IE 3301 Worked Examples

Teaching · IE 3301 Engineering Probability · Worked examples

Each section has one idea, a short explanation and something to try. The worked example is folded underneath, and it shows its work the same way every time:

Write out the equation.
List the variables: the terms in that equation.
Fill in the equation with the numbers.
Simplify, and box the answer.

Distributions: mean, variance and area
The central limit theorem
Covariance and correlation
Hypothesis tests, errors and power
Linear regression
A designed experiment

1. Distributions: mean, variance and area

Calculate the mean, variance and standard deviation of a distribution
Find a probability as a sum (discrete) or an area (continuous)
Recognize the common distributions and what they model

A discrete variable counts and has a pmf (bars that sum to 1); a continuous one measures and has a pdf (a curve with area 1), where probability is area. The mean is the balance point, $\mu = \sum x\,p(x)$ or $\int x f(x)\,dx$ , and the variance is $E[X^2] - \mu^2$ .

Try it. Pick a distribution and move its parameters. Shade a probability, or draw random samples and watch their average settle on the mean.

Shade P(X ≤ x) up to x

Worked example: defective parts in a box (discrete)

$X$ = 0, 1, 2 or 3 defective parts, with probabilities 0.50, 0.30, 0.15 and 0.05.

Equations: $\mu = \sum x\,p(x)$ , $\sigma^2 = \sum x^2 p(x) - \mu^2$ .
Variables: $x = 0, 1, 2, 3$ ; $p(x) = 0.50, 0.30, 0.15, 0.05$ (they sum to 1).
Fill in: $\mu = 0 + 0.30 + 0.30 + 0.15$ ; $\sigma^2 = (0.30 + 0.60 + 0.45) - 0.75^2$ .
Answer: $\mu = 0.75,\ \sigma^2 = 0.7875,\ \sigma \approx 0.887$

The mean is the balance point; the band is one standard deviation either side.

Worked examples: probability as area (continuous)

A. $f(x) = 3x^2$ on 0 to 1. Equation $P(X \le a) = \int_0^a f(x)\,dx$ ; fill in $\int_0^{0.5} 3x^2\,dx = 0.5^3$ ; answer 0.125. Its mean is $\int_0^1 3x^3\,dx = 0.75$ .

B. Pump life is exponential with mean 200 h. Equation $P(T > t) = e^{-t/\text{mean}}$ ; fill in $e^{-300/200}$ ; answer $P(T > 300) \approx 0.223$

The shaded areas are the answers.

The common distributions at a glance

Distribution	Models	pmf or pdf	Mean	Variance
Binomial( $n, p$ )	successes in $n$ trials	$\binom{n}{x}p^x(1-p)^{n-x}$	$np$	$np(1-p)$
Geometric( $p$ )	trials until the first success	$(1-p)^{x-1}p$	$1/p$	$(1-p)/p^2$
Poisson( $\lambda$ )	events in an interval	$e^{-\lambda}\lambda^x/x!$	$\lambda$	$\lambda$
Uniform( $a, b$ )	equally likely in a range	$1/(b-a)$	$(a+b)/2$	$(b-a)^2/12$
Exponential( $\lambda$ )	time between random events	$\lambda e^{-\lambda x}$	$1/\lambda$	$1/\lambda^2$
Normal( $\mu, \sigma^2$ )	sums and averages	$\frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2}$	$\mu$	$\sigma^2$

2. The central limit theorem

State the central limit theorem
Find the standard error of a sample mean, σ/√n
Explain when to use t instead of z

Average enough independent observations and the average is approximately normal, whatever the population looks like, with mean $\mu$ and standard error $\sigma/\sqrt n$ . That is why the normal is everywhere, and why $z = (\bar x - \mu)/(\sigma/\sqrt n)$ works. When $\sigma$ is estimated from the sample, use t instead: it has heavier tails, and approaches the normal as $n$ grows (try it in the explorer above).

Try it. Pick a population that is nothing like a bell curve, then raise the sample size. The averages turn normal anyway.

Sample size n

Worked example: bolts longer than 50.8 mm

Lengths are normal, mean 50 mm, SD 0.4 mm.

Equation: $z = (x - \mu)/\sigma$ .
Variables: $x = 50.8,\ \mu = 50,\ \sigma = 0.4$ .
Fill in: $z = 0.8/0.4 = 2$ ; look up $P(Z > 2)$ .
Answer: about 2.3% of bolts

The t distribution's heavier tails: its 95% critical value is 2.26 at 9 degrees of freedom, against 1.96 for z.

3. Covariance and correlation

Find marginal distributions from a joint table
Calculate covariance and correlation
Check whether two variables are independent

A joint table gives the probability of each pair of values. Add up a row or column for one variable on its own (the marginal). $\operatorname{Cov}(X,Y) = E[XY] - E[X]E[Y]$ is positive when the two tend to be high together; correlation $\rho = \operatorname{Cov}/(\sigma_X\sigma_Y)$ scales it to between −1 and 1. Independent means every cell equals its row total times its column total.

Try it. Move probability around the table and watch the covariance and correlation respond. Can you make them independent?

