By the end of this chapter, you will be able to:
These skills will help you make informed decisions and solve real-world problems confidently in your trade.
Welding professionals often collect and analyze data on weld quality, defect rates, and production parameters to improve their processes. Presenting this data accurately enables better decision-making and quality control in fabrication projects. This chapter focuses on methods to present welding-related data effectively by understanding different types of variables and the organization of data sets.
Data presentation is fundamental in welding operations to monitor and control quality. Whether measuring weld bead dimensions or counting defect occurrences, knowing how to organize and present data ensures clear communication and analysis. In the Kenyan welding industry, effective data presentation supports compliance with standards and enhances productivity.
Continuous variables represent measurements that can take any value within a range and are common in welding, such as weld bead length or temperature.
The general formula for calculating the mean of continuous data is:
$$ \bar{x} = \frac{\Sigma x_i}{n} $$
where \(x_i\) are individual measurements and \(n\) is the total number of measurements.
Example 1: A welder measures the thickness of five metal sheets as 4.5 mm, 4.7 mm, 4.6 mm, 4.8 mm, and 4.7 mm. Find the average thickness.
Given: thickness values \(x_1=4.5\) mm, \(x_2=4.7\) mm, \(x_3=4.6\) mm, \(x_4=4.8\) mm, \(x_5=4.7\) mm, \(n=5\)
$$ \bar{x} = \frac{4.5 + 4.7 + 4.6 + 4.8 + 4.7}{5} $$
$$ = \frac{23.3}{5} $$
$$ = 4.66 \text{ mm} $$
Answer: 4.66 mm
Example 2: During a welding process, the temperature readings recorded every minute are 150°C, 152°C, 149°C, 153°C, 151°C, and 150°C. Calculate the mean temperature.
Given: \(x_i = 150, 152, 149, 153, 151, 150\) °C, \(n=6\)
$$ \bar{x} = \frac{150 + 152 + 149 + 153 + 151 + 150}{6} $$
$$ = \frac{905}{6} $$
$$ = 150.83^\circ C $$
Answer: 150.83°C
Example 3: A welding machine’s wire feed speed is measured six times as 9.8, 10.0, 9.9, 10.1, 9.7, and 10.2 m/min. Calculate the average wire feed speed.
Given: \(x_i = 9.8, 10.0, 9.9, 10.1, 9.7, 10.2\) m/min, \(n=6\)
$$ \bar{x} = \frac{9.8 + 10.0 + 9.9 + 10.1 + 9.7 + 10.2}{6} $$
$$ = \frac{59.7}{6} $$
$$ = 9.95 \text{ m/min} $$
Answer: 9.95 m/min
Example 4: The weld bead width in millimeters for seven welds are 5.1, 5.0, 5.2, 5.3, 5.1, 5.4, and 5.2. Find the mean weld bead width.
Given: \(x_i = 5.1, 5.0, 5.2, 5.3, 5.1, 5.4, 5.2\) mm, \(n=7\)
$$ \bar{x} = \frac{5.1 + 5.0 + 5.2 + 5.3 + 5.1 + 5.4 + 5.2}{7} $$
$$ = \frac{36.3}{7} $$
$$ = 5.19 \text{ mm} $$
Answer: 5.19 mm
Example 5: A temperature sensor records the following temperatures (°C) during a welding procedure: 145, 148, 147, 146, 149, 150, 148, 147. Calculate the average temperature.
Given: \(x_i = 145, 148, 147, 146, 149, 150, 148, 147\) °C, \(n=8\)
$$ \bar{x} = \frac{145 + 148 + 147 + 146 + 149 + 150 + 148 + 147}{8} $$
$$ = \frac{1180}{8} $$
$$ = 147.5^\circ C $$
Answer: 147.5°C
Discrete variables count occurrences or events and are whole numbers, such as the number of welding defects or weld passes.
The frequency \(f_i\) of each discrete value is used to calculate the mean:
$$ \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} $$
where \(x_i\) are values and \(f_i\) their frequencies.
Example 1: A welding inspector counts the number of defects in 5 welds: 0, 1, 2, 1, and 3. Find the mean number of defects.
