---
title: "Descriptive Statistics"
book: "High School Mathematics"
subject: math
language: en
chapter: 17
exercises: 10
source: https://one-course.com/books/math/2/en/chapter/17-descriptive-statistics
---

# Chapter 17 — Descriptive Statistics

Statistics compresses a list of numbers into a few meaningful summaries: where the data sits ([mean](#def-g11-stat-mean), [median](#def-g11-stat-median)), and how spread out it is ([quartiles](#def-g11-stat-quartiles), [variance](#def-g11-stat-variance), [standard deviation](#def-g11-stat-variance)). This chapter works everything out on one running example. The same summaries reappear for random variables in [Chapter 18](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#ch-g11-prob), and their probabilistic laws in [Chapter 34](https://one-course.com/books/math/2/en/chapter/34-sums-of-random-variables-and-the-law-of-large-numbers#ch-g12-sums).

Throughout the chapter, our running dataset is the marks (out of $20$) of a class of $12$ students:

$$
8,\ 9,\ 10,\ 10,\ 11,\ 12,\ 12,\ 12,\ 13,\ 14,\ 16,\ 17 .
$$

## 17.1 Mean and median

**Definition 17.1 (Mean).**

The *mean* of the values $x_1, x_2, \dots, x_n$ is

$$
\bar x = \frac{x_1 + x_2 + \dots + x_n}{n}.
$$

If the value $x_i$ appears with [frequency](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-series) $n_i$ (with $n_1 + \dots + n_k = n$), then $\bar x = \frac{n_1 x_1 + \dots + n_k x_k}{n}$.

**Definition 17.2 (Median).**

A *median* of a sorted dataset is a value that splits it in half: at least half the values are $\leq$ it, and at least half are $\geq$ it. In practice, sort the $n$ values; if $n$ is odd, the median is the middle value; 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.

**Example 17.3.**

For the running dataset, the sum is $144$, so $\bar x = \frac{144}{12} =
12$. The two middle values ($6$th and $7$th of the sorted list) are both $12$, so the [median](#def-g11-stat-median) is $12$ too. [Mean](#def-g11-stat-mean) and [median](#def-g11-stat-median) need not coincide: the [median](#def-g11-stat-median) ignores *how far* the extreme values are, the [mean](#def-g11-stat-mean) does not. Replacing the top mark $17$ by $20$ moves the [mean](#def-g11-stat-mean) to $12.25$ but leaves the [median](#def-g11-stat-median) at $12$.

## 17.2 Quartiles

**Definition 17.4 (Quartiles, interquartile range).**

The *first quartile* $Q_1$ is the smallest value such that at least a quarter of the data is $\leq Q_1$; the *third quartile* $Q_3$ is the smallest value such that at least three quarters of the data is $\leq Q_3$. The *interquartile range* is $Q_3 - Q_1$: the width of the central half of the data.

**Example 17.5.**

$n = 12$: a quarter is $3$, so $Q_1$ is the $3$rd sorted value: $Q_1 = 10$. Three quarters is $9$, so $Q_3$ is the $9$th value: $Q_3 = 13$. The [interquartile range](#def-g11-stat-quartiles) is $3$: the central half of the class sits within $3$ marks.

![The running dataset as a dot plot, its mean (red), and the five-number summary above it (orange): minimum, Q_1, median, Q_3, maximum. The box contains the central half of the data.](https://one-course.com/images/onecourse/chapters/math-2/g11-stat/fig-2f4ea9327614.svg)

*The running dataset as a dot plot, its [mean](#def-g11-stat-mean) (red), and the five-number summary above it (orange): [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema), $Q_1$, [median](#def-g11-stat-median), $Q_3$, [maximum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema). The box contains the central half of the data.*

## 17.3 Variance and standard deviation

**Definition 17.6 (Variance, standard deviation).**

The *variance* of the dataset is the [mean](#def-g11-stat-mean) of the squared deviations from the [mean](#def-g11-stat-mean):

$$
V = \frac{1}{n}\sum_{i=1}^{n} (x_i - \bar x)^2,
$$

and the *standard deviation* is $\sigma = \sqrt{V}$, expressed in the same unit as the data.

**Proposition 17.7 (Shortcut formula).**

The [variance](#def-g11-stat-variance) is the [mean](#def-g11-stat-mean) of the squares minus the square of the [mean](#def-g11-stat-mean):

$$
V = \frac{1}{n}\sum_{i=1}^{n} x_i^2 \;-\; \bar x^{\,2}.
$$

**Proof.** Expand each squared deviation: $(x_i - \bar x)^2 = x_i^2 - 2\bar x\, x_i + \bar x^2$. Summing and dividing by $n$:

$$
V = \frac1n \sum x_i^2 - 2\bar x \cdot \underbrace{\frac1n \sum
x_i}_{=\ \bar x} + \frac1n \cdot n\bar x^2
= \frac1n\sum x_i^2 - 2\bar x^2 + \bar x^2
= \frac1n\sum x_i^2 - \bar x^2 .
$$

