Chapter 8 · Solutions
Practice Challenges, Worked Answers ✅
Full solutions to the five "Measures of Central Tendency" 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
import pandas as pd
print("✅ Ready.")
✅ Ready.
CHALLENGE 1 · THE THREE AVERAGES
Compute mean, median, mode
For x = [4, 8, 6, 5, 8, 9, 8, 3], compute the mean, median, and mode.
In [2]:
x = pd.Series([4, 8, 6, 5, 8, 9, 8, 3])
print(f"Sorted : {sorted(x)}")
print(f"Mean = {x.mean():.2f}")
print(f"Median = {x.median():.1f}")
print(f"Mode = {x.mode().iloc[0]}")
Sorted : [3, 4, 5, 6, 8, 8, 8, 9] Mean = 6.38 Median = 7.0 Mode = 8
Answer: Mean = 6.375, Median = 7.0 (average of the two middle values 6 and 8), Mode = 8 (appears three times).
CHALLENGE 2 · THE OUTLIER TEST
Which average changes more?
Start with [10, 12, 11, 13, 12]. Add the value 100. Recompute the mean and median, and say which one the outlier affected more.
In [3]:
base = pd.Series([10, 12, 11, 13, 12])
with_out = pd.concat([base, pd.Series([100])], ignore_index=True)
print(f"Before: mean {base.mean():.1f}, median {base.median():.1f}")
print(f"After : mean {with_out.mean():.1f}, median {with_out.median():.1f}")
Before: mean 11.6, median 12.0 After : mean 26.3, median 12.0
Answer: the mean jumped from 11.6 to about 26.3, while the median moved only from 12 to 12. The outlier hammered the mean and barely touched the median, so the median is the better summary here.
CHALLENGE 3 · SKEW DIRECTION
Read the skew from mean vs median
A class's exam scores have mean = 72 and median = 78. Is the distribution left-skewed, right-skewed, or symmetric? What does that say about the scores?
In [4]:
mean, median = 72, 78
if mean < median: skew = "LEFT-skewed (negative): a tail of low scores pulls the mean down"
elif mean > median: skew = "RIGHT-skewed (positive): a tail of high scores pulls the mean up"
else: skew = "symmetric"
print(f"mean {mean} < median {median} -> {skew}")
mean 72 < median 78 -> LEFT-skewed (negative): a tail of low scores pulls the mean down
Answer: Since mean (72) < median (78), the data is left-skewed: a few low scorers stretch the left tail and drag the mean below the median. Most students did fairly well.
CHALLENGE 4 · GEOMETRIC MEAN
Average a growth rate
A company's revenue grows 20% in year 1, 5% in year 2, and falls 10% in year 3. What is the average annual growth rate? (Hint: use the geometric mean of the growth factors.)
In [5]:
factors = np.array([1.20, 1.05, 0.90]) # +20%, +5%, -10%
gm = np.exp(np.mean(np.log(factors))) # geometric mean of the factors
print(f"Average annual growth factor = {gm:.4f}")
print(f"Average annual growth rate = {(gm-1)*100:.1f}%")
print(f"(The naive arithmetic average would wrongly give {(factors.mean()-1)*100:.1f}%.)")
Average annual growth factor = 1.0428 Average annual growth rate = 4.3% (The naive arithmetic average would wrongly give 5.0%.)
Answer: Average annual growth ≈ 4.3%. Growth compounds, so you multiply the factors and take the cube root (the geometric mean), not the arithmetic average.
CHALLENGE 5 · GROUPED DATA
Mean from a frequency table
A survey groups response times into bins with midpoints [10, 20, 30, 40] seconds and frequencies [2, 5, 8, 5]. Estimate the mean response time.
In [6]:
mid = np.array([10, 20, 30, 40])
freq = np.array([2, 5, 8, 5])
mean = np.sum(mid * freq) / np.sum(freq)
print(f"People : {freq.sum()}")
print(f"Estimated mean time: {mean:.1f} seconds")
People : 20 Estimated mean time: 28.0 seconds
Answer: Estimated mean ≈ 28.5 seconds, using Σ(midpoint × frequency) / Σ(frequency). With only grouped counts, the class midpoints stand in for the raw values.
🎉 Nicely done!
You computed all three averages, watched an outlier wreck the mean, read skew from mean vs median, used the geometric mean for growth, and averaged a frequency table. That is the full toolkit for describing the center of a dataset.
Statistics, Data Science and AI: A Visual Handbook · © 2026 John Fisher