Source: Weisberg, Applied Linear Regression, Ch. 1; STAT 5302 HW1 (Problem 1.5)
Difficulty: Introductory to Intermediate | Prerequisites: Basic probability, understanding of the normal distribution bell curve.
Tags: normal distribution, z-score, quantile, percentile, CDF, pnorm, qnorm, prediction interval, confidence interval for a new observation, STAT 5302
Most of applied regression rests on the assumption that errors (and therefore responses) are normally distributed. Before you can check that assumption or use it, you need to be fluent with normal-distribution mechanics: turning a raw value into a z-score, looking up tail probabilities, finding quantiles, and constructing prediction intervals for individual observations. Problem 1.5 in this homework drills exactly these skills using mothers' heights with a known population mean and variance.
If you know a variable is normally distributed with a given mean and variance, you can compute the probability of landing in any range by standardising to z-scores and using the standard normal CDF. Quantiles work in reverse: given a probability, find the cutoff value. A prediction interval for a single new observation is wider than a confidence interval for the mean because it accounts for individual variability, not just sampling uncertainty.
Normal distribution (Gaussian distribution)
A continuous probability distribution defined by two parameters: the mean μ (centre) and the variance σ² (spread). Its density is the familiar bell curve, symmetric about μ. In simple terms, most values cluster near the average, and extreme values become increasingly rare.
Z-score (standard score)
The number of standard deviations a value x sits from the mean: z = (x − μ) / σ. Standardising converts any normal variable to the standard normal N(0, 1). Think of it as: how many "ruler-lengths" (σ) away from the centre are you?
Cumulative distribution function (CDF)
Φ(z) = P(Z ≤ z) for the standard normal. It gives the probability of falling at or below a given value. In R: pnorm(x, mean, sd).
Quantile (inverse CDF)
The value x_p such that P(X ≤ x_p) = p. It answers: "Below what value does p proportion of the distribution fall?" In R: qnorm(p, mean, sd).
Percentile
The quantile expressed as a percentage. The 25th percentile is the 0.25 quantile.
Median
The 0.50 quantile. For a normal distribution, the median equals the mean.
First quartile (Q1)
The 0.25 quantile. 25% of the distribution lies below this value.
Prediction interval
An interval intended to contain a single future observation with a stated probability (e.g. 95%). Unlike a confidence interval for the mean, it must account for individual variability (σ²) on top of any estimation uncertainty. Think of it as: where will the next person land, as opposed to where is the average?
All of Problem 1.5 assumes mheight ~ N(μ = 62.5, σ² = 5.5). Note: the population variance is 5.5, so σ = √5.5 ≈ 2.345.
P(X < 68): Standardise: z = (68 − 62.5) / √5.5 ≈ 2.35. Then P(X < 68) = Φ(2.35). In R: pnorm(68, 62.5, sqrt(5.5)).
P(60 < X < 65): Compute P(X < 65) − P(X < 60). Standardise both bounds, look up each, and subtract. In R: pnorm(65, 62.5, sqrt(5.5)) - pnorm(60, 62.5, sqrt(5.5)).
You need v such that P(μ − v < X < μ + v) = 0.95.
By symmetry of the normal distribution, this means P(X < μ + v) = 0.975.
Standardise: (μ + v − μ) / σ = v / σ must equal z₀.₉₇₅ = 1.96.
So v = 1.96 × σ = 1.96 × √5.5 ≈ 4.60.
In R: qnorm(0.975, 0, 1) * sqrt(5.5).
The 0.25 quantile (Q1) is the value below which 25% of the distribution falls.
Q1 = μ + σ × z₀.₂₅ = 62.5 + √5.5 × (−0.6745) ≈ 60.92.
In R: qnorm(0.25, 62.5, sqrt(5.5)).
The median of a normal distribution is simply μ. Here, median = 62.5.
The first quartile is the same as the 0.25 quantile, so Q1 ≈ 60.92.
A 95% prediction interval for a randomly selected mother's height, when μ and σ² are known, is: μ ± z₀.₉₇₅ × σ.
This gives 62.5 ± 1.96 × √5.5 = 62.5 ± 4.60, or approximately (57.90, 67.10).
This is wider than a confidence interval for the mean would be, because it captures where an individual falls, not just the centre.