Given values and frequencies:
| Defects \(x_i\) | Frequency \(f_i\) |
|---|---|
| 0 | 1 |
| 1 | 2 |
| 2 | 1 |
| 3 | 1 |
Calculate:
$$ \bar{x} = \frac{(0 \times 1) + (1 \times 2) + (2 \times 1) + (3 \times 1)}{1 + 2 + 1 + 1} $$
$$ = \frac{0 + 2 + 2 + 3}{5} $$
$$ = \frac{7}{5} $$
$$ = 1.4 $$
Answer: 1.4 defects per weld
Example 2: A welder records the number of weld passes needed to complete 6 joints: 3, 4, 3, 5, 4, 3. Calculate the average number of passes.
Frequencies:
| Passes \(x_i\) | Frequency \(f_i\) |
|---|---|
| 3 | 3 |
| 4 | 2 |
| 5 | 1 |
Calculate:
$$ \bar{x} = \frac{(3 \times 3) + (4 \times 2) + (5 \times 1)}{3 + 2 + 1} $$
$$ = \frac{9 + 8 + 5}{6} $$
$$ = \frac{22}{6} $$
$$ = 3.67 $$
Answer: 3.67 passes per joint
Example 3: The number of welding defects found in 8 sections are: 0, 1, 0, 2, 1, 3, 0, 1. Find the mean defects per section.
Frequencies:
| Defects \(x_i\) | Frequency \(f_i\) |
|---|---|
| 0 | 3 |
| 1 | 3 |
| 2 | 1 |
| 3 | 1 |
Calculate:
$$ \bar{x} = \frac{(0 \times 3) + (1 \times 3) + (2 \times 1) + (3 \times 1)}{3 + 3 + 1 + 1} $$
$$ = \frac{0 + 3 + 2 + 3}{8} $$
$$ = \frac{8}{8} $$
$$ = 1 $$
Answer: 1 defect per section
Example 4: A welding manager records the number of rejected welds per day over a week: 2, 3, 2, 4, 3, 1, 2. Calculate the average rejections per day.
Frequencies:
| Rejected welds \(x_i\) | Frequency \(f_i\) |
|---|---|
| 1 | 1 |
| 2 | 3 |
| 3 | 2 |
| 4 | 1 |
Calculate:
$$ \bar{x} = \frac{(1 \times 1) + (2 \times 3) + (3 \times 2) + (4 \times 1)}{1 + 3 + 2 + 1} $$
$$ = \frac{1 + 6 + 6 + 4}{7} $$
$$ = \frac{17}{7} $$
$$ = 2.43 $$
Answer: 2.43 rejected welds per day
Example 5: During inspection, the number of welding defects per batch of 10 batches were: 1, 0, 2, 1, 1, 3, 0, 2, 1, 0. Calculate the average defects per batch.
Frequencies:
| Defects \(x_i\) | Frequency \(f_i\) |
|---|---|
| 0 | 3 |
| 1 | 4 |
| 2 | 2 |
| 3 | 1 |
Calculate:
$$ \bar{x} = \frac{(0 \times 3) + (1 \times 4) + (2 \times 2) + (3 \times 1)}{3 + 4 + 2 + 1} $$
$$ = \frac{0 + 4 + 4 + 3}{10} $$
$$ = \frac{11}{10} $$
$$ = 1.1 $$
Answer: 1.1 defects per batch
Grouped data involves organizing continuous or discrete data into classes or intervals for easier analysis, such as grouping weld bead thickness ranges.
The mean for grouped data is calculated by:
$$ \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} $$
where \(x_i\) is the midpoint of each class interval and \(f_i\) is the frequency.
Example 1: Weld bead thickness measurements are grouped into intervals with frequencies:
| Thickness (mm) | Frequency |
|---|---|
| 4.0 - 4.4 | 3 |
| 4.5 - 4.9 | 5 |
| 5.0 - 5.4 | 2 |
Calculate the mean thickness.
Calculate midpoints:
\(x_1 = \frac{4.0 + 4.4}{2} = 4.2\) mm
\(x_2 = \frac{4.5 + 4.9}{2} = 4.7\) mm
\(x_3 = \frac{5.0 + 5.4}{2} = 5.2\) mm
Calculate:
$$ \bar{x} = \frac{(3 \times 4.2) + (5 \times 4.7) + (2 \times 5.2)}{3 + 5 + 2} $$
$$ = \frac{12.6 + 23.5 + 10.4}{10} $$
$$ = \frac{46.5}{10} $$
$$ = 4.65 \text{ mm} $$
Answer: 4.65 mm
Example 2: Grouped data of welding temperature readings and frequencies:
| Temperature (°C) | Frequency |
|---|---|
| 140 - 144 | 4 |
| 145 - 149 | 7 |
| 150 - 154 | 5 |
Calculate the mean temperature.
