Source: Final Exam Practice Problems and Solutions
Tags: normal distribution, Gaussian, z-score, probability, confidence interval, confidence level, sample size, z-alpha/2, standard deviation, mean, ENGR 216, PHYS 216
The normal distribution describes how many real-world measurements cluster around a mean. You standardise any normal variable to the Z-distribution to look up probabilities. Confidence intervals use sample data plus the known (or estimated) population standard deviation to bracket where the true mean likely falls, and the required sample size grows with the square of the precision you need.
Normal distribution (Gaussian distribution)
A symmetric, bell-shaped probability distribution fully described by its mean (μ) and standard deviation (σ). About 68% of values fall within ±1σ, 95% within ±2σ, 99.7% within ±3σ.
Z-score (standard normal variable)
Z = (X − μ) / σ. Converts any normally distributed variable to a standard normal with mean 0, standard deviation 1.
Cumulative distribution function (CDF)
P(X ≤ x), the probability that a random variable takes a value less than or equal to x. For normal distributions, this is Φ(z).
z_{α/2} (z-alpha-over-two)
The z-value such that the area in both tails combined equals α. Equivalently, Φ(z_{α/2}) = 1 − α/2. Used to build symmetric confidence intervals.
Confidence interval (CI)
A range of values, computed from sample data, that is expected to contain the true population parameter with a stated probability (confidence level).
Confidence level
The probability (e.g. 95%) that the interval-construction procedure will capture the true parameter. It is a property of the method, not of any single interval.
Sample size (N)
The number of observations. Larger N produces narrower confidence intervals for the same confidence level.
The standard approach for any "find the probability" question:
Identify μ and σ
Standardise: Z = (X − μ) / σ
Use the Z-table (or CDF) to find the probability
Given: μ = 15.0 kips, σ = 1.25 kips.
(a) P(X ≤ 18)
Z = (18 − 15) / 1.25 = 2.4, Φ(2.4) = 0.9918
(b) P(10 ≤ X ≤ 12)
Z₁ = (10 − 15) / 1.25 = −4.0, Z₂ = (12 − 15) / 1.25 = −2.4
P = Φ(−2.4) − Φ(−4.0) = 0.0082 − 0.0000 = 0.0082
(c) P(|X − 15| ≤ 2σ)
This is asking for the probability within ±2 standard deviations = 0.9544 (the well-known 95.44% rule).
Given: μ = 200 m, σ = 30 m. Damage if altitude < 100 m.
Z = (100 − 200) / 30 = −3.33, Φ(−3.33) = 0.0004
Very small probability, roughly 4 in 10,000.
Given: 10% exceed 10.256 Ω, 5% are below 9.671 Ω.
Set up two equations:
P(X > 10.256) = 0.10, so P(X ≤ 10.256) = 0.90, meaning (10.256 − μ)/σ = 1.28
P(X < 9.671) = 0.05, meaning (9.671 − μ)/σ = −1.645
From the first: μ + 1.28σ = 10.256
From the second: μ − 1.645σ = 9.671
Subtract: 2.925σ = 0.585, so σ = 0.2, μ = 10.256 − 1.28(0.2) = 10.0
Given: μ = 12 lb, σ = 3.5 lb. Want c such that 99% of packages are at least 1 lb under c, i.e. P(X ≤ c − 1) = 0.99.
z = 2.33 (for 99th percentile)
c − 1 = μ + zσ = 12 + 2.33(3.5) = 20.155
c = 21.155 lb
To find z_{α/2} for a given α:
The area in each tail is α/2
Look up the z-value where Φ(z) = 1 − α/2
α | α/2 | z_{α/2} |
|---|---|---|
0.0055 | 0.00275 | 2.775 |
0.09 | 0.045 | 1.695 |
0.663 | 0.3315 | 0.435 |
The interval is: x̄ ± z_{α/2} · (σ / √N)
The confidence level equals 1 − α. Some reverse lookups:
Interval x̄ ± 2.81(σ/√N): z_{α/2} = 2.81, α = 0.005, confidence = 99.5%
Interval x̄ ± 1.44(σ/√N): z_{α/2} = 1.44, α = 0.15, confidence = 85%
99.7% confidence: z_{α/2} = 2.96
75% confidence: z_{α/2} = 1.15
Given: σ = 0.75 (known population std dev).
