5.2 Engineering Data Analysis I — Descriptive Statistics & Probability Distributions

5.2 Engineering Data Analysis I — Descriptive Statistics & Probability Distributions

Engineering Data Analysis is topic G of the Mathematics subject in the PRC Enhanced Table of Specifications (PRBEE Resolution No. 40, s. 2024). At 5.00% of the total examination and 20 of the 100 Mathematics items, it is the largest single Mathematics topic — larger than Calculus 1, Calculus 2 or Differential Equations individually. It entered the syllabus through CHED Memorandum Order No. 88, s. 2017, which made Engineering Data Analysis a required BSEE course.

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1. Measures of Central Tendency

The RMS entry is not incidental: the same statistic that engineers call "effective value" is a descriptive statistic, and the REE examination asks for it in both languages.

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2. Measures of Dispersion

Range $= x_{\max} - x_{\min}$ — quick but discards all interior information.

Variance and standard deviation — note the divisor carefully:

$$\sigma^2 = \frac{\sum (x_i - \mu)^2}{N} \quad \text{(population)} \qquad\qquad s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1} \quad \text{(sample)}$$

The $(n-1)$ divisor is Bessel's correction; it makes $s^2$ an unbiased estimator of $\sigma^2$. Applying the population formula to sample data is the most frequent arithmetic error in this topic, and the examiners supply both values as distractors.

Coefficient of variation expresses dispersion as a dimensionless ratio, letting you compare the spread of a 220 V measurement set against a 69 kV set:

$$CV = \frac{s}{\bar{x}} \times 100\%$$

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3. Discrete Probability Distributions

Binomial Distribution

Fixed $n$ independent trials, constant success probability $p$, $q = 1-p$:

$$P(X = k) = \binom{n}{k} p^{k} q^{\,n-k}, \qquad \mu = np, \qquad \sigma^2 = npq$$

Use it for: the number of defective breakers in a batch of 20, the number of successful reclosures out of 5 attempts.

Poisson Distribution

Events occurring independently at a constant average rate $\lambda$ over an interval of time, length or area:

$$P(X = k) = \frac{\lambda^{k} e^{-\lambda}}{k!}, \qquad \mu = \sigma^2 = \lambda$$

The defining signature — mean equals variance — is itself an examinable fact. Use it for: faults per 100 circuit-km per year, customer interruptions per feeder per month, lightning strikes per tower per year.

Poisson as a Limit of the Binomial

When $n$ is large and $p$ small with $np = \lambda$ moderate, the binomial converges to the Poisson. This is why rare-event reliability problems are Poisson problems.

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4. Continuous Probability Distributions

Exponential Distribution

Models the time between Poisson events, i.e. time to failure at constant hazard rate $\lambda$:

$$f(t) = \lambda e^{-\lambda t}, \qquad R(t) = P(T > t) = e^{-\lambda t}, \qquad \text{MTBF} = \frac{1}{\lambda}$$

The exponential distribution is memoryless: a component that has survived 10 000 hours has the same remaining-life distribution as a new one. This is the assumption behind constant-failure-rate reliability blocks.

Normal (Gaussian) Distribution

$$z = \frac{x - \mu}{\sigma}$$

The $z = 1.96$ and $z = 1.645$ values recur in every acceptance-limit and confidence-interval item, so commit them to memory.

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Solved Board Exam Examples

Example 1: Sample Standard Deviation of Insulation Test Readings

Five megger readings on a motor winding give 480, 495, 510, 470 and 505 MΩ. Compute the sample standard deviation.

Solution. The mean is

$$\bar{x} = \frac{480 + 495 + 510 + 470 + 505}{5} = \frac{2460}{5} = 492 \text{ M}\Omega$$

Deviations and their squares: $(-12)^2 = 144$, $(3)^2 = 9$, $(18)^2 = 324$, $(-22)^2 = 484$, $(13)^2 = 169$; the sum is $1130$.

$$s = \sqrt{\frac{1130}{5 - 1}} = \sqrt{282.5} = \boxed{16.81 \text{ M}\Omega}$$

Dividing by 5 instead of 4 would give 15.03 MΩ — the classic distractor.

Example 2: Poisson Fault Rate on a Distribution Feeder

A 13.8 kV feeder experiences an average of 3 faults per year. What is the probability of exactly 2 faults in a given year, and of no faults at all?

Solution. With $\lambda = 3$:

$$P(X = 2) = \frac{3^{2} e^{-3}}{2!} = \frac{9(0.049787)}{2} = \boxed{0.224}$$ $$P(X = 0) = \frac{3^{0} e^{-3}}{0!} = e^{-3} = \boxed{0.0498}$$

A fault-free year on this feeder occurs only about once per twenty years.

Example 3: Normal Acceptance Limit on Transformer Losses

Distribution transformer no-load losses are normally distributed with $\mu = 150$ W and $\sigma = 12$ W. A purchase specification rejects any unit exceeding 174 W. What fraction is rejected?

Solution.

$$z = \frac{174 - 150}{12} = 2.00$$

From the standard normal table, $P(Z \le 2.00) = 0.9772$, so

$$P(X > 174) = 1 - 0.9772 = \boxed{0.0228 \text{ or } 2.28\%}$$

About 23 units in every 1 000 fail the loss guarantee.