Chapter 39 — Normal Distribution Challenge Solutions

Chapter 39 · Solutions

The Normal Distribution, 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
print("ready")

ready

CHALLENGE 1 · EMPIRICAL RULE

Reading 68 - 95 - 99.7

Adult heights are Normal(170 cm, 8 cm). Without a calculator, what fraction of people are between 162 and 178 cm? Between 154 and 186 cm?

In [2]:

mu, sd = 170, 8
print(f"162 to 178 is mu +/- 1 sigma -> about 68%   (exact {(stats.norm.cdf(178,mu,sd)-stats.norm.cdf(162,mu,sd))*100:.1f}%)")
print(f"154 to 186 is mu +/- 2 sigma -> about 95%   (exact {(stats.norm.cdf(186,mu,sd)-stats.norm.cdf(154,mu,sd))*100:.1f}%)")

162 to 178 is mu +/- 1 sigma -> about 68%   (exact 68.3%)
154 to 186 is mu +/- 2 sigma -> about 95%   (exact 95.4%)

Answer: 162 to 178 is μ ± 1σ, so about 68%; 154 to 186 is μ ± 2σ, so about 95%. The empirical rule lets you answer these in your head, no z-table required.

CHALLENGE 2 · z-SCORE

How many standard deviations?

On a test with mean 75 and standard deviation 5, a student scores 85. Find the z-score and the percentile.

In [3]:

z = (85 - 75) / 5
print(f"z = (85 - 75)/5 = {z:.1f}")
print(f"percentile = P(Z < 2) = {stats.norm.cdf(z)*100:.1f}%")

z = (85 - 75)/5 = 2.0
percentile = P(Z < 2) = 97.7%

Answer: z = (85 − 75)/5 = 2.0, so the score is two standard deviations above the mean. That places the student at the 97.7th percentile, higher than about 98% of test-takers.

CHALLENGE 3 · COMPARE VIA z

Apples to oranges

Ana scored 88 on Exam A (mean 80, sd 4). Ben scored 90 on Exam B (mean 85, sd 10). Whose result is more impressive relative to peers?

In [4]:

zA = (88 - 80) / 4
zB = (90 - 85) / 10
print(f"Ana: z = {zA:.2f}")
print(f"Ben: z = {zB:.2f}")
print("Ana is stronger relative to her exam" if zA>zB else "Ben is stronger")

Ana: z = 2.00
Ben: z = 0.50
Ana is stronger relative to her exam

Answer: Ana's z = 2.0 versus Ben's z = 0.5. Although Ben's raw score is higher, Ana is far more exceptional relative to her exam (two standard deviations above the mean versus half of one). The z-score, not the raw number, is the fair comparison.

CHALLENGE 4 · TAIL PROBABILITY

Beyond the mean

IQ scores are Normal(100, 15). What fraction of people score above 130 (the 'gifted' threshold)?

In [5]:

mu, sd = 100, 15
z = (130 - mu)/sd
print(f"z = (130-100)/15 = {z:.1f}")
print(f"P(X > 130) = 1 - P(Z < 2) = {(1-stats.norm.cdf(z))*100:.2f}%")

z = (130-100)/15 = 2.0
P(X > 130) = 1 - P(Z < 2) = 2.28%

Answer: 130 is z = 2 above the mean, so P(X > 130) = 1 − Φ(2) ≈ 2.28%. By the empirical rule, about 2.5% lie beyond +2σ, matching the exact 2.28%.

CHALLENGE 5 · STANDARDIZE A DATASET

The ML preprocessing step

Standardize the small dataset [10, 12, 14, 16, 18] to z-scores (mean 0, std 1), the way a model would before training.

In [6]:

x = np.array([10, 12, 14, 16, 18])
z = (x - x.mean()) / x.std()
print(f"mean = {x.mean()}, std = {x.std():.3f}")
print(f"z-scores = {z.round(3)}")
print(f"check: mean {z.mean():.3f}, std {z.std():.3f}")

mean = 14.0, std = 2.828
z-scores = [-1.414 -0.707  0.     0.707  1.414]
check: mean 0.000, std 1.000

Answer: the mean is 14 and the standard deviation is √8 ≈ 2.83, giving z-scores [−1.41, −0.71, 0, 0.71, 1.41]. The standardized data has mean 0 and std 1, exactly what StandardScaler produces, and the most common feature-preprocessing step in machine learning.

🎉 Nicely done

You applied the empirical rule, computed and compared z-scores, found a tail probability, and standardized a dataset. The z-score, how many standard deviations from the mean, ties together percentiles, comparisons, anomaly detection, and ML preprocessing.