Midpoints:
\(x_1 = \frac{140 + 144}{2} = 142\) °C
\(x_2 = \frac{145 + 149}{2} = 147\) °C
\(x_3 = \frac{150 + 154}{2} = 152\) °C
Calculate:
$$ \bar{x} = \frac{(4 \times 142) + (7 \times 147) + (5 \times 152)}{4 + 7 + 5} $$
$$ = \frac{568 + 1029 + 760}{16} $$
$$ = \frac{2357}{16} $$
$$ = 147.31^\circ C $$
Answer: 147.31°C
Example 3: Weld bead width grouped data:
| Width (mm) | Frequency |
|---|---|
| 5.0 - 5.4 | 6 |
| 5.5 - 5.9 | 3 |
| 6.0 - 6.4 | 1 |
Calculate the mean bead width.
Midpoints:
\(x_1 = \frac{5.0 + 5.4}{2} = 5.2\) mm
\(x_2 = \frac{5.5 + 5.9}{2} = 5.7\) mm
\(x_3 = \frac{6.0 + 6.4}{2} = 6.2\) mm
Calculate:
$$ \bar{x} = \frac{(6 \times 5.2) + (3 \times 5.7) + (1 \times 6.2)}{6 + 3 + 1} $$
$$ = \frac{31.2 + 17.1 + 6.2}{10} $$
$$ = \frac{54.5}{10} $$
$$ = 5.45 \text{ mm} $$
Answer: 5.45 mm
Example 4: Grouped data on number of weld defects per batch:
| Defects | Frequency |
|---|---|
| 0 - 1 | 8 |
| 2 - 3 | 5 |
| 4 - 5 | 2 |
Calculate mean defects per batch.
Midpoints:
\(x_1 = \frac{0 + 1}{2} = 0.5\)
\(x_2 = \frac{2 + 3}{2} = 2.5\)
\(x_3 = \frac{4 + 5}{2} = 4.5\)
Calculate:
$$ \bar{x} = \frac{(8 \times 0.5) + (5 \times 2.5) + (2 \times 4.5)}{8 + 5 + 2} $$
$$ = \frac{4 + 12.5 + 9}{15} $$
$$ = \frac{25.5}{15} $$
$$ = 1.7 $$
Answer: 1.7 defects per batch
Example 5: Grouped data of welding time in minutes with frequencies:
| Time (min) | Frequency |
|---|---|
| 30 - 34 | 3 |
| 35 - 39 | 4 |
| 40 - 44 | 3 |
Calculate the mean welding time.
Midpoints:
\(x_1 = \frac{30 + 34}{2} = 32\) min
\(x_2 = \frac{35 + 39}{2} = 37\) min
\(x_3 = \frac{40 + 44}{2} = 42\) min
Calculate:
$$ \bar{x} = \frac{(3 \times 32) + (4 \times 37) + (3 \times 42)}{3 + 4 + 3} $$
$$ = \frac{96 + 148 + 126}{10} $$
$$ = \frac{370}{10} $$
$$ = 37 \text{ minutes} $$
Answer: 37 minutes
Ungrouped data consists of raw data points collected directly from measurements or counts without classification.
The mean is calculated by:
$$ \bar{x} = \frac{\Sigma x_i}{n} $$
where \(x_i\) are individual data points.
Example 1: A welding operator records weld times (minutes) for 5 welds: 15, 17, 16, 18, 15. Find the average weld time.
Given: \(x_i = 15, 17, 16, 18, 15\), \(n=5\)
$$ \bar{x} = \frac{15 + 17 + 16 + 18 + 15}{5} $$
$$ = \frac{81}{5} $$
$$ = 16.2 \text{ minutes} $$
Answer: 16.2 minutes
Example 2: The number of welding defects found in 6 inspections are: 2, 3, 1, 4, 2, 3. Calculate the mean defects per inspection.
