–2.0\): \(x_4 = 1.9\)
• \(2.1–2.3\): \(x_5 = 2.2\)

Step 2: Multiply each midpoint by its frequency.

• \(6 \times 1.0 = 6.0\)
• \(14 \times 1.3 = 18.2\)
• \(20 \times 1.6 = 32.0\)
• \(7 \times 1.9 = 13.3\)
• \(3 \times 2.2 = 6.6\)

Step 3: Sum the products and frequencies.

• \(\Sigma f x = 6.0 + 18.2 + 32.0 + 13.3 + 6.6 = 76.1\)
• \(\Sigma f = 6 + 14 + 20 + 7 + 3 = 50\)

Step 4: Calculate the mean.

$$ \bar{x} = \frac{76.1}{50} = 1.522 $$

Answer: \(1.52\) s

Example 2:

A batch of temperature sensor readings from a cold storage facility is grouped as follows:

Calculate the mean temperature.

Given: Class intervals and frequencies.

Step 1: Midpoints:

• \(2–4\): \(3\)
• \(5–7\): \(6\)
• \(8–10\): \(9\)
• \(11–13\): \(12\)
• \(14–16\): \(15\)

Step 2: Multiply and sum.

• \(8 \times 3 = 24\)
• \(12 \times 6 = 72\)
• \(20 \times 9 = 180\)
• \(7 \times 12 = 84\)
• \(3 \times 15 = 45\)

\(\Sigma f x = 24 + 72 + 180 + 84 + 45 = 405\)

\(\Sigma f = 8 + 12 + 20 + 7 + 3 = 50\)

Step 3: Mean

$$ \bar{x} = \frac{405}{50} = 8.10 $$

Answer: \(8.10^\circ\)C

Example 3:

A contractor tests the compressive strength (in MPa) of concrete cubes, recording the following grouped results:

Find the mean compressive strength.

Midpoints: \(22, 27, 32, 37, 42\)

Products: \(5 \times 22 = 110\), \(14 \times 27 = 378\), \(18 \times 32 = 576\), \(8 \times 37 = 296\), \(5 \times 42 = 210\)

\(\Sigma f x = 110 + 378 + 576 + 296 + 210 = 1570\)

\(\Sigma f = 5 + 14 + 18 + 8 + 5 = 50\)

Mean:

$$ \bar{x} = \frac{1570}{50} = 31.4 $$

Answer: \(31.4\) MPa

1.1.2 Ungrouped Data

Ungrouped data is a raw list of individual measurements, such as manually recorded voltages or equipment downtime minutes. The mean, median, mode, and standard deviation can be calculated directly from the list.

The mean is calculated as:$$ \bar{x} = \frac{\Sigma x}{n} $$

The standard deviation is:$$ \sigma = \sqrt{\frac{\Sigma (x - \bar{x})^2}{n}} $$

Worked Examples

Example 1:

A technician measures the resistance (in ohms) of six control panel relays: \(5, 6, 8, 4, 10, 3\).

Mean:

$$ \bar{x} = \frac{5 + 6 + 8 + 4 + 10 + 3}{6} = \frac{36}{6} = 6 $$

Standard deviation:

Deviations: \((5-6)^2 = 1\), \((6-6)^2 = 0\), \((8-6)^2 = 4\), \((4-6)^2 = 4\), \((10-6)^2 = 16\), \((3-6)^2 = 9\)

Sum: \(1 + 0 + 4 + 4 + 16 + 9 = 34\)

$$ \sigma = \sqrt{\frac{34}{6}} = \sqrt{5.6667} = 2.382 $$

Answer: Mean = \(6\) ohms, Standard deviation = \(2.38\) ohms

Example 2:

A supervisor records the number of faults detected in a PLC each day for a week: \(2, 3, 3, 4, 5, 2, 1\).

Mean:

$$ \bar{x} = \frac{2 + 3 + 3 + 4 + 5 + 2 + 1}{7} = \frac{20}{7} = 2.857 $$

Standard deviation:

Deviations: \((2-2.857)^2 = 0.734\), \((3-2.857)^2 = 0.020\), \((3-2.857)^2 = 0.020\), \((4-2.857)^2 = 1.306\), \((5-2.857)^2 = 4.592\), \((2-2.857)^2 = 0.734\), \((1-2.857)^2 = 3.448\)

Sum: \(0.734 + 0.020 + 0.020 + 1.306 + 4.592 + 0.734 + 3.448 = 10.854\)

$$ \sigma = \sqrt{\frac{10.854}{7}} = \sqrt{1.551} = 1.245 $$

Answer: Mean = \(2.86\) faults, Standard deviation = \(1.25\) faults

Example 3:

A quality inspector measures the thickness (mm) of five control panel enclosures: \(1.2, 1.5, 1.3, 1.4, 1.6\).

Mean:

$$ \bar{x} = \frac{1.2 + 1.5 + 1.3 + 1.4 + 1.6}{5} = \frac{7.0}{5} = 1.4 $$

Standard deviation:

\((1.2-1.4)^2 = 0.04\), \((1.5-1.4)^2 = 0.01\), \((1.3-1.4)^2 = 0.01\), \((1.4-1.4)^2 = 0.00\), \((1.6-1.4)^2 = 0.04\)

Sum: \(0.04 + 0.01 + 0.01 + 0.00 + 0.04 = 0.10\)

$$ \sigma = \sqrt{\frac{0.10}{5}} = \sqrt{0.02} = 0.141 $$

Answer: Mean = \(1.4\) mm, Standard deviation = \(0.14\) mm

Chapter Summary

This chapter explored the foundational concepts of statistics as applied in engineering mathematics, beginning with the classification of data into grouped and ungrouped forms. It examined methods of data collection, emphasizing the importance of sampling, common errors encountered, and various sampling techniques along with their limitations. The process of tabulating data was covered, including the creation of class intervals, determination of class boundaries, construction of frequency tables, and calculation of cumulative frequency. Diagrammatic and graphical presentation methods such as histograms, frequency polygons, bar charts, pie charts, and curves were discussed to illustrate effective communication of statistical information. Measures of central tendency, namely mean, mode, and median, were calculated and interpreted to summarize data sets. The chapter concluded with an analysis of measures of dispersion, focusing on variance and standard deviation, to assess the spread and reliability of engineering data. By mastering these statistical tools, students are equipped to analyze, interpret, and present data in civil engineering contexts with precision and clarity.