---
title: "Descriptive Statistics"
book: "High School Mathematics"
subject: math
language: en
chapter: 8
exercises: 9
source: https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics
---

# Chapter 8 — Descriptive Statistics

Statistics summarizes a large collection of numbers — grades, salaries, temperatures — by a few well-chosen indicators. This chapter introduces the two families of indicators: those locating the *center* of the data ([mean](#def-g10-stats-mean), [median](#def-g10-stats-median)) and those measuring its *spread* ([range](#def-g10-stats-quartiles), [quartiles](#def-g10-stats-quartiles)). A deeper treatment, including the standard deviation, comes in [Chapter 17](https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics#ch-g11-stat).

## 8.1 Data and frequencies

**Definition 8.1 (Statistical series).**

A *statistical series* is a list of values $x_1, x_2, \dots, x_N$ observed on a population of size $N$. When a value $x$ appears $n$ times, $n$ is its *count* and the quotient

$$
f = \frac{n}{N}
$$

is its *frequency* (often expressed as a percentage). The frequencies of all values add up to $1$.

**Example 8.2.**

The grades of $20$ students on a $5$-point quiz:

| grade $x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| --- | --- | --- | --- | --- | --- | --- |
| count $n$ | $1$ | $2$ | $4$ | $6$ | $5$ | $2$ |
| [frequency](#def-g10-stats-series) $f$ | $0.05$ | $0.10$ | $0.20$ | $0.30$ | $0.25$ | $0.10$ |

Check: $1 + 2 + 4 + 6 + 5 + 2 = 20$ and the frequencies sum to $1$.

![The bar chart of the quiz grades: one bar per value, with height equal to the count.](https://one-course.com/images/onecourse/chapters/math-2/g10-stats/fig-4e073399710b.svg)

*The bar chart of the quiz grades: one bar per value, with height equal to the count.*

## 8.2 Measures of center

**Definition 8.3 (Mean).**

The *mean* of the series is

$$
\bar x = \frac{x_1 + x_2 + \dots + x_N}{N}.
$$

If the values $v_1, \dots, v_k$ appear with counts $n_1, \dots, n_k$, the same number is computed as the *weighted mean*

$$
\bar x = \frac{n_1 v_1 + n_2 v_2 + \dots + n_k v_k}{n_1 + n_2 + \dots + n_k}.
$$

**Example 8.4.**

For the quiz of [Example 8.2](#ex-g10-stats-frequencies):

$$
\bar x = \frac{1 \times 0 + 2 \times 1 + 4 \times 2 + 6 \times 3
+ 5 \times 4 + 2 \times 5}{20}
= \frac{0 + 2 + 8 + 18 + 20 + 10}{20}
= \frac{58}{20} = 2.9 .
$$

**Definition 8.5 (Median).**

The *median* is a value that splits the *sorted* series in two halves: at least half the values are $\leq$ the median, and at least half are $\geq$ it. In practice, sort the $N$ values;

- if $N$ is odd, the median is the middle value, in position $\frac{N+1}{2}$ ;
- if $N$ is even, take the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of the two middle values, in positions $\frac N2$ and $\frac N2 + 1$ .

**Example 8.6.**

Sorted series of $7$ house prices (in thousands): $120$, $150$, $160$, $180$, $210$, $240$, $900$. The [median](#def-g10-stats-median) is the $4$th value, $180$. The [mean](#def-g10-stats-mean) is $\frac{1960}{7} = 280$ — larger than $6$ of the $7$ prices! One extreme value ($900$) pulls the [mean](#def-g10-stats-mean) far more than the [median](#def-g10-stats-median): the [median](#def-g10-stats-median) is *robust*, the [mean](#def-g10-stats-mean) is not.

## 8.3 Measures of spread

**Definition 8.7 (Range, quartiles).**

For a sorted series:

- the *range* is the difference between the largest and smallest values;
- the *first quartile* $Q_1$ is the smallest value such that at least a quarter ( $25\%$ ) of the values are $\leq Q_1$ ; the *third quartile* $Q_3$ is the smallest value such that at least three quarters ( $75\%$ ) of the values are $\leq Q_3$ ;
- the *interquartile range* $Q_3 - Q_1$ measures the spread of the central half of the data.