In R: qnorm(c(0.025, 0.975), 62.5, sqrt(5.5)).
z = \frac{x - \mu}{\sigma}P(a < X < b) = \Phi\!\left(\frac{b - \mu}{\sigma}\right) - \Phi\!\left(\frac{a - \mu}{\sigma}\right)x_p = \mu + \sigma \cdot z_p \quad \text{(quantile at probability } p\text{)}\text{95\% prediction interval (known } \sigma\text{):} \quad \mu \pm 1.96\,\sigmaManufacturers use prediction intervals to set tolerance limits: if a bolt diameter is normally distributed, the 99% prediction interval tells you the range within which virtually all individual bolts will fall, which drives quality control thresholds. Normal quantiles also underpin percentile-based growth charts in paediatrics, where a child's height or weight is plotted against population percentiles.
Students often confuse prediction intervals with confidence intervals. A CI for the mean narrows as n grows (more data, more precision about μ). A prediction interval for a single new observation does not shrink to zero because σ² remains no matter how much data you collect.
Students sometimes forget that σ² = 5.5 means σ = √5.5 ≈ 2.345, not 5.5. The variance is the square of the standard deviation. Plugging in 5.5 directly as the SD will give wrong answers.
Students occasionally think the median of any distribution is the mean. That is only true for symmetric distributions. For the normal distribution it holds, but do not generalise carelessly.
Students sometimes mix up pnorm (CDF: value to probability) and qnorm (inverse CDF: probability to value) in R. If you want a quantile, you give a probability to qnorm. If you want a probability, you give a value to pnorm.
⚠️ Know the difference between σ and σ². The homework gives the variance (5.5), not the standard deviation. Read the problem carefully.
⚠️ Be prepared to compute probabilities both by hand (z-table) and in R (pnorm, qnorm). Exams may allow one or both.
⚠️ The prediction interval for a new observation is a frequent exam question. Be ready to explain why it is wider than a CI for the mean.
⚠️ Symmetric-interval problems (find v such that P covers 95%) come up regularly. The key insight is splitting α/2 into each tail.
True or False: For a normal distribution, the mean, median, and mode are all equal. (True.)
Fill in the blank: If X ~ N(62.5, 5.5), then σ = ______. (√5.5 ≈ 2.345.)
True or False: A 95% prediction interval for a new observation will always be wider than a 95% CI for the mean (given the same data). (True.)
Fill in the blank: To find P(X > 68), you compute 1 − ______(68, 62.5, √5.5) in R. (pnorm.)
True or False: The 0.25 quantile of N(62.5, 5.5) is larger than 62.5. (False: it is below the mean.)
Q: Assume mheight ~ N(62.5, 5.5). What is the probability that a randomly selected mother is shorter than 68 inches?
A: z = (68 − 62.5) / √5.5 ≈ 2.35. From a z-table or R, Φ(2.35) ≈ 0.9906. So P(X < 68) ≈ 0.991.
Q: What is the probability that a mother's height is between 60 and 65 inches?
A: P(60 < X < 65) = Φ((65 − 62.5)/√5.5) − Φ((60 − 62.5)/√5.5) = Φ(1.066) − Φ(−1.066) ≈ 0.8569 − 0.1431 = 0.714.
Q: Find v such that P(62.5 − v < X < 62.5 + v) = 0.95.
A: You need the central 95% of the distribution, so each tail has 2.5%. v = z₀.₉₇₅ × σ = 1.96 × √5.5 ≈ 4.60 inches.
Q: What is the 0.25 quantile (first quartile) of mheight under this model?
A: Q1 = μ + σ × z₀.₂₅ = 62.5 + √5.5 × (−0.6745) ≈ 62.5 − 1.58 = 60.92 inches.
Q: Give a 95% prediction interval for the height of a randomly selected mother.
A: μ ± 1.96σ = 62.5 ± 1.96 × √5.5 = 62.5 ± 4.60, i.e. approximately (57.90, 67.10).
Normal distribution calculations feed directly into regression diagnostics. When you check Q-Q plots of residuals later in the course, you are comparing observed quantiles against theoretical normal quantiles, exactly the operation you practise here. Prediction intervals for individual observations reappear as prediction intervals for new responses in regression (with the added uncertainty from estimating the regression line). The z-to-t transition matters too: once σ² is estimated from data, you replace z-quantiles with t-quantiles, and the prediction interval widens slightly.
normal distribution, Gaussian, bell curve, z-score, standardisation, CDF, pnorm, qnorm, quantile function, inverse normal, percentile, quartile, Q1, median, prediction interval, tolerance interval, confidence interval vs prediction interval, N(62.5 5.5), mheight, alr4 Heights data, Weisberg Applied Linear Regression Ch 1, STAT 5302