	Y = 0	Y = 1	Y = 2	total
X = 0 no overtime
X = 1 overtime
total				1

Worked example: overtime and defects

The explorer starts on this table: $X = 1$ for an overtime shift, $Y$ = defects found.

Equations: $\operatorname{Cov} = E[XY] - E[X]E[Y]$ , $\rho = \operatorname{Cov}/(\sigma_X\sigma_Y)$ .
Variables: $E[X] = 0.5,\ \sigma_X^2 = 0.25$ ; $E[Y] = 0.85,\ \sigma_Y^2 = 1.35 - 0.85^2 = 0.6275$ ; $E[XY] = (1)(1)(0.20) + (1)(2)(0.20) = 0.60$ .
Fill in: $\operatorname{Cov} = 0.60 - (0.5)(0.85)$ ; $\rho = 0.175/\sqrt{0.25 \times 0.6275}$ .
Answer: $\operatorname{Cov} = 0.175,\ \rho \approx 0.44$ . Not independent: $0.30 \ne 0.5 \times 0.4$ .

Correlation is not causation: overtime might cause defects, or a rush order might cause both. A designed experiment (section 6) is how to tell.

4. Hypothesis tests, errors and power

State the hypotheses and find the critical value
Decide with the test statistic or the confidence interval
Explain Type I and Type II errors, the p-value and power

The critical values are the bounds of the confidence interval; the rejection region is everything outside. Reject $H_0$ if the test statistic falls outside them, or, the same thing, if the confidence interval misses $\mu_0$ . For a two-sided test, put $\alpha/2$ in each tail.

	H₀ is true	H₀ is false
Reject H₀	Type I error (probability α)	correct: power, 1 − β
Fail to reject H₀	correct	Type II error (probability β)

Try it. Move the true mean away from 50 and watch the power grow. Then see what a bigger sample or a looser α does.

True mean μ Sample size n σ α

Worked example: is the bolt machine off target?

Ten bolts (mm): 50.3, 49.8, 50.6, 50.9, 50.2, 50.7, 49.9, 50.5, 50.8, 50.4. Target 50 mm, α = 0.05.

Hypotheses: $H_0: \mu = 50$ , $H_1: \mu \ne 50$ .
Equation: $t = (\bar x - \mu_0)/(s/\sqrt n)$ (σ unknown, so t).
Variables: $\bar x = 50.41,\ s = 0.3665,\ n = 10$ ; critical value $t_{0.025,\,9} = 2.262$ .
Fill in: $t = 0.41/(0.3665/\sqrt{10}) = 0.41/0.1159$ .
Answer: $t = 3.54 > 2.262$ : reject H₀ (p ≈ 0.006). Same by the interval: $50.41 \pm 2.262(0.1159) = (50.15, 50.67)$ misses 50.

The rejection region holds 5% of the t distribution; the observed t lands inside it.

5. Linear regression

Fit a least-squares line
Interpret the slope and R²
Check the residuals

Least squares picks the line that makes the squared vertical distances to the points (the residuals) as small as possible: $b_1 = S_{xy}/S_{xx}$ , $b_0 = \bar y - b_1\bar x$ . $R^2$ is the share of the variation the line explains.

Try it. Click the chart to add points. Then try to beat least squares with your own slope.

Your slope (through the means)

Worked example: reactor temperature and yield

Equations: $b_1 = S_{xy}/S_{xx}$ , $b_0 = \bar y - b_1\bar x$ , $R^2 = 1 - \text{SSE}/\text{SST}$ .
Variables: $\bar x = 185,\ \bar y = 69.61,\ S_{xx} = 4200,\ S_{xy} = 873.5$ .
Fill in: $b_1 = 873.5/4200$ ; $b_0 = 69.61 - 0.208(185)$ .
Answer: $\hat y = 31.14 + 0.208x,\ R^2 = 0.987$ . The slope test gives $t \approx 21 > 2.447$ : temperature matters.

The line and its residuals; use it only inside the data, 150 to 220 °C.

6. A designed experiment

Lay out a 2×2 factorial design
Calculate main effects and the interaction
Read the same results as a regression

Regression on data you happened to collect shows association; a designed experiment sets the factors, so it can show cause. Run every combination of two factors at two levels (coded −1 and +1). An effect is the average at + minus the average at −, and the interaction is the same for the product column.

Try it. Set the yield at each of the four runs. Make the lines parallel to kill the interaction, or cross them.

Run 1: A low, B low Run 2: A high, B low Run 3: A low, B high Run 4: A high, B high

Worked example: temperature and time

Equation: effect = (average where the column is +) − (average where it is −).
Variables: yields 60, 72, 64, 84 at (A, B) = (−, −), (+, −), (−, +), (+, +).
Fill in: A: $(72+84)/2 - (60+64)/2$ ; B: $(64+84)/2 - (60+72)/2$ ; AB: $(60+84)/2 - (72+64)/2$ .
Answer: A = 16, B = 8, AB = 4; as a regression, $\hat y = 70 + 8x_1 + 4x_2 + 2x_1x_2$ .

Parallel lines would mean no interaction; these spread apart.

More worked material: statistics basics and linear regression and the basics of simulation, in Python.

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