Data presentation and probability form the foundation for analyzing uncertainties and measurements in mechatronics technology. In Kenyan industries such as manufacturing automation and robotics, understanding how to organize and interpret data is crucial for quality control, system diagnostics, and predictive maintenance. This chapter focuses on how continuous and discrete variables are handled, and how grouped and ungrouped data sets are presented to facilitate probability calculations relevant to mechatronics applications.

1.1 Data presentation probability

Data presentation in mechatronics involves organizing measurements and observations to reveal patterns and support decision-making. Probability quantifies the likelihood of events, essential for assessing system reliability and error rates. The nature of the data, continuous or discrete, determines the presentation method, while grouping data helps simplify large data sets for analysis.

1.1.1 Continuous variables

Continuous variables represent measurements that can take any value within a range. Examples in mechatronics include sensor voltage outputs, temperature readings, and motor speeds, which vary smoothly over intervals.

The probability density function (PDF) describes the distribution of continuous variables. The area under the PDF curve between two points equals the probability that the variable falls within that range.

The mean \(\mu\) and variance \(\sigma^2\) summarize the central tendency and spread of the continuous data.

The normal distribution is common in mechatronics for noise and error modeling.

$$ f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$

Worked Examples

Example 1: A temperature sensor in a factory records readings normally distributed with mean \(\mu = 50^\circ C\) and standard deviation \(\sigma = 2^\circ C\). Find the probability that a reading is between 48°C and 52°C.

Given: \(\mu = 50\), \(\sigma = 2\), interval \(48 \leq x \leq 52\)

Use standard normal variable: \(Z = \frac{X-\mu}{\sigma}\)

$$ Z_1 = \frac{48 - 50}{2} = -1 $$

$$ Z_2 = \frac{52 - 50}{2} = 1 $$

Probability \(P(48 \leq X \leq 52) = P(-1 \leq Z \leq 1)\)

From standard normal tables, \(P(-1 \leq Z \leq 1) = 0.6826\)

Answer: 68.26\%

Example 2: The speed of a conveyor belt motor follows a normal distribution with mean 1500 rpm and variance 100 rpm\(^2\). Calculate the probability that speed exceeds 1520 rpm.

Given: \(\mu = 1500\), \(\sigma = \sqrt{100} = 10\), find \(P(X > 1520)\)

$$ Z = \frac{1520 - 1500}{10} = 2 $$

Probability \(P(X > 1520) = P(Z > 2)\)

From tables, \(P(Z > 2) = 0.0228\)

Answer: 2.28\%

Example 3: A sensor voltage output is uniformly distributed between 0 V and 5 V. Find the probability that a reading is between 1 V and 3 V.

Given: Uniform distribution between \(a=0\), \(b=5\), find \(P(1 \leq X \leq 3)\)

$$ f(x) = \frac{1}{b, a} = \frac{1}{5 - 0} = 0.2 $$

Probability \(P = f(x) \times (3 - 1) = 0.2 \times 2 = 0.4\)

Answer: 40\%

Example 4: The lifetime of a robotic actuator is exponentially distributed with mean lifetime \(\mu = 1000\) hours. Find the probability it fails before 800 hours.

Given: \(\mu = 1000\), failure before \(t = 800\)

Exponential PDF:

$$ f(t) = \lambda e^{-\lambda t} $$

Mean lifetime \(\mu = \frac{1}{\lambda}\), so \(\lambda = \frac{1}{1000} = 0.001\)

CDF:

$$ P(T \leq t) = 1 - e^{-\lambda t} $$

Calculate:

$$ P(T \leq 800) = 1 - e^{-0.001 \times 800} = 1 - e^{-0.8} = 1 - 0.4493 = 0.5507 $$

Answer: 55.07\%

Example 5: The noise voltage in a sensor follows a normal distribution with zero mean and standard deviation 0.05 V. What is the probability that noise voltage is less than -0.1 V?

Given: \(\mu = 0\), \(\sigma = 0.05\), find \(P(X < -0.1)\)

Calculate \(Z\):

$$ Z = \frac{-0.1 - 0}{0.05} = -2 $$

Probability \(P(Z < -2) = 0.0228\)

Answer: 2.28\%

1.1.2 Discrete variable

Discrete variables take distinct, separate values, often counts of events or objects. In mechatronics, examples include the number of defective units in a batch, the count of sensor triggers, or the number of faults detected during a test cycle.

Probability mass functions (PMF) define the probability for each discrete value.

Common discrete distributions include the binomial, Poisson, and geometric distributions.

The sum of probabilities for all possible discrete outcomes equals 1.

