Industrial Mechatronics Technology professionals frequently collect and analyze data from sensors, control systems, and production processes to optimize machine performance and ensure quality. Understanding how to present data effectively is crucial for diagnosing system behaviors, monitoring outputs, and making informed decisions in automated industrial environments. This chapter focuses on the statistical presentation of different types of data encountered in industrial mechatronics, facilitating accurate interpretation and communication of findings.

8.1 Data presentation

Data presentation is foundational in industrial mechatronics for summarizing measurements and observations from automated systems. Whether monitoring continuous sensor outputs or counting discrete events, choosing the correct data representation method enhances clarity and supports precise analysis. This section covers continuous and discrete variables, as well as grouped and ungrouped data formats common in industrial contexts.

8.1.1 Continuous variables

Continuous variables represent quantities measured on a scale that can take any value within a range, such as temperature, pressure, or motor speed. These variables are essential in mechatronics for controlling processes with high precision.

The general formula to represent continuous data distribution is the probability density function (PDF), but for data presentation, frequency distributions and histograms are commonly used.

$$ f(x) = \frac{\text{Number of observations in interval}}{\text{Total observations} \times \text{Interval width}} $$

Worked Examples

Example 1: A temperature sensor in an automated kiln records 50 readings between 200°C and 300°C. The data is grouped in intervals of 10°C. Calculate the frequency density for the interval 230°C to 240°C if 5 readings fall in this range.

Given:
Number of observations in interval = 5
Total observations = 50
Interval width = 10°C

$$ f(x) = \frac{5}{50 \times 10} $$

$$ f(x) = \frac{5}{500} = 0.01 \text{ per } °C $$

Answer: 0.01 per °C

Example 2: A conveyor belt speed sensor measures speeds continuously from 1.0 m/s to 2.0 m/s. If 20 readings fall between 1.3 m/s and 1.5 m/s out of 100 total readings, calculate the frequency density for this interval.

Given:
Number of observations in interval = 20
Total observations = 100
Interval width = 1.5 - 1.3 = 0.2 m/s

$$ f(x) = \frac{20}{100 \times 0.2} $$

$$ f(x) = \frac{20}{20} = 1 \text{ per } m/s $$

Answer: 1 per m/s

Example 3: A pressure sensor outputs readings between 100 kPa and 200 kPa. The readings are grouped into intervals of 20 kPa. If the frequency density for 140 kPa to 160 kPa is 0.015 per kPa, and total observations are 80, find the number of observations in this interval.

Given:
Frequency density, \(f(x) = 0.015\) per kPa
Total observations, \(N = 80\)
Interval width = 20 kPa

Formula for frequency density:$$ f(x) = \frac{\text{Frequency}}{N \times \text{Interval width}} $$

Rearranged for frequency:$$ \text{Frequency} = f(x) \times N \times \text{Interval width} $$

Substitute values:$$ \text{Frequency} = 0.015 \times 80 \times 20 $$

$$ \text{Frequency} = 0.015 \times 1600 = 24 $$

Answer: 24 observations

Kenyan Context Example

Example 4: At Bamburi Cement Ltd, a temperature sensor in the kiln records 60 readings between 1200°C and 1400°C, grouped in intervals of 20°C. If 9 readings fall in the interval 1260°C to 1280°C, calculate the frequency density for this interval.

Given:
Number of observations in interval = 9
Total observations = 60
Interval width = 20°C

$$ f(x) = \frac{9}{60 \times 20} $$$$ f(x) = \frac{9}{1200} = 0.0075 \text{ per } ^\circ C $$

Answer: 0.0075 per °C

8.1.2 Discrete variable

Discrete variables take specific separated values, often counts of events or items like the number of defective units or faults detected. These variables are vital in quality control and maintenance scheduling in mechatronics systems.