The same identity, for random variables, is [Proposition 18.13](https://one-course.com/books/math/2/en/chapter/18-probability-and-random-variables#prop-g11-prob-konig). ∎

**Example 17.8.**

For the running dataset, the deviations from $\bar x = 12$ are $-4$, $-3$, $-2$, $-2$, $-1$, $0$, $0$, $0$, $1$, $2$, $4$, $5$; their squares sum to

$$
16 + 9 + 4 + 4 + 1 + 0 + 0 + 0 + 1 + 4 + 16 + 25 = 80 .
$$

Hence

$$
V = \frac{80}{12} = \frac{20}{3} \approx 6.67,
\qquad
\sigma = \sqrt{20/3} \approx 2.6 \text{ marks}.
$$

Roughly speaking, a typical mark sits about $2.6$ marks away from the [mean](#def-g11-stat-mean).

**Proposition 17.9 (Change of units).**

If every value is transformed by $y_i = a x_i + b$, then

$$
\bar y = a \bar x + b,
\qquad
V_y = a^2\, V_x,
\qquad
\sigma_y = \abs{a}\,\sigma_x .
$$

**Proof.** $\bar y = \frac1n \sum (a x_i + b) = a\left(\frac1n\sum x_i\right) +
\frac1n \cdot nb = a\bar x + b$. Then $y_i - \bar y = a(x_i - \bar x)$: each deviation is scaled by $a$ (the shift $b$ cancels), so each squared deviation is scaled by $a^2$, and so is their [mean](#def-g11-stat-mean). Taking square roots gives $\sigma_y = \abs a\,\sigma_x$. ∎

**Example 17.10.**

Temperatures recorded in degrees Celsius with [mean](#def-g11-stat-mean) $20$ and [standard deviation](#def-g11-stat-variance) $3$ convert to Fahrenheit by $F = 1.8\,C + 32$: the [mean](#def-g11-stat-mean) becomes $1.8 \times 20 + 32 = 68$, the [standard deviation](#def-g11-stat-variance) $1.8 \times 3 = 5.4$. Shifting a dataset does not change its spread; rescaling it rescales the spread.

**Method 17.11 (Summarizing and comparing datasets).**

To compare two datasets, put side by side: [mean](#def-g11-stat-mean) (or [median](#def-g11-stat-median)) for *position*, [standard deviation](#def-g11-stat-variance) (or [interquartile range](#def-g11-stat-quartiles)) for *spread*. The pair ([median](#def-g11-stat-median), [interquartile range](#def-g11-stat-quartiles)) is robust to extreme values; the pair ([mean](#def-g11-stat-mean), [standard deviation](#def-g11-stat-variance)) uses every value and feeds [probability](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-distribution) theory.

**Example 17.12.**

Two archers each shoot $10$ arrows; scores average $8.5$ for both, but the [standard deviations](#def-g11-stat-variance) are $0.5$ and $2.1$. Same level on average — but the first archer is far more consistent.

## 17.4 Exercises

**Exercise 17.1 ★.**

Compute the [mean](#def-g11-stat-mean), [median](#def-g11-stat-median), [quartiles](#def-g11-stat-quartiles) and [interquartile range](#def-g11-stat-quartiles) of the dataset

$$
5,\ 7,\ 7,\ 8,\ 9,\ 10,\ 11,\ 14 .
$$

**Solution of Exercise 17.1.**

$n = 8$, sum $71$: [mean](#def-g11-stat-mean) $\bar x = \frac{71}{8} = 8.875$. [Median](#def-g11-stat-median): [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of the $4$th and $5$th values, $\frac{8 + 9}{2} = 8.5$. [Quartiles](#def-g11-stat-quartiles): a quarter of $8$ is $2$, so $Q_1$ is the $2$nd value, $7$; three quarters is $6$, so $Q_3$ is the $6$th value, $10$. [Interquartile range](#def-g11-stat-quartiles): $10 - 7 = 3$.

**Exercise 17.2 ★.**

Compute the [variance](#def-g11-stat-variance) and [standard deviation](#def-g11-stat-variance) of the dataset $2,\ 4,\ 4,\ 4,\ 5,\ 5,\ 7,\ 9$ ([mean](#def-g11-stat-mean) first, then the shortcut formula of [Proposition 17.7](#prop-g11-stat-shortcut)).

**Solution of Exercise 17.2.**

Sum $40$, so $\bar x = 5$. Sum of squares: $4 + 16 + 16 + 16 + 25 + 25 + 49 + 81 = 232$. By the shortcut formula,

$$
V = \frac{232}{8} - 5^2 = 29 - 25 = 4,
\qquad
\sigma = 2 .
$$

**Exercise 17.3 ★.**

A die is rolled $50$ times; the outcomes are summarized by frequencies:

| outcome | 1 | 2 | 3 | 4 | 5 | 6 |
| --- | --- | --- | --- | --- | --- | --- |
| [frequency](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-series) | 6 | 9 | 8 | 10 | 9 | 8 |

Compute the [mean](#def-g11-stat-mean) outcome and the [median](#def-g11-stat-median).