The expected value \(E(X)\) and variance \(\sigma^2\) summarize the distribution.

$$ P(X = x) = p_x, \quad \sum p_x = 1 $$

Worked Examples

Example 1: A quality control inspector finds that the number of defective parts per batch follows a binomial distribution with \(n=10\) parts and probability of defect \(p=0.1\). Find the probability of exactly 2 defective parts.

Given: \(n=10\), \(x=2\), \(p=0.1\)

Binomial formula:

$$ P(X=x) = \binom{n}{x} p^x (1-p)^{n-x} $$

Calculate:

$$ P(2) = \binom{10}{2} (0.1)^2 (0.9)^8 = 45 \times 0.01 \times 0.4305 = 0.1937 $$

Answer: 19.37\%

Example 2: The number of faults detected in a robotic arm per hour follows a Poisson distribution with mean \(\lambda = 3\). Find the probability that exactly 5 faults occur in one hour.

Given: \(\lambda=3\), \(x=5\)

Poisson formula:

$$ P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!} $$

Calculate:

$$ P(5) = \frac{3^5 e^{-3}}{5!} = \frac{243 \times 0.0498}{120} = 0.1008 $$

Answer: 10.08\%

Example 3: A sensor triggers randomly with probability \(p=0.2\) each time a signal is sent. What is the probability that the first trigger occurs on the third signal?

Given: \(p=0.2\), \(x=3\)

Geometric distribution:

$$ P(X = x) = (1-p)^{x-1} p $$

Calculate:

$$ P(3) = (1-0.2)^{2} \times 0.2 = (0.8)^2 \times 0.2 = 0.128 $$

Answer: 12.8\%

Example 4: In a sample of 15 sensors, each has a 5% chance of failure. Find the probability that no sensor fails.

Given: \(n=15\), \(p=0.05\), \(x=0\)

Binomial:

$$ P(0) = \binom{15}{0} (0.05)^0 (0.95)^{15} = 1 \times 1 \times 0.4633 = 0.4633 $$

Answer: 46.33\%

Example 5: A mechatronics system detects faults independently with probability 0.1 each cycle. What is the probability that a fault is detected within the first 4 cycles?

Given: \(p=0.1\), \(x \leq 4\)

Calculate \(P(X \leq 4) = 1 - P(X > 4)\)

\(P(X > 4)\) means no fault in first 4 cycles:

$$ P(\text{no fault in 4 cycles}) = (1-0.1)^4 = 0.9^4 = 0.6561 $$

So,

$$ P(X \leq 4) = 1 - 0.6561 = 0.3439 $$

Answer: 34.39\%

1.1.3 Grouped data

Grouped data arises when continuous or discrete data are organized into classes or intervals to simplify analysis. In mechatronics, sensor readings or test outcomes may be grouped into ranges for pattern detection or fault categorization.

Frequency distributions summarize grouped data by showing the number of observations in each class.

Relative frequency expresses the proportion of total observations per class.

Histograms graphically display grouped data distributions.

Class midpoints and widths are used to estimate statistical measures like mean and variance.

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

Mean estimate:

$$ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} $$

where \(f_i\) is class frequency and \(x_i\) is class midpoint.

Worked Examples

Example 1: A batch of sensor voltages is grouped into classes: 0-2 V (5 readings), 2-4 V (12 readings), 4-6 V (8 readings). Find the estimated mean voltage.

Given:

Calculate mean:

$$ \bar{x} = \frac{(5 \times 1) + (12 \times 3) + (8 \times 5)}{5 + 12 + 8} = \frac{5 + 36 + 40}{25} = \frac{81}{25} = 3.24\,V $$

Answer: 3.24 V

Example 2: Grouped data of motor speeds (rpm) with frequencies: 1400-1449 (6), 1450-1499 (9), 1500-1549 (15). Calculate the relative frequency of the 1500-1549 class.

Total frequency:

$$ 6 + 9 + 15 = 30 $$

Relative frequency:

$$ \frac{15}{30} = 0.5 $$

Answer: 50\%

Example 3: For the grouped data below, find the estimated variance.

Step 1: Calculate mean \(\bar{x}\):

$$ \bar{x} = \frac{(4 \times 15) + (10 \times 25) + (6 \times 35)}{4 + 10 + 6} = \frac{60 + 250 + 210}{20} = \frac{520}{20} = 26 $$

Step 2: Calculate variance \(\sigma^2\):

$$ \sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i} $$

Calculate each term:

$$ (15 - 26)^2 = 121, \quad (25 - 26)^2 = 1, \quad (35 - 26)^2 = 81 $$

Multiply by frequencies:

$$ 4 \times 121 = 484, \quad 10 \times 1 = 10, \quad 6 \times 81 = 486 $$

Sum:

$$ 484 + 10 + 486 = 980 $$

Calculate variance:

$$ \sigma^2 = \frac{980}{20} = 49 $$

Answer: 49 units\(^2\)

Example 4: A histogram shows the following class intervals and frequencies for sensor error percentages: 0-5% (3), 5-10% (7), 10-15% (10). Calculate the estimated mean error.