**Method 8.8 (Finding the quartiles).**

For a series of $N$ sorted values:

1. compute $\frac N4$ ; if it is not an [integer](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) round it *up* to the next [integer](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) ; $Q_1$ is the value in that position;
2. compute $\frac{3N}{4}$ ; round up if needed; $Q_3$ is the value in that position.

**Example 8.9.**

Take the $11$ sorted values

$$
2,\ 3,\ 5,\ 5,\ 6,\ 7,\ 8,\ 9,\ 9,\ 10,\ 12 .
$$

[Median](#def-g10-stats-median): position $\frac{11+1}{2} = 6$, so the [median](#def-g10-stats-median) is $7$. $Q_1$: $\frac{11}{4} = 2.75$, round up to $3$; the $3$rd value is $5$. $Q_3$: $\frac{33}{4} = 8.25$, round up to $9$; the $9$th value is $9$. [Range](#def-g10-stats-quartiles): $12 - 2 = 10$; [interquartile range](#def-g10-stats-quartiles): $9 - 5 = 4$.

![The box plot of : the box spans the interquartile interval (Q_1, Q_3) and contains the central half of the data; the whiskers reach the extreme values.](https://one-course.com/images/onecourse/chapters/math-2/g10-stats/fig-4aab36cc4a94.svg)

*The box plot of [Example 8.9](#ex-g10-stats-quartiles): the box spans the interquartile [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $\intcc{Q_1}{Q_3}$ and contains the central half of the data; the whiskers reach the extreme values.*

**Example 8.10 (Comparing two groups).**

Two classes take the same test (scores out of $20$):

|  | [median](#def-g10-stats-median) | $Q_1$ | $Q_3$ |
| --- | --- | --- | --- |
| class A | $12$ | $10$ | $14$ |
| class B | $12$ | $6$ | $17$ |

Same [median](#def-g10-stats-median), very different spreads: class A is homogeneous (half its scores in $\intcc{10}{14}$), class B is heterogeneous (its central half spans $\intcc{6}{17}$). Indicators of center alone never tell the whole story.

**Remark 8.11.**

Neither the [mean](#def-g10-stats-mean) nor the [median](#def-g10-stats-median) is “better”: the [mean](#def-g10-stats-mean) uses every value (and is needed to compute totals), the [median](#def-g10-stats-median) resists extreme values. A serious summary of a series gives at least one indicator of center *and* one of spread.

## 8.4 Exercises

**Exercise 8.1 ★.**

Here are the numbers of books read in a year by $15$ students:

$$
0,\ 1,\ 1,\ 2,\ 2,\ 2,\ 3,\ 3,\ 4,\ 4,\ 5,\ 6,\ 7,\ 8,\ 12 .
$$

Compute the [mean](#def-g10-stats-mean), the [median](#def-g10-stats-median) and the [range](#def-g10-stats-quartiles) of this series.

**Solution of Exercise 8.1.**

Sum: $0+1+1+2+2+2+3+3+4+4+5+6+7+8+12 = 60$, so the [mean](#def-g10-stats-mean) is $\frac{60}{15} = 4$. The series is already sorted and $N = 15$ is odd: the [median](#def-g10-stats-median) is the $8$th value, $3$. [Range](#def-g10-stats-quartiles): $12 - 0 = 12$.

**Exercise 8.2 ★.**

A die was rolled $50$ times:

| outcome | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
| --- | --- | --- | --- | --- | --- | --- |
| count | $7$ | $9$ | $8$ | $10$ | $6$ | $10$ |

Compute the [frequency](#def-g10-stats-series) of each outcome and the [mean](#def-g10-stats-mean) outcome.

