Chapter 38 — Continuous Distributions Challenge Solutions

Chapter 38 · Solutions

Continuous Distributions, Worked Answers ✅

Full solutions to the five challenges. Try them yourself first, then compare.

Statistics, Data Science and AI: A Visual Handbook · John Fisher · 2026

⚙️ Setup¶

In [1]:

import numpy as np
from scipy import stats
rng = np.random.default_rng(381)
print("ready")

ready

CHALLENGE 1 · UNIFORM

Flat probability

A bus is equally likely to arrive at any minute in a 10-minute window, X ~ Uniform(0, 10). Find P(X < 3) and the mean wait.

In [2]:

print(f"P(X < 3)  = 3/10 = {stats.uniform.cdf(3, 0, 10):.3f}")
print(f"mean      = (a+b)/2 = {stats.uniform.mean(0, 10):.1f}")

P(X < 3)  = 3/10 = 0.300
mean      = (a+b)/2 = 5.0

Answer: for a uniform, probability is proportional to length, so P(X < 3) = 3/10 = 0.30, and the mean is the midpoint (0 + 10)/2 = 5 minutes.

CHALLENGE 2 · EXPONENTIAL

Waiting times

Buses arrive on average every 10 minutes (exponential). Find P(wait > 15 min) and P(wait < 5 min).

In [3]:

scale = 10        # mean = 1/lambda = 10
print(f"P(wait > 15) = e^(-15/10) = {1 - stats.expon.cdf(15, scale=scale):.4f}")
print(f"P(wait < 5)  = 1 - e^(-5/10) = {stats.expon.cdf(5, scale=scale):.4f}")

P(wait > 15) = e^(-15/10) = 0.2231
P(wait < 5)  = 1 - e^(-5/10) = 0.3935

Answer: P(wait > 15) = e⁻¹·⁵ ≈ 0.223, and P(wait < 5) = 1 − e⁻⁰·⁵ ≈ 0.393. The exponential has a long right tail, so long waits, while less likely, are far from rare.

CHALLENGE 3 · NORMAL

Probabilities under the bell

IQ scores are modeled as X ~ Normal(100, 15). Find P(X < 115) and P(85 < X < 115).

In [4]:

mu, sd = 100, 15
print(f"P(X < 115)        = {stats.norm.cdf(115, mu, sd):.4f}")
p_mid = stats.norm.cdf(115, mu, sd) - stats.norm.cdf(85, mu, sd)
print(f"P(85 < X < 115)   = {p_mid:.4f}")

P(X < 115)        = 0.8413
P(85 < X < 115)   = 0.6827

Answer: 115 is one standard deviation above the mean, so P(X < 115) = Φ(1) ≈ 0.8413. The range 85 to 115 is exactly μ ± 1σ, so P(85 < X < 115) ≈ 0.6827, the famous 68% from the empirical rule (Chapter 39).

CHALLENGE 4 · QUANTILES

Reading the CDF backward

For the same IQ distribution N(100, 15), find the 90th percentile, the score that 90% of people fall below.

In [5]:

mu, sd = 100, 15
p90 = stats.norm.ppf(0.90, mu, sd)
print(f"90th percentile = {p90:.2f}")

90th percentile = 119.22

Answer: the 90th percentile is μ + z₀.⁹₀·σ = 100 + 1.2816(15) ≈ 119.2. The percent-point function (inverse CDF) turns a probability back into a value, the reverse of asking "what fraction is below x".

CHALLENGE 5 · FIT A GAUSSIAN

Maximum likelihood

Given a sample of measurements, estimate the parameters of a normal distribution by maximum likelihood.

In [6]:

sample = rng.normal(50, 12, size=300)
mu_hat, sigma_hat = sample.mean(), sample.std()
print(f"MLE mu_hat    = {mu_hat:.2f}")
print(f"MLE sigma_hat = {sigma_hat:.2f}")
print("Assuming Gaussian errors, minimizing squared error == maximizing this likelihood.")

MLE mu_hat    = 48.98
MLE sigma_hat = 12.52
Assuming Gaussian errors, minimizing squared error == maximizing this likelihood.

Answer: the maximum-likelihood estimates are simply the sample mean and sample standard deviation (here near 50 and 12, the values we generated from). The deeper point: assuming normally distributed errors makes least-squares fitting the maximum-likelihood choice, which is why squared-error loss is everywhere in regression and deep learning.

🎉 Nicely done

You worked with the uniform, exponential, and normal, read probabilities and percentiles off the CDF, and fit a Gaussian by maximum likelihood. Areas, not heights, are probabilities, and the normal is the distribution you will reach for most often.