Midpoints: 2.5, 7.5, 12.5

$$ \bar{x} = \frac{(3 \times 2.5) + (7 \times 7.5) + (10 \times 12.5)}{3 + 7 + 10} = \frac{7.5 + 52.5 + 125}{20} = \frac{185}{20} = 9.25\% $$

Answer: 9.25\%

Example 5: The grouped data below represents the number of faults detected in batches:

Calculate the relative frequency of batches with 2-3 faults.

Total frequency:

$$ 8 + 14 + 8 = 30 $$

Relative frequency:

$$ \frac{14}{30} = 0.467 $$

Answer: 46.7\%

1.1.4 Ungrouped data

Ungrouped data consists of raw observations or measurements listed individually without classification. In mechatronics, ungrouped data may be sensor readings collected during a test run or the number of faults recorded in each cycle.

Statistical measures such as mean, median, mode, variance, and standard deviation are calculated directly from ungrouped data.

The mean \(\bar{x}\) is the arithmetic average of all data points.

Variance \(\sigma^2\) measures data spread around the mean.

Data visualization methods include dot plots and stem-and-leaf plots.

$$ \bar{x} = \frac{\sum x_i}{n}, \quad \sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} $$

Worked Examples

Example 1: Calculate the mean of the sensor temperature readings: 45, 48, 50, 47, 46°C.

Given: \(x_i = 45, 48, 50, 47, 46\), \(n=5\)

$$ \bar{x} = \frac{45 + 48 + 50 + 47 + 46}{5} = \frac{236}{5} = 47.2^\circ C $$

Answer: 47.2°C

Example 2: Find the variance of the voltage readings: 2.1, 2.3, 2.4, 2.2, 2.5 V.

Calculate mean:

$$ \bar{x} = \frac{2.1 + 2.3 + 2.4 + 2.2 + 2.5}{5} = \frac{11.5}{5} = 2.3\,V $$

Calculate squared deviations:

$$ (2.1 - 2.3)^2 = 0.04, \quad (2.3 - 2.3)^2 = 0, \quad (2.4 - 2.3)^2 = 0.01, \quad (2.2 - 2.3)^2 = 0.01, \quad (2.5 - 2.3)^2 = 0.04 $$

Sum:

$$ 0.04 + 0 + 0.01 + 0.01 + 0.04 = 0.10 $$

Variance:

$$ \sigma^2 = \frac{0.10}{5} = 0.02\,V^2 $$

Answer: 0.02 V\(^2\)

Example 3: Calculate the median of the following fault counts: 3, 5, 2, 6, 4.

Order data: 2, 3, 4, 5, 6

Median is the middle value:

Answer: 4

Example 4: Find the mode of sensor error percentages: 1%, 2%, 3%, 2%, 4%, 2%, 5%.

Frequency of 2% is highest (3 times).

Answer: 2%

Example 5: Calculate the standard deviation of the following distances measured by a robot: 100 mm, 105 mm, 98 mm, 102 mm, 95 mm.

Step 1: Mean

$$ \bar{x} = \frac{100 + 105 + 98 + 102 + 95}{5} = \frac{500}{5} = 100\,mm $$

Step 2: Squared deviations

$$ (100 - 100)^2 = 0, \quad (105 - 100)^2 = 25, \quad (98 - 100)^2 = 4, \quad (102 - 100)^2 = 4, \quad (95 - 100)^2 = 25 $$

Sum:

$$ 0 + 25 + 4 + 4 + 25 = 58 $$

Variance:

$$ \sigma^2 = \frac{58}{5} = 11.6\,mm^2 $$

Standard deviation:

$$ \sigma = \sqrt{11.6} = 3.41\,mm $$

Answer: 3.41 mm

Chapter Summary

This chapter covered the presentation of data and the fundamentals of probability, beginning with the graphical representation of continuous variables using histograms and line graphs, and discrete variables through bar and pie charts. It explored grouped and ungrouped data presentations, including cumulative frequency and ogive curves. The chapter then examined measures of central tendency, detailing the calculation of mean, mode, and median for both grouped and ungrouped data sets. Measures of dispersion such as standard deviation, variance, range, quartile deviation, and mean deviation were introduced to describe data variability. Probability concepts were discussed with emphasis on scenarios involving replacement and without replacement. Finally, the chapter presented key probability distribution functions including binomial, Poisson, and normal distributions, which are essential for modeling and analyzing real-world engineering data. This comprehensive coverage equips students with the statistical tools necessary for data analysis and decision making in engineering contexts.