**Solution of Exercise 8.2.**

Frequencies: $\frac{7}{50} = 0.14$, $\frac{9}{50} = 0.18$, $\frac{8}{50} = 0.16$, $\frac{10}{50} = 0.20$, $\frac{6}{50} = 0.12$, $\frac{10}{50} = 0.20$ (sum $1$). [Mean](#def-g10-stats-mean) outcome:

$$
\bar x = \frac{7 \times 1 + 9 \times 2 + 8 \times 3 + 10 \times 4
+ 6 \times 5 + 10 \times 6}{50}
= \frac{7 + 18 + 24 + 40 + 30 + 60}{50}
= \frac{179}{50} = 3.58 .
$$

**Exercise 8.3 ★.**

Find the [median](#def-g10-stats-median), $Q_1$ and $Q_3$ of the sorted series

$$
1,\ 2,\ 4,\ 4,\ 5,\ 6,\ 6,\ 7,\ 8,\ 9,\ 9,\ 11 \qquad (N = 12),
$$

and draw the corresponding box plot.

**Solution of Exercise 8.3.**

$N = 12$ (even): the [median](#def-g10-stats-median) is the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of the $6$th and $7$th values, both equal to $6$: [median](#def-g10-stats-median) $6$.

$Q_1$: $\frac{12}{4} = 3$, so $Q_1$ is the $3$rd value: $4$.

$Q_3$: $\frac{3 \times 12}{4} = 9$, so $Q_3$ is the $9$th value: $8$.

Box plot: box from $4$ to $8$ with the [median](#def-g10-stats-median) bar at $6$, whiskers from $1$ to $11$.

**Exercise 8.4 ★.**

The [mean](#def-g10-stats-mean) of $24$ test scores is $11.5$. A $25$th student takes the test and scores $19$. What is the new [mean](#def-g10-stats-mean) of the class?

**Solution of Exercise 8.4.**

The $24$ scores total $24 \times 11.5 = 276$. With the new score the total is $276 + 19 = 295$ for $25$ students: new [mean](#def-g10-stats-mean) $\frac{295}{25} = 11.8$.

**Exercise 8.5 ★★.**

A student has a [mean](#def-g10-stats-mean) of $12$ after $5$ tests, all with the same weight. What score on the $6$th test would raise the [mean](#def-g10-stats-mean) to $13$? Is a [mean](#def-g10-stats-mean) of $14$ reachable (scores are out of $20$)?

**Solution of Exercise 8.5.**

Current total: $5 \times 12 = 60$. A [mean](#def-g10-stats-mean) of $13$ over $6$ tests requires a total of $78$, so the $6$th score must be $78 - 60 = 18$. A [mean](#def-g10-stats-mean) of $14$ would require $84 - 60 = 24 > 20$: not reachable.

**Exercise 8.6 ★★.**

Invent a series of $7$ nonnegative values whose [median](#def-g10-stats-median) is $10$ and whose [mean](#def-g10-stats-mean) is larger than $20$; then one whose [median](#def-g10-stats-median) is $10$ and whose [mean](#def-g10-stats-mean) is smaller than $6$. Why can the [mean](#def-g10-stats-mean) of such a series never be smaller than $\frac{40}{7}$? What do these examples show?

**Solution of Exercise 8.6.**

[Mean](#def-g10-stats-mean) larger than $20$: for instance $1,\ 2,\ 3,\ 10,\ 40,\ 50,\ 60$ — [median](#def-g10-stats-median) $10$ (the $4$th sorted value), [mean](#def-g10-stats-mean) $\frac{166}{7} \approx 23.7$.

[Mean](#def-g10-stats-mean) smaller than $6$: for instance $0,\ 0,\ 0,\ 10,\ 10,\ 10,\ 10$ — [median](#def-g10-stats-median) $10$, [mean](#def-g10-stats-mean) $\frac{40}{7} \approx 5.7$.

The [mean](#def-g10-stats-mean) cannot go below $\frac{40}{7}$: for the [median](#def-g10-stats-median) (the $4$th sorted value) to be $10$, the values in positions $4$ to $7$ must all be $\geq 10$, so the total is at least $40$ and the [mean](#def-g10-stats-mean) at least $\frac{40}{7}$ — the second example is extremal.

These examples show that the [mean](#def-g10-stats-mean) and the [median](#def-g10-stats-median) are largely independent: knowing one says little about the other, which is why a good summary reports both.

**Exercise 8.7 ★★.**

In a company, the $9$ employees earn $1800$ each per month and the director earns $12000$.

1. Compute the [mean](#def-g10-stats-mean) and the [median](#def-g10-stats-median) salary.
2. Which indicator best describes a “typical” salary here? Why?

