Statistics and probability are essential in the auto body repair industry to analyze data, assess quality control, and make informed decisions based on measurements and observed outcomes. For an Auto Body Technician in Kenya, understanding how to present and interpret data effectively supports tasks such as estimating repair times, managing inventory, and evaluating repair success rates. This chapter focuses on methods of data presentation relevant to continuous and discrete variables, as well as grouped and ungrouped data, providing the foundation for further statistical analysis and probability applications.

4.1 Data presentation

Data presentation is the first step in analyzing information collected during auto body repair processes. Accurate presentation helps technicians identify patterns, make comparisons, and communicate findings clearly. In Kenya’s automotive workshops, data can range from measurements of panel thickness to counts of repaired vehicles, each requiring appropriate presentation techniques.

4.1.1 Continuous variables

Continuous variables take any value within a range and are measurable quantities, such as paint thickness or repair time. Presenting continuous data typically involves organizing measurements to reveal trends or distribution.

Worked Examples

Example 1: A technician measures paint thickness on 5 car doors: 0.15 mm, 0.18 mm, 0.20 mm, 0.16 mm, 0.19 mm. Find the mean paint thickness.

Given: Paint thickness values \(0.15, 0.18, 0.20, 0.16, 0.19\) mm

$$\text{Mean} = \frac{\Sigma x_i}{n}$$

$$= \frac{0.15 + 0.18 + 0.20 + 0.16 + 0.19}{5}$$

$$= \frac{0.88}{5}$$

$$= 0.176 \text{ mm}$$

Answer: 0.176 mm

Example 2: Find the range of the paint thickness data in Example 1.

Given: Maximum \(0.20\) mm, minimum \(0.15\) mm

$$\text{Range} = \text{Max} - \text{Min}$$

$$= 0.20 - 0.15$$

$$= 0.05 \text{ mm}$$

Answer: 0.05 mm

Example 3: The repair times (in hours) for 6 vehicles are: 3.5, 4.0, 3.8, 4.2, 3.9, 4.1. Calculate the mean repair time.

Given: Repair times \(3.5, 4.0, 3.8, 4.2, 3.9, 4.1\) hours

$$\text{Mean} = \frac{3.5 + 4.0 + 3.8 + 4.2 + 3.9 + 4.1}{6}$$

$$= \frac{23.5}{6}$$

$$= 3.9167 \text{ hours}$$

Answer: 3.9167 hours

Example 4: A technician records vehicle speeds during a test: 60.2, 59.8, 60.5, 60.0, 59.7 km/h. Calculate the median speed.

Arrange data: 59.7, 59.8, 60.0, 60.2, 60.5

Median is middle value:

Answer: 60.0 km/h

Example 5: Calculate the variance of the repair times in Example 3.

Given: Repair times \(3.5, 4.0, 3.8, 4.2, 3.9, 4.1\), mean \(3.9167\)

$$\text{Variance} = \frac{\Sigma (x_i - \bar{x})^2}{n}$$

Calculate each squared difference:

\((3.5 - 3.9167)^2 = 0.1736\)

\((4.0 - 3.9167)^2 = 0.0069\)

\((3.8 - 3.9167)^2 = 0.0136\)

\((4.2 - 3.9167)^2 = 0.0803\)

\((3.9 - 3.9167)^2 = 0.0003\)

\((4.1 - 3.9167)^2 = 0.0337\)

Sum: \(0.1736 + 0.0069 + 0.0136 + 0.0803 + 0.0003 + 0.0337 = 0.3084\)

$$\text{Variance} = \frac{0.3084}{6} = 0.0514$$

Answer: 0.0514 hours²

4.1.2 Discrete variable

Discrete variables are countable, such as the number of dents on a vehicle or the number of vehicles repaired daily. Presenting discrete data often involves frequency distributions or bar charts.

Frequency tables summarize how many times each value occurs, aiding in identifying modes or common occurrences.

Worked Examples

Example 1: Count the number of dents on 8 different vehicles: 2, 3, 1, 4, 2, 3, 1, 2. Construct a frequency table.