The frequency of each discrete value is presented using frequency tables or bar charts.

$$ f_i = \text{Number of occurrences of discrete value } x_i $$

Worked Examples

Example 1: In a shift, a machine recorded defects per hour as follows: 0 defects (3 hours), 1 defect (5 hours), 2 defects (2 hours). Find the total number of defects.

Given:
\(f_0 = 3\), \(f_1 = 5\), \(f_2 = 2\)

Total defects =$$ 0 \times 3 + 1 \times 5 + 2 \times 2 $$

$$ = 0 + 5 + 4 = 9 $$

Answer: 9 defects

Example 2: A sensor counts the number of faults per day over 10 days: 1, 0, 2, 1, 3, 0, 1, 2, 2, 1. Prepare a frequency table.

Values: 0, 1, 2, 3
Count frequencies:

• 0 faults: 2 days
• 1 fault: 4 days
• 2 faults: 3 days
• 3 faults: 1 day

Frequency table:
| Faults | Frequency |
|---|---|
| 0 | 2 |
| 1 | 4 |
| 2 | 3 |
| 3 | 1 |

Example 3: Calculate the mean number of faults per day from the frequency table in Example 2.

Given:
Faults \(x_i\) and frequencies \(f_i\) as above
Total days \(N = 10\)

Mean \( \bar{x} = \frac{\Sigma f_i x_i}{N} \)

$$ \bar{x} = \frac{(0 \times 2) + (1 \times 4) + (2 \times 3) + (3 \times 1)}{10} $$

$$ = \frac{0 + 4 + 6 + 3}{10} = \frac{13}{10} = 1.3 \text{ faults/day} $$

Answer: 1.3 faults/day

Kenyan Context Example

Example 4: At Kenya Power's Nairobi substation, the number of circuit breaker trips per day over a week is: 0, 1, 2, 0, 1, 3, 1. Prepare a frequency table.

Values: 0, 1, 2, 3
Count frequencies:

• 0 trips: 2 days
• 1 trip: 3 days
• 2 trips: 1 day
• 3 trips: 1 day

Frequency table:
| Trips | Frequency |
|---|---|
| 0 | 2 |
| 1 | 3 |
| 2 | 1 |
| 3 | 1 |

8.1.3 Grouped data

Grouped data organizes measurements into intervals or classes, allowing easier summarization of large datasets from sensors or production batches. This method is commonly used in industrial reporting for parameters like voltage levels or cycle times.

The midpoint \(x_m\) of each group is used for statistical calculations.

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

Worked Examples

Example 1: A voltage sensor records the following grouped data:

Calculate the midpoint for each voltage class.

Calculate midpoints:
200-220:$$ x_m = \frac{200 + 220}{2} = 210 $$

220-240:$$ x_m = \frac{220 + 240}{2} = 230 $$

240-260:$$ x_m = \frac{240 + 260}{2} = 250 $$

260-280:$$ x_m = \frac{260 + 280}{2} = 270 $$

Answer: Midpoints are 210 V, 230 V, 250 V, 270 V respectively.

Example 2: Using the data from Example 1, calculate the mean voltage.

Given: frequencies \(f_i\) and midpoints \(x_{m_i}\)

Mean \( \bar{x} = \frac{\Sigma f_i x_{m_i}}{\Sigma f_i} \)

$$ \bar{x} = \frac{4 \times 210 + 6 \times 230 + 10 \times 250 + 5 \times 270}{4 + 6 + 10 + 5} $$

$$ = \frac{840 + 1380 + 2500 + 1350}{25} = \frac{6070}{25} = 242.8 \text{ V} $$

Answer: 242.8 V

Example 3: Calculate the variance of the voltage data in Example 1.