**Solution of Exercise 8.7.**

*1.* [Mean](#def-g10-stats-mean): $\frac{9 \times 1800 + 12000}{10} = \frac{16200 + 12000}{10} = 2820$. [Median](#def-g10-stats-median): sorted, the $10$ salaries are nine values $1800$ then $12000$; the [median](#def-g10-stats-median) is the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of the $5$th and $6$th values, both $1800$: [median](#def-g10-stats-median) $1800$.

*2.* The [median](#def-g10-stats-median) ($1800$) describes the typical salary: it is what $9$ employees out of $10$ actually earn. The [mean](#def-g10-stats-mean) ($2820$) is pulled up by the single high salary and matches nobody’s payslip.

**Exercise 8.8 ★★.**

A class of $12$ boys has a [mean](#def-g10-stats-mean) height of $168$ cm; the $18$ girls of the same class have a [mean](#def-g10-stats-mean) height of $163$ cm. Compute the [mean](#def-g10-stats-mean) height of the whole class. (Careful: it is *not* the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $168$ and $163$ — weight the means by the group sizes.)

**Solution of Exercise 8.8.**

Total height: $12 \times 168 + 18 \times 163 = 2016 + 2934 = 4950$ cm for $30$ students, so the [mean](#def-g10-stats-mean) is $\frac{4950}{30} = 165$ cm. It is closer to the girls’ [mean](#def-g10-stats-mean) because they are more numerous (weighted [mean](#def-g10-stats-mean), weights $12$ and $18$).

**Exercise 8.9 ★★★.**

A series of $N$ values $x_1, \dots, x_N$ has [mean](#def-g10-stats-mean) $\bar x$. Each value is transformed into $y_i = a x_i + b$ for fixed numbers $a$ and $b$.

1. Show that the [mean](#def-g10-stats-mean) of the new series is $a \bar x + b$ .
2. The temperatures of a week, in degrees Celsius, have [mean](#def-g10-stats-mean) $20$ . What is their [mean](#def-g10-stats-mean) in degrees Fahrenheit ( $F = 1.8\,C + 32$ )?

**Solution of Exercise 8.9.**

*1.* The new [mean](#def-g10-stats-mean) is

$$
\bar y = \frac{y_1 + \dots + y_N}{N}
= \frac{(a x_1 + b) + \dots + (a x_N + b)}{N}
= \frac{a(x_1 + \dots + x_N) + Nb}{N}
= a \bar x + b .
$$

*2.* With $a = 1.8$ and $b = 32$: [mean](#def-g10-stats-mean) $1.8 \times 20 + 32 = 68$ degrees Fahrenheit.

## 8.5 Problem: The average person does not exist

**Problem 8.1.**

Weekend problem — mean against median: outliers, grading curves, the warehouse on the road, and the cockpit designed for nobody

In 1950 the US Air Force measured $4\,000$ pilots on ten body dimensions and built the cockpit for the *average man*. It fitted — as one lieutenant discovered by counting — essentially nobody. Statistics summarizes crowds with single numbers, and each summary tells the truth about something and lies about something else. This problem takes the [mean](#def-g10-stats-mean) and the [median](#def-g10-stats-median) ([Definition 8.3](#def-g10-stats-mean), [Definition 8.5](#def-g10-stats-median)) to court, lets each show what it is best at, and closes the cockpit case.

**Part I — Two summaries on trial.**

1. A street sells seven houses, in thousands of euros: $200$ , $210$ , $220$ , $230$ , $240$ , $250$ — and one mansion at $1\,200$ . Compute the [mean](#def-g10-stats-mean) and the [median](#def-g10-stats-median) price. Which number describes “a house in this street”?
2. Remove the mansion and recompute both. By how much did each summary move? State the moral about outliers.
3. A census table: out of $100$ families, $20$ have no child, $25$ have one, $30$ have two, $15$ have three, $10$ have four. Compute the [mean](#def-g10-stats-mean) number of children. No family “has $1.7$ children” — so what real quantity does the [mean](#def-g10-stats-mean) encode? (Multiply it by $100$ .)
4. For the fifteen marks $4, 6, 7, 8, 9, 10, 10, 11, 12, 12, 13, 14, 15, 17, 19$ : find the [median](#def-g10-stats-median) and the [quartiles](#def-g10-stats-quartiles) $Q_1$ , $Q_3$ ( [Method 8.8](#met-g10-stats-quartiles) ).
5. For the same series, give the [range](#def-g10-stats-quartiles) and the [interquartile range](#def-g10-stats-quartiles) . What does each measure?