Given: Dents per vehicle

Answer: Frequency table as above

Example 2: Calculate the mode of the dents data in Example 1.

Given frequencies: 1 dent (2), 2 dents (3), 3 dents (2), 4 dents (1)

Mode is the value with highest frequency: 2 dents

Answer: 2 dents

Example 3: The number of vehicles repaired per day over 5 days is: 6, 7, 5, 8, 7. Find the mean number of vehicles repaired per day.

Given: \(6, 7, 5, 8, 7\)

$$\text{Mean} = \frac{6 + 7 + 5 + 8 + 7}{5}$$

$$= \frac{33}{5}$$

$$= 6.6 \text{ vehicles}$$

Answer: 6.6 vehicles

Example 4: Determine the variance of the vehicles repaired per day in Example 3.

Given: Mean \(6.6\), values \(6, 7, 5, 8, 7\)

Calculate squared differences:

\((6 - 6.6)^2 = 0.36\)

\((7 - 6.6)^2 = 0.16\)

\((5 - 6.6)^2 = 2.56\)

\((8 - 6.6)^2 = 1.96\)

\((7 - 6.6)^2 = 0.16\)

Sum: \(0.36 + 0.16 + 2.56 + 1.96 + 0.16 = 5.2\)

$$\text{Variance} = \frac{5.2}{5} = 1.04$$

Answer: 1.04 vehicles²

Example 5: Calculate the probability that a randomly selected vehicle from the 8 in Example 1 has exactly 3 dents.

Given: Frequency of 3 dents = 2, total vehicles = 8

$$P(3 \text{ dents}) = \frac{2}{8} = 0.25$$

Answer: 0.25 or 25%

4.1.3 Grouped data

Grouped data organizes continuous or discrete data into intervals or classes, which is useful for large datasets like vehicle repair costs or paint thickness ranges. Grouping simplifies data analysis and presentation.

Calculating class midpoints, frequencies, and cumulative frequencies is essential for further analysis such as estimating the mean or median.

Worked Examples

Example 1: Repair costs (in Ksh) for 15 vehicles are grouped as:

Calculate the class midpoints.

Midpoint formula:

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

• For 5000 - 7999:

$$\frac{5000 + 7999}{2} = \frac{12999}{2} = 6499.5 \text{ Ksh}$$

• For 8000 - 10999:

$$\frac{8000 + 10999}{2} = \frac{18999}{2} = 9499.5 \text{ Ksh}$$

• For 11000 - 13999:

$$\frac{11000 + 13999}{2} = \frac{24999}{2} = 12499.5 \text{ Ksh}$$

• For 14000 - 16999:

$$\frac{14000 + 16999}{2} = \frac{30999}{2} = 15499.5 \text{ Ksh}$$

Answer: Midpoints are 6499.5, 9499.5, 12499.5, 15499.5 Ksh

Example 2: Estimate the mean repair cost using the midpoints and frequencies from Example 1.

Given:

Formula:

$$\text{Mean} = \frac{\Sigma f_i x_i}{\Sigma f_i}$$

Calculate numerator:

$$4 \times 6499.5 = 25998$$

$$6 \times 9499.5 = 56997$$

$$3 \times 12499.5 = 37498.5$$

$$2 \times 15499.5 = 30999$$

Sum numerator:

$$25998 + 56997 + 37498.5 + 30999 = 151492.5$$

Sum denominator:

$$4 + 6 + 3 + 2 = 15$$

Calculate mean:

$$\text{Mean} = \frac{151492.5}{15} = 10099.5 \text{ Ksh}$$

Answer: 10099.5 Ksh

Example 3: Calculate the cumulative frequency for the data in Example 1.

Answer: Cumulative frequencies are 4, 10, 13, 15

Example 4: A technician measures paint thickness in ranges with frequencies:

Calculate the midpoint of the second class.

Given: Lower limit = 0.15 mm, upper limit = 0.19 mm

$$\text{Midpoint} = \frac{0.15 + 0.19}{2} = \frac{0.34}{2} = 0.17 \text{ mm}$$

Answer: 0.17 mm

Example 5: Estimate the mean paint thickness using the data in Example 4.

