Source: ENGR/PHYS 216 Final Exam Practice Problems
Tags: normal distribution, z-score, confidence interval, moving average, finite difference, smoothing, velocity, sample size, ENGR 216, Texas A&M
This set covers three interrelated areas: smoothing noisy data with moving averages and estimating velocity via finite differences, working with the normal (Gaussian) distribution to find probabilities and standard deviations, and constructing confidence intervals for population means. These topics connect measurement and statistics to practical engineering decisions like "how many samples do I need?"
Three-point moving average
A smoothing technique that replaces each data point with the average of itself and its immediate neighbours: x̄ᵢ = (xᵢ₋₁ + xᵢ + xᵢ₊₁) / 3
Reduces noise but slightly lags or smears sharp features. Cannot be applied to the first or last data point.
Forward finite difference
An approximation of the derivative (velocity) using the current point and the next point: v(tᵢ) ≈ [x(tᵢ₊₁) − x(tᵢ)] / Δt
Other variants include backward difference and central difference.
Normal (Gaussian) distribution
The bell curve. Defined by two parameters: mean (μ) and standard deviation (σ). Many natural measurements follow this distribution. The probability density function is symmetric about the mean.
Z-score
The number of standard deviations a value lies from the mean: z = (x − μ) / σ
Converts any normal distribution to the standard normal (μ = 0, σ = 1).
Confidence interval (CI)
A range of values within which the true population mean is expected to fall, at a stated confidence level. For a known population standard deviation: CI = x̄ ± z* × (σ / √n)
Margin of error (E)
The half-width of the confidence interval: E = z* × (σ / √n)
When experimental data is noisy, you smooth first, then differentiate. Differentiating noisy data amplifies the noise, so smoothing is essential.
Worked example (Problem 6): Time and position data with Δt = 0.1 s. Smooth with three-point moving average, then find velocity at t = 0.5 s using a forward finite difference.
Step 1 – Compute smoothed positions:
x̄(0.5) = (13 + 15 + 19) / 3 = 15.667 m
x̄(0.6) = (15 + 19 + 21) / 3 = 18.333 m
Step 2 – Forward finite difference at t = 0.5:
v(0.5) = [x̄(0.6) − x̄(0.5)] / (0.6 − 0.5) = (18.333 − 15.667) / 0.1 = 26.7 m/s
If you know the population mean and the percentage of data falling within a specified range, you can work backwards to find σ.
Worked example (Problem 7): Watermelon radii: 75.4% have radii between 15.0 cm and 22.0 cm. Population mean μ = 18.5 cm.
Step 1 – Note that the interval is symmetric about the mean (18.5 − 15.0 = 3.5, 22.0 − 18.5 = 3.5).
Step 2 – For a symmetric interval: P(−z < Z < z) = 0.754, so Φ(z) = (1 + 0.754)/2 = 0.877.
Step 3 – Look up or compute: z = Φ⁻¹(0.877) = 1.16.
Step 4 – Since z = 3.5/σ: σ = 3.5 / 1.16 = 3.02 cm.
⚠️ When the interval is not symmetric about the mean, you cannot use this shortcut. You would need to solve a system with two different z-scores.
The normal distribution also lets you find threshold values for a given percentage.
Worked example (Problem 8): Air filter catches 90% of particles. Mean diameter = 0.5 μm, σ = 0.2 μm. Find the largest particle that passes through.
The filter catches the largest 90%, so the 10% smallest pass through. The largest that passes is the 10th percentile.
z = Φ⁻¹(0.10) = −1.282 x = μ + z × σ = 0.5 + (−1.282)(0.2) = 0.244 μm
A confidence interval estimates where the true population mean lies, given sample data.
Formula (known σ): CI = x̄ ± z* × (σ / √n)
Common z* values:
90% CI: z* = 1.645
95% CI: z* = 1.960
99% CI: z* = 2.576
Worked example (Problem 9): Zinc concentration: x̄ = 2.6 g/ml, σ = 0.3 g/ml, n = 36. Find 99% CI.
E = 2.576 × 0.3 / √36 = 2.576 × 0.05 = 0.129
Lower bound = 2.6 − 0.129 = 2.47 g/ml Upper bound = 2.6 + 0.129 = 2.73 g/ml
Worked example (Problem 11): Automobile km: x̄ = 23,500, σ = 3,900, n = 100. Find 99% CI.
E = 2.576 × 3900 / √100 = 2.576 × 390 = 1004.6
Lower bound = 23,500 − 1005 = 22,495 km/year Upper bound = 23,500 + 1005 = 24,505 km/year
If you want a confidence interval with a specific margin of error E, you can solve for the required sample size n.
Rearranging E = z* × σ / √n gives:
n = (z* × σ / E)²
Always round up to the next whole integer, since you cannot take a fraction of a measurement.
Worked example (Problem 10): σ = 0.3 g/ml, 95% confidence, maximum error E = 0.05 g/ml.
n = (1.960 × 0.3 / 0.05)² = (11.76)² = 138.3
Round up: n = 139 measurements
Three-point moving average: x̄ᵢ = (xᵢ₋₁ + xᵢ + xᵢ₊₁) / 3
Forward finite difference (velocity): v(tᵢ) = [x(tᵢ₊₁) − x(tᵢ)] / Δt
Z-score: z = (x − μ) / σ
Confidence interval (known σ): x̄ − z*(σ/√n) ≤ μ ≤ x̄ + z*(σ/√n)
Required sample size: n = (z* × σ / E)²
⚠️ Forward finite difference uses the next point: [x(i+1) − x(i)] / Δt. Backward uses the previous. Central uses both. Know which one the problem asks for.
⚠️ Always smooth before differentiating, never the reverse. Differentiation amplifies noise.
⚠️ The z* value depends on the confidence level. For a 99% CI, z* = 2.576 (not 2.33, which is for a one-tailed 99% test).
⚠️ When computing required sample size, always round UP. Rounding down would give a margin of error slightly larger than desired.
⚠️ These confidence interval formulas assume σ is known. If only the sample standard deviation s is available and n is small, you would use a t-distribution instead (though this exam typically uses known σ).
Q: What is the smoothed position at t = 0.4 s given raw positions of 7, 13, and 15 at t = 0.3, 0.4, and 0.5?
A: x̄(0.4) = (7 + 13 + 15) / 3 = 11.667 m.
Q: 75.4% of a normally distributed population falls within a symmetric range of 3.5 units on either side of the mean. What is σ?
A: Φ(3.5/σ) = 0.877, so 3.5/σ = 1.16, giving σ = 3.02.
Q: A 95% confidence interval has a margin of error of 0.05, and σ = 0.3. How many measurements are needed?
A: n = (1.96 × 0.3 / 0.05)² = 138.3, round up to 139.
Q: Why do we divide by √n in the confidence interval formula?
A: The standard error σ/√n represents the uncertainty in the sample mean, not in individual measurements. More measurements reduce the uncertainty in the mean.
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