**Part II — Grading curves and merged classes.**

6. From [Exercise 8.9](#exo-g10-stats-9) : shifting every value by $+b$ shifts [mean](#def-g10-stats-mean) , [median](#def-g10-stats-median) and [quartiles](#def-g10-stats-quartiles) by $+b$ ; scaling by $a$ scales them by $a$ . What happens to the [range](#def-g10-stats-quartiles) and the [interquartile range](#def-g10-stats-quartiles) under a shift? Under a scaling? Justify.
7. A test has [mean](#def-g10-stats-mean) $9$ (out of $20$ ) and the teacher wants [mean](#def-g10-stats-mean) $12$ . Two curves are proposed: add $3$ points to everyone, or multiply every mark by $\frac43$ . Compute the fate of a student at $6$ and one at $18$ under each curve. Which curve favors whom — and which has a ceiling problem?
8. Class A ( $20$ students) has [mean](#def-g10-stats-mean) $12$ ; class B ( $30$ students) has [mean](#def-g10-stats-mean) $10$ . Compute the [mean](#def-g10-stats-mean) of the $50$ students together, and explain why it is not $11$ (the weighted-average lesson of the [harmonic-mean](#def-g10-stats-mean) weekend problem of the Middle School volume, in the classroom).
9. Show that [medians](#def-g10-stats-median) refuse this game: compute the [medians](#def-g10-stats-median) of $A = \{1, 2, 3\}$ and $B = \{0, 10\}$ , then the [median](#def-g10-stats-median) of the merged list, and compare with the size-weighted average of the two [medians](#def-g10-stats-median) .
10. Figure skating: five judges score $5.2$ , $5.3$ , $5.4$ , $5.5$ , $9.9$ . Compute the [mean](#def-g10-stats-mean) , the [median](#def-g10-stats-median) , and the *trimmed [mean](#def-g10-stats-mean)* (drop the lowest and highest, average the rest). Why do most judged sports trim?

**Part III — Reading distributions.**

11. Two factories both produce bolts of [mean](#def-g10-stats-mean) length $50$ mm. Factory A: $Q_1 = 49.9$ , $Q_3 = 50.1$ ; factory B: $Q_1 = 48$ , $Q_3 = 52$ . You need interchangeable bolts: which factory, and which statistic decided?
12. At a clinic, the [median](#def-g10-stats-median) wait is $10$ minutes but the [mean](#def-g10-stats-mean) wait is $25$ . What shape of data does this pair betray? Construct a five-value dataset with exactly this [median](#def-g10-stats-median) and [mean](#def-g10-stats-mean) .
13. “Your baby is at the 75th percentile for height” — what does that sentence [mean](#def-g10-stats-mean) ? And can every child of a town be “above average”? Above the [median](#def-g10-stats-median) ? Explain the difference.
14. Class A: min $4$ , $Q_1\,8$ , [median](#def-g10-stats-median) $11$ , $Q_3\,13$ , max $17$ . Class B: min $2$ , $Q_1\,9$ , [median](#def-g10-stats-median) $11$ , $Q_3\,12$ , max $19$ . Compare: typical mark, homogeneity of the middle half, wildness of the extremes.
15. The journalist’s checklist: a headline says “average salary at TechCorp: $5\,000$ euros”. List the three questions this problem has taught you to ask before believing the headline says anything about a typical employee.