Midpoints:

• 0.10 - 0.14: \( \frac{0.10 + 0.14}{2} = 0.12 \) mm

• 0.15 - 0.19: 0.17 mm (from Example 4)

• 0.20 - 0.24: \( \frac{0.20 + 0.24}{2} = 0.22 \) mm

Frequencies: 5, 8, 7

Calculate numerator:

$$5 \times 0.12 = 0.6$$

$$8 \times 0.17 = 1.36$$

$$7 \times 0.22 = 1.54$$

Sum numerator:

$$0.6 + 1.36 + 1.54 = 3.5$$

Sum denominator:

$$5 + 8 + 7 = 20$$

Calculate mean:

$$\text{Mean} = \frac{3.5}{20} = 0.175 \text{ mm}$$

Answer: 0.175 mm

4.1.4 Ungrouped data

Ungrouped data refers to raw data points recorded individually without classification into intervals. It is useful for small datasets or initial data collection before summarizing.

Statistics such as mean, median, mode, variance, and standard deviation can be directly calculated from ungrouped data.

Worked Examples

Example 1: The thickness of five vehicle panels in mm is: 0.16, 0.18, 0.15, 0.19, 0.17. Calculate the mean thickness.

Given: \(0.16, 0.18, 0.15, 0.19, 0.17\) mm

$$\text{Mean} = \frac{0.16 + 0.18 + 0.15 + 0.19 + 0.17}{5}$$

$$= \frac{0.85}{5}$$

$$= 0.17 \text{ mm}$$

Answer: 0.17 mm

Example 2: Calculate the median of the data in Example 1.

Arrange data: 0.15, 0.16, 0.17, 0.18, 0.19

Median is middle value:

Answer: 0.17 mm

Example 3: Find the mode of the following dent counts on 10 vehicles: 2, 3, 2, 4, 3, 2, 5, 3, 3, 4.

Given: Dent counts

Count frequencies:

• 2 dents: 3 times

• 3 dents: 4 times

• 4 dents: 2 times

• 5 dents: 1 time

Mode is 3 dents (highest frequency)

Answer: 3 dents

Example 4: Calculate the variance of the panel thickness data in Example 1.

Given: Mean \(0.17\), values \(0.16, 0.18, 0.15, 0.19, 0.17\)

Calculate squared differences:

\((0.16 - 0.17)^2 = 0.0001\)

\((0.18 - 0.17)^2 = 0.0001\)

\((0.15 - 0.17)^2 = 0.0004\)

\((0.19 - 0.17)^2 = 0.0004\)

\((0.17 - 0.17)^2 = 0\)

Sum:

$$0.0001 + 0.0001 + 0.0004 + 0.0004 + 0 = 0.001$$

$$\text{Variance} = \frac{0.001}{5} = 0.0002$$

Answer: 0.0002 mm²

Example 5: Calculate the standard deviation of the data in Example 1.

Given Variance \(0.0002\),

$$\text{Standard Deviation} = \sqrt{0.0002} = 0.01414 \text{ mm}$$

Answer: 0.01414 mm

Chapter Summary

This chapter explored various methods of data presentation including histograms and line graphs for continuous variables, as well as bar and pie graphs for discrete variables. It examined grouped and ungrouped data, highlighting tools such as cumulative frequency curves and ogives to effectively display data distributions. Measures of central tendency including mean, mode, and median were calculated for both grouped and ungrouped data sets to summarize data characteristics. The chapter also covered measures of dispersion, focusing on standard deviation and variance, essential for understanding data variability in different contexts. Probability concepts were introduced with scenarios involving replacement and without replacement, illustrating fundamental probability calculations. The study of probability distribution functions included detailed treatment of binomial and Poisson distributions, explaining their applications in discrete event modeling. Finally, the chapter concluded with the normal distribution, emphasizing its importance in representing continuous data and its widespread use in statistical analysis.