(a) 95% CI, N = 20, x̄ = 4.85
z_{0.025} = 1.96
CI = 4.85 ± 1.96(0.75/√20) = 4.85 ± 0.33 = [4.52, 5.18]
(b) 98% CI, N = 16, x̄ = 4.56
z_{0.01} = 2.33
CI = 4.56 ± 2.33(0.75/√16) = 4.56 ± 0.44 = [4.12, 5.00]
(c) Required N for 95% CI width of 0.40
Width = 2 × z_{α/2} · σ/√N, so 0.40 = 2(1.96)(0.75)/√N
√N = 2(1.96)(0.75)/0.40 = 7.35, N = 54.02, round up to N = 54
(d) Required N for 99% CI with margin 0.2
0.2 = z_{0.005} · σ/√N = 2.576 × 0.75/√N
√N = 2.576(0.75)/0.2 = 9.66, N = 93.3, round up to N = 94
Given: 95% CI for cough syrup alcohol content is [7.8, 9.4], N = 50.
(a) A 90% CI from the same data would be narrower. Lower confidence means a smaller z-value, which shrinks the interval.
(b) "We can be highly confident μ is between 7.8 and 9.4" is correct. 95% confidence is quite high.
(c) "Doubling the sample size gives CI [8.2, 9.0]" is incorrect. Doubling N divides the margin of error by √2 (roughly 1.414), not by 2. You cannot predict the exact new interval because a new sample may have a different x̄.
Z-score: Z = (X − μ) / σ
Confidence interval (σ known): x̄ ± z_{α/2} · (σ / √N)
Required sample size for target margin E: N = (z_{α/2} · σ / E)², round up
Required sample size for target width W: N = (2 · z_{α/2} · σ / W)², round up
⚠️ Always round sample size up to the next whole number. You cannot take a fractional observation.
⚠️ Doubling sample size does not halve the interval width. It divides the margin by √2.
⚠️ A confidence interval is about the procedure, not the specific interval. Saying "there is a 95% probability μ is in this interval" is technically incorrect once the interval is fixed. The correct statement is about the long-run success rate of the method.
⚠️ For reverse-normal problems (finding μ and σ from percentiles), set up two simultaneous equations using the Z-values for the given percentiles.
⚠️ Know the common z-values: 1.645 (90%), 1.96 (95%), 2.33 (98%), 2.576 (99%).
Q: A parachute opens at a normally distributed altitude with μ = 200 m, σ = 30 m. What is the probability it opens below 100 m?
A: Z = (100 − 200)/30 = −3.33. Φ(−3.33) ≈ 0.0004. About a 0.04% chance.
Q: If 10% of resistors exceed 10.256 Ω and 5% are below 9.671 Ω, what are μ and σ?
A: Set up μ + 1.28σ = 10.256 and μ − 1.645σ = 9.671. Solving gives μ = 10.0 Ω, σ = 0.2 Ω.
Q: You have a 95% CI. If you wanted a 90% CI from the same data, would it be wider or narrower?
A: Narrower. A lower confidence level uses a smaller z-value, reducing the margin of error.
Q: How does doubling the sample size affect the width of a confidence interval?
A: It reduces the margin of error by a factor of √2 (approximately 1.414), not by half.
Q: What sample size is needed to estimate a mean within ±0.2 at 99% confidence if σ = 0.75?
A: N = (2.576 × 0.75 / 0.2)² = 93.3, round up to 94.
normal distribution, Gaussian, bell curve, z-score, z-table, standard normal, CDF, cumulative distribution, confidence interval, CI, confidence level, z-alpha/2, critical value, sample size calculation, margin of error, interval width, population mean, known sigma, ENGR 216 Module 4, ENGR 216 Module 5