Step 1: Calculate mean \( \bar{x} = 242.8 \) V (from Example 2)
Step 2: Use variance formula for grouped data:

$$ \sigma^2 = \frac{\Sigma f_i (x_{m_i} - \bar{x})^2}{\Sigma f_i} $$

Calculate squared deviations:
\((210 - 242.8)^2 = ( -32.8 )^2 = 1075.84\)
\((230 - 242.8)^2 = ( -12.8 )^2 = 163.84\)
\((250 - 242.8)^2 = 7.2^2 = 51.84\)
\((270 - 242.8)^2 = 27.2^2 = 739.84\)

Multiply by frequencies:
\(4 \times 1075.84 = 4303.36\)
\(6 \times 163.84 = 983.04\)
\(10 \times 51.84 = 518.4\)
\(5 \times 739.84 = 3699.2\)

Sum:$$ 4303.36 + 983.04 + 518.4 + 3699.2 = 9504 $$

Divide by total frequency \(N=25\):$$ \sigma^2 = \frac{9504}{25} = 380.16 \text{ V}^2 $$

Answer: 380.16 V²

Kenyan Context Example

Example 4: At Kenya Breweries Ltd, the following grouped data shows the alcohol content (ABV %) of batches:

Calculate the mean ABV.

Midpoints:
4.0-4.5: \(x_m = \frac{4.0+4.5}{2} = 4.25\)
4.5-5.0: \(x_m = \frac{4.5+5.0}{2} = 4.75\)
5.0-5.5: \(x_m = \frac{5.0+5.5}{2} = 5.25\)
5.5-6.0: \(x_m = \frac{5.5+6.0}{2} = 5.75\)

Mean:$$ \bar{x} = \frac{8 \times 4.25 + 12 \times 4.75 + 18 \times 5.25 + 7 \times 5.75}{8+12+18+7} $$$$ = \frac{34 + 57 + 94.5 + 40.25}{45} = \frac{225.75}{45} = 5.017 \text{\% ABV} $$

Answer: 5.017% ABV

8.1.4 Ungrouped data

Ungrouped data refers to raw individual data points collected from sensors or measurements before summarization. This form is essential for detailed analysis and is often the starting point before data grouping or statistical calculations.

Measures such as mean, median, and mode are calculated directly from ungrouped data.

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

Worked Examples

Example 1: A robotic arm records the following cycle times in seconds: 12, 15, 14, 13, 16. Calculate the mean cycle time.

Given: \(x_i = \{12, 15, 14, 13, 16\}\), \(n=5\)

$$ \bar{x} = \frac{12 + 15 + 14 + 13 + 16}{5} $$

$$ = \frac{70}{5} = 14 \text{ seconds} $$

Answer: 14 seconds

Example 2: Find the median of the cycle times in Example 1.

Step 1: Arrange data in ascending order: 12, 13, 14, 15, 16
Step 2: Identify middle value (3rd value since \(n=5\))

Answer: Median = 14 seconds

Example 3: Calculate the mode of the following sensor fault counts recorded over 7 days: 2, 3, 2, 1, 4, 2, 3.

Given: data set \( \{2, 3, 2, 1, 4, 2, 3\} \)

Count frequencies:

• 1 occurs 1 time
• 2 occurs 3 times
• 3 occurs 2 times
• 4 occurs 1 time

Mode is the value with the highest frequency.

Answer: Mode = 2 faults

Chapter Summary

This chapter covered the presentation of data using various graphical methods suited to different types of variables, including histograms and line graphs for continuous data, bar and pie charts for discrete data, as well as techniques for grouped and ungrouped data such as cumulative frequency and ogive curves. It then explored measures of central tendency, detailing how to calculate the mean, mode, and median for both grouped and ungrouped data sets. The chapter proceeded to measures of dispersion, explaining the concepts and calculations of standard deviation, variance, range, quartile deviation, and mean deviation to describe data variability. Different types of probability were examined, distinguishing between scenarios with and without replacement. Finally, the chapter introduced key probability distribution functions, focusing on the binomial, Poisson, and normal distributions, which model various types of random phenomena in engineering contexts. These topics collectively provide essential tools for analyzing and interpreting data relevant to engineering projects and decision-making.