Engineering Mathematics II for Automotive Mechatronics professionals in Kenya requires a solid grasp of statistical tools to analyze data and make informed decisions about vehicle diagnostics, quality control, and reliability assessments. This chapter focuses on applying statistics and probability to interpret data sets commonly encountered in automotive workshops, manufacturing, and research. Understanding measures of central tendency and dispersion enables technicians and engineers to summarize data effectively and assess variability, which is critical for troubleshooting and optimizing automotive systems.

2.1 Measures of central tendency mean, mode and median

Measures of central tendency summarize a data set by identifying a central or typical value. In automotive mechatronics, these measures help interpret sensor readings, engine performance data, and diagnostic test results. The three main measures are mean, median, and mode, each useful depending on data type and distribution. Accurate calculation and interpretation of these measures assist technicians in identifying normal operating conditions and detecting anomalies.

2.1.1 Mean (Arithmetic Mean)

The mean or arithmetic mean is the sum of all data values divided by the number of values. It provides a measure of the average value in a data set.

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

where \( \bar{x} \) is the mean, \( x_i \) are the data values, and \( n \) is the number of values.

Worked Examples

Example 1: An automotive technician records the fuel consumption (litres per 100 km) of 5 vehicles as 8.2, 7.8, 8.5, 7.9, and 8.1. Find the mean fuel consumption.

Given: \( x_1 = 8.2 \), \( x_2 = 7.8 \), \( x_3 = 8.5 \), \( x_4 = 7.9 \), \( x_5 = 8.1 \), \( n = 5 \)

$$ \bar{x} = \frac{8.2 + 7.8 + 8.5 + 7.9 + 8.1}{5} $$$$ = \frac{40.5}{5} $$$$ = 8.1 \text{ litres/100 km} $$

Answer: 8.1 litres per 100 km

Example 2: The engine temperatures (°C) recorded at different intervals are 90, 92, 89, 91, 90, 93. Calculate the mean engine temperature.

Given: \( x_i = 90, 92, 89, 91, 90, 93 \), \( n=6 \)

$$ \bar{x} = \frac{90 + 92 + 89 + 91 + 90 + 93}{6} $$$$ = \frac{545}{6} $$$$ = 90.83^\circ C $$

Answer: 90.83 °C

Example 3: A sensor records voltage values (in volts) as 12.5, 12.7, 12.6, 12.8, 12.4, 12.7, 12.9. Find the mean voltage.

Given: \( x_i = 12.5, 12.7, 12.6, 12.8, 12.4, 12.7, 12.9 \), \( n=7 \)

$$ \bar{x} = \frac{12.5 + 12.7 + 12.6 + 12.8 + 12.4 + 12.7 + 12.9}{7} $$$$ = \frac{88.6}{7} $$$$ = 12.66 \text{ volts} $$

Answer: 12.66 volts

2.1.2 Median

The median is the middle value in an ordered data set and is useful when data contains outliers or is skewed. It divides the data into two halves.

• For an odd number of values, the median is the middle value after sorting.
• For an even number of values, the median is the average of the two middle values.

Worked Examples

Example 1: Determine the median of the engine RPM readings: 1500, 1600, 1700, 1800, 1900.

Given: \( n=5 \) (odd), sorted data: 1500, 1600, 1700, 1800, 1900

Median is the 3rd value:

Answer: 1700 RPM

Example 2: Find the median of the battery voltages: 12.1, 12.3, 12.5, 12.7, 12.9, 13.0.

Given: \( n=6 \) (even), sorted data: 12.1, 12.3, 12.5, 12.7, 12.9, 13.0

Median is average of 3rd and 4th values:

$$ \text{Median} = \frac{12.5 + 12.7}{2} = \frac{25.2}{2} = 12.6 \text{ volts} $$

Answer: 12.6 volts

Example 3: Find the median of sensor error percentages: 2.1, 1.9, 3.0, 2.5, 2.8, 1.7, 2.3.

Sorted data: 1.7, 1.9, 2.1, 2.3, 2.5, 2.8, 3.0 (n=7 odd)

Median is 4th value:

Answer: 2.3 %

2.1.3 Mode

The mode is the most frequently occurring value in a data set. It is helpful in identifying the most common measurement or fault occurrence in automotive diagnostics.

Worked Examples

Example 1: Find the mode of the tire pressure readings (psi): 30, 32, 30, 31, 33, 30, 31.

Frequencies:

• 30 psi: 3 times
• 31 psi: 2 times
• 32 psi: 1 time
• 33 psi: 1 time

Mode is 30 psi.

Answer: 30 psi

Example 2: In a batch of spark plugs, the number of defects per plug is: 0, 1, 0, 2, 1, 0, 3, 1.

Frequencies:

• 0 defects: 3 times
• 1 defect: 3 times
• 2 defects: 1 time
• 3 defects: 1 time

Modes are 0 and 1 (bimodal).

Answer: 0 defects and 1 defect

Example 3: Find the mode of the coolant temperature readings (°C): 85, 87, 85, 88, 90, 87, 85, 87.

Frequencies:

• 85 °C: 3 times
• 87 °C: 3 times
• 88 °C: 1 time
• 90 °C: 1 time

Modes are 85 °C and 87 °C (bimodal).

Answer: 85 °C and 87 °C

Chapter Summary

This chapter began by exploring measures of central tendency, focusing on the mean, mode, and median as key indicators of data location. It then examined measures of dispersion, which describe the spread of data, followed by detailed calculations of variance and standard deviation to quantify variability. The concept of probability was defined to establish the foundation for understanding uncertainty, and the fundamental laws of probability were presented to govern event outcomes. The chapter further introduced expectation, variance, and standard deviation in the context of random variables, providing tools for analysis of both discrete and continuous types. Various probability distributions were studied, including the binomial, Poisson, and normal distributions, each with distinct characteristics and applications. The mean, variance, and standard deviation of these probability distributions were calculated to understand their behavior. Finally, the chapter demonstrated practical applications of probability distributions in solving engineering problems related to uncertainty and risk assessment.