Given: \(x_i = 2, 3, 1, 4, 2, 3\), \(n=6\)
$$ \bar{x} = \frac{2 + 3 + 1 + 4 + 2 + 3}{6} $$
$$ = \frac{15}{6} $$
$$ = 2.5 $$
Answer: 2.5 defects per inspection
Example 3: Times (seconds) taken to complete weld beads: 120, 125, 130, 128, 122, 126. Calculate the average time.
Given: \(x_i = 120, 125, 130, 128, 122, 126\), \(n=6\)
$$ \bar{x} = \frac{120 + 125 + 130 + 128 + 122 + 126}{6} $$
$$ = \frac{751}{6} $$
$$ = 125.17 \text{ seconds} $$
Answer: 125.17 seconds
Example 4: Weld bead heights (mm) measured as: 3.1, 3.0, 3.2, 3.3, 3.1, 3.4, 3.2. Find the average height.
Given: \(x_i = 3.1, 3.0, 3.2, 3.3, 3.1, 3.4, 3.2\), \(n=7\)
$$ \bar{x} = \frac{3.1 + 3.0 + 3.2 + 3.3 + 3.1 + 3.4 + 3.2}{7} $$
$$ = \frac{22.3}{7} $$
$$ = 3.19 \text{ mm} $$
Answer: 3.19 mm
Example 5: A welder records the number of weld passes for 8 joints: 2, 3, 3, 2, 4, 3, 2, 3. Calculate the mean passes.
Given: \(x_i = 2, 3, 3, 2, 4, 3, 2, 3\), \(n=8\)
$$ \bar{x} = \frac{2 + 3 + 3 + 2 + 4 + 3 + 2 + 3}{8} $$
$$ = \frac{22}{8} $$
$$ = 2.75 $$
Answer: 2.75 passes
A welder measures the diameter of 6 rods as 12.4 mm, 12.6 mm, 12.5 mm, 12.7 mm, 12.5 mm, and 12.8 mm. Calculate the average diameter. (5 marks)
The number of defects found in 7 welds are 1, 0, 2, 3, 1, 0, 2. Find the mean number of defects. (5 marks)
Given the grouped data for welding current (Amps) and frequencies:
| Current (A) | Frequency |
|---|---|
| 100 - 104 | 4 |
| 105 - 109 | 6 |
| 110 - 114 | 5 |
Calculate the mean welding current. (6 marks)
Create a free account to open more of this chapter.
Free: practical guides, quick cards, workplace scenarios and more.
Create a free accountThis chapter covered the presentation of data using different graphical methods suited for continuous and discrete variables, including histograms, line graphs, bar graphs, and pie charts, as well as techniques for grouped and ungrouped data such as cumulative frequency curves and ogives. 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 discuss measures of dispersion, focusing on the calculation of standard deviation and variance, again distinguishing between grouped and ungrouped data. In the probability section, methods for calculating probabilities with and without replacement were explained. The chapter also introduced probability distribution functions, specifically the binomial and Poisson distributions, highlighting their applications. Finally, the normal distribution was examined as a fundamental continuous probability distribution, emphasizing its properties and relevance in statistical analysis. This comprehensive coverage equips learners with essential skills for applying statistics and probability in practical contexts.
A welder records the lengths (in mm) of 10 welded joints as follows: 48, 50, 52, 47, 49, 51, 53, 48, 50, 49. Calculate the mean length of the welded joints. (3 marks)
The number of welding defects found in 15 inspections are: 0, 1, 2, 0, 3, 1, 0, 2, 1, 0, 3, 2, 1, 0, 1. Construct a frequency table and calculate the mode number of defects. (4 marks)
A welding workshop recorded the diameters (in mm) of 10 steel rods as follows: 12, 14, 15, 12, 13, 14, 16, 15, 14, 13. Construct a frequency table and draw a bar graph to represent this discrete data. (4 marks)
The tensile strength (in MPa) of 15 welded joints was measured and grouped as follows: 400–420 (3 joints), 421–440 (5 joints), 441–460 (4 joints), 461–480 (3 joints). Draw a histogram to represent this continuous grouped data. (4 marks)
At the start of this chapter we promised you would be able to:
Tick each one you can genuinely do.
So, are you there yet?
You're competent when you can confidently do 50% or more of what this chapter promised.
Sign in to record how you're doing.