**Part IV — The cockpit case.**

16. Build your own evidence: construct five house prices with [mean](#def-g10-stats-mean) $300$ and [median](#def-g10-stats-median) $200$ (thousands of euros).
17. The [mean](#def-g10-stats-mean) is the balance point: prove from the definition that the deviations from the [mean](#def-g10-stats-mean) always cancel, $\sum (x_i - \bar x) = 0$ , and verify it on your dataset of question 16.
18. The [median](#def-g10-stats-median) is the errand-minimizer: shops sit at kilometers $1$ , $3$ , $4$ , $9$ , $13$ of a road, and a warehouse must minimize the *total distance* to the five shops. Compute that total for a warehouse at the [median](#def-g10-stats-median) ( $4$ ), at the [mean](#def-g10-stats-mean) ( $6$ ), and at $5$ . Who wins? (This is general: the [median](#def-g10-stats-median) minimizes total absolute distance — test any other spot.)
19. The cockpit, resolved: suppose a pilot has a $30\,\%$ chance of being “close to average” on one body dimension, independently across $10$ dimensions. Estimate the chance of being close to average on *all ten* — and reconcile the result with what the lieutenant found in $4\,000$ pilots. What did the Air Force build instead of the average cockpit?
20. Finale — the user’s guide, one line each: when totals are what matters (a budget, a harvest), use the …; when the typical individual matters (salaries, waiting times), use the …; when fairness of spread matters, report the …; when contamination threatens, use the …. Close the case with the problem’s title.

**Solution of Problem 8.1.**

**1.** [Mean](#def-g10-stats-mean): $\frac{2\,550}{7} \approx 364$ thousand euros; [median](#def-g10-stats-median): the fourth of the seven ordered prices, $230$. The [median](#def-g10-stats-median) describes the street — six of the seven houses cost nowhere near $364$.

**2.** Without the mansion: [mean](#def-g10-stats-mean) $\frac{1\,350}{6} = 225$, [median](#def-g10-stats-median) $\frac{220 + 230}{2} = 225$. The [mean](#def-g10-stats-mean) fell by about $139$; the [median](#def-g10-stats-median) by $5$. One extreme value can drag the [mean](#def-g10-stats-mean) anywhere; the [median](#def-g10-stats-median) barely notices — it is *robust*.

**3.** [Mean](#def-g10-stats-mean) $= \frac{0 \times 20 + 1 \times 25 + 2 \times
30 + 3 \times 15 + 4 \times 10}{100} = \frac{170}{100} = 1.7$. No family has $1.7$ children, but $1.7 \times 100 = 170$ *is* the exact total number of children: the [mean](#def-g10-stats-mean) encodes the total, redistributed equally — its true talent.

**4.** $N = 15$: [median](#def-g10-stats-median) $=$ 8th value $= 11$; $Q_1 =$ 4th value $= 8$; $Q_3 =$ 12th value $= 14$.

**5.** [Range](#def-g10-stats-quartiles) $19 - 4 = 15$: the full spread, extremes included. [Interquartile range](#def-g10-stats-quartiles) $14 - 8 = 6$: the width of the middle half — spread without the extremes’ noise.

**6.** A shift $+b$ moves *every* value, so differences of values — [range](#def-g10-stats-quartiles), IQR — are unchanged: $(x_i + b) - (x_j + b) = x_i - x_j$. A scaling by $a$ multiplies all differences by $a$: [range](#def-g10-stats-quartiles) and IQR are scaled by $\abs a$.

**7.** Additive curve: $6 \to 9$ and $18 \to 21$ — the weak student gains $50\,\%$, the strong one $17\,\%$, and $21$ bursts the $20$-point scale only mildly. Multiplicative: $6 \to 8$ and $18 \to 24$ — ratios preserved, but $24$ is impossible on a $20$-scale: the ceiling problem. Additive curves flatter the bottom; multiplicative ones explode the top.

**8.** Total marks: $20 \times 12 + 30 \times 10 = 540$, so the merged [mean](#def-g10-stats-mean) is $\frac{540}{50} = 10.8$: class B’s thirty students outweigh class A’s twenty — means merge by *weighted* average, never by simple average.

**9.** [Median](#def-g10-stats-median) of $A$: $2$; of $B$: $5$. Merged list $\{0, 1, 2, 3, 10\}$: [median](#def-g10-stats-median) $2$. The size-weighted average of the [medians](#def-g10-stats-median) would be $\frac{3 \times 2 + 2 \times 5}{5} = 3.2
\neq 2$: [medians](#def-g10-stats-median) carry no total, so no merging formula exists — one must re-sort the whole list.

**10.** [Mean](#def-g10-stats-mean): $\frac{31.3}{5} = 6.26$. [Median](#def-g10-stats-median): $5.4$. Trimmed [mean](#def-g10-stats-mean): $\frac{5.3 + 5.4 + 5.5}{3} = 5.4$. One enthusiastic (or corrupt) judge moved the [mean](#def-g10-stats-mean) by almost a point; the [median](#def-g10-stats-median) and the trimmed [mean](#def-g10-stats-mean) shrugged — hence the trimming rules of judged sports.

**11.** Factory A: middle half within $0.2$ mm; factory B: within $4$ mm. Same [mean](#def-g10-stats-mean), opposite reliability: buy from A. The [interquartile range](#def-g10-stats-quartiles) decided — means alone compare centers, never consistency.

**12.** [Median](#def-g10-stats-median) far below [mean](#def-g10-stats-mean) betrays a *long right tail*: most waits short, a few catastrophic. Example: $\{5, 5, 10, 10, 95\}$ — [median](#def-g10-stats-median) $10$, [mean](#def-g10-stats-mean) $\frac{125}{5} = 25$.

**13.** It means $75\,\%$ of babies of that age are shorter (and $25\,\%$ taller): a rank, not a size. “Every child above the [mean](#def-g10-stats-mean)” is impossible — if all values exceeded the [mean](#def-g10-stats-mean), their average would exceed the [mean](#def-g10-stats-mean), which is itself: contradiction. But *almost* all can be: one very short child can hold the [mean](#def-g10-stats-mean) below everyone else. Above the *[median](#def-g10-stats-median)*: never more than half, by definition. Ranks cannot be flattered; averages can.

**14.** Typical mark: same, [median](#def-g10-stats-median) $11$ for both. Homogeneity of the middle: B tighter (IQR $3$ against $5$). Extremes: B wilder ([range](#def-g10-stats-quartiles) $17$ against $13$). B is a class of similar students plus a few outliers; A is evenly spread.

**15.** (1) Is $5\,000$ the [mean](#def-g10-stats-mean) or the [median](#def-g10-stats-median) — and may I have the [median](#def-g10-stats-median)? (2) What is the spread ([quartiles](#def-g10-stats-quartiles)), so I know how typical “typical” is? (3) Who was counted — sample size and selection (the how-data-lies weekend problem of the Middle School volume’s poll lesson)?

**16.** For instance $100, 150, 200, 250, 800$: [median](#def-g10-stats-median) $200$, [mean](#def-g10-stats-mean) $\frac{1\,500}{5} = 300$.

**17.** $\sum (x_i - \bar x) = \sum x_i - N \bar x =
N\bar x - N\bar x = 0$. Check: $-200 - 150 - 100 - 50 + 500 =
0$. The [mean](#def-g10-stats-mean) is the point where the deviations balance — the data’s center of gravity (the centroid weekend problem of the Middle School volume’s $G$, in one dimension).

**18.** At the [median](#def-g10-stats-median) $4$: $3 + 1 + 0 + 5 + 9 = 18$ km. At the [mean](#def-g10-stats-mean) $6$: $5 + 3 + 2 + 3 + 7 = 20$ km. At $5$: $4 + 2 + 1 + 4 + 8 = 19$ km. The [median](#def-g10-stats-median) wins — moving away from it always passes more shops than it approaches, so the total can only grow.

**19.** $0.3^{10} \approx 6 \times 10^{-6}$: about six pilots *per million*. In $4\,000$ pilots the expected count is $0.02$ — zero, as lieutenant Daniels found. The Air Force stopped designing for the average man and made everything adjustable: seats, pedals, straps — design for the *spread*, not the center.

**20.** Totals: the *[mean](#def-g10-stats-mean)*. Typical individual: the *[median](#def-g10-stats-median)*. Fairness of spread: the *[quartiles](#def-g10-stats-quartiles)* (and IQR). Contamination: the *trimmed [mean](#def-g10-stats-mean)*. And the closing line: the average person does not exist — but the [median](#def-g10-stats-median) person almost does.