**Solution of Exercise 17.3.**

[Mean](#def-g11-stat-mean):

$$
\bar x = \frac{1 \times 6 + 2 \times 9 + 3 \times 8 + 4 \times 10
+ 5 \times 9 + 6 \times 8}{50}
= \frac{181}{50} = 3.62 .
$$

[Median](#def-g11-stat-median): the cumulative frequencies are $6, 15, 23, 33, \dots$, so the $25$th and $26$th sorted outcomes are both $4$: the [median](#def-g11-stat-median) is $4$.

**Exercise 17.4 ★★.**

A class of $15$ students has [mean](#def-g11-stat-mean) mark $11.2$. A $16$th student joins and scores $18$. What is the new class [mean](#def-g11-stat-mean)?

**Solution of Exercise 17.4.**

The total of the $15$ marks is $15 \times 11.2 = 168$; with the newcomer, $\frac{168 + 18}{16} = \frac{186}{16} = 11.625$.

**Exercise 17.5 ★★.**

The [mean](#def-g11-stat-mean) of the dataset $6, 8, x, 12, 14$ is $10$. Find $x$, then compute the [standard deviation](#def-g11-stat-variance) of the completed dataset.

**Solution of Exercise 17.5.**

$\frac{6 + 8 + x + 12 + 14}{5} = 10$ gives $40 + x = 50$, so $x = 10$. The deviations of $6, 8, 10, 12, 14$ from the [mean](#def-g11-stat-mean) $10$ are $-4, -2, 0, 2, 4$, with squares summing to $40$: $V = \frac{40}{5} = 8$ and $\sigma = 2\sqrt2 \approx 2.83$.

**Exercise 17.6 ★.**

Marks with [mean](#def-g11-stat-mean) $12.4$ and [standard deviation](#def-g11-stat-variance) $2.5$ are rescaled by $y = \frac{x}{2} + 4$. Give the new [mean](#def-g11-stat-mean) and the new [standard deviation](#def-g11-stat-variance).

**Solution of Exercise 17.6.**

By [Proposition 17.9](#prop-g11-stat-affine) with $a = \frac12$, $b = 4$: new [mean](#def-g11-stat-mean) $\frac{12.4}{2} + 4 = 10.2$; new [standard deviation](#def-g11-stat-variance) $\frac{2.5}{2} = 1.25$.

**Exercise 17.7 ★★.**

Group A ($20$ values) has [mean](#def-g11-stat-mean) $10$; group B ($30$ values) has [mean](#def-g11-stat-mean) $15$. Compute the [mean](#def-g11-stat-mean) of the $50$ values pooled together. Is it the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) $12.5$? Why not?

**Solution of Exercise 17.7.**

The pooled [mean](#def-g11-stat-mean) is the *weighted* average

$$
\frac{20 \times 10 + 30 \times 15}{50} = \frac{650}{50} = 13,
$$

not $12.5$: group B contributes more values, so it pulls the pooled [mean](#def-g11-stat-mean) towards its own.

**Exercise 17.8 ★★.**

Two production lines fill $1$-kg sugar bags. [Samples](https://one-course.com/books/math/2/en/chapter/9-probability-and-sampling#def-g10-proba-sample) give: line 1: [mean](#def-g11-stat-mean) $1.002$ kg, [standard deviation](#def-g11-stat-variance) $0.006$ kg; line 2: [mean](#def-g11-stat-mean) $1.001$ kg, [standard deviation](#def-g11-stat-variance) $0.019$ kg. Which line should be recalibrated first, and why?

**Solution of Exercise 17.8.**

Both lines are nearly on target on average ($1.002$ and $1.001$ kg), but line 2’s [standard deviation](#def-g11-stat-variance) is three times larger: its bags vary much more, producing far more noticeably under- and over-filled bags. Line 2 should be dealt with first — consistency, not just the [mean](#def-g11-stat-mean), is what needs fixing.

**Exercise 17.9 ★★.**

A dataset of $n = 10$ values has $\sum x_i = 70$ and $\sum x_i^2 = 540$. Compute its [mean](#def-g11-stat-mean), [variance](#def-g11-stat-variance) and [standard deviation](#def-g11-stat-variance).

**Solution of Exercise 17.9.**

$\bar x = \frac{70}{10} = 7$; $V = \frac{540}{10} - 7^2 = 54 - 49 = 5$; $\sigma = \sqrt5 \approx 2.24$.

**Exercise 17.10 ★★★.**

The dataset $1,\ 2,\ 2,\ 3,\ 3,\ 3,\ 4,\ 4,\ 5,\ 23$ contains an *outlier*.

1. Compute the [mean](#def-g11-stat-mean) and [median](#def-g11-stat-median) with and without the value $23$ .
2. Compute the [interquartile range](#def-g11-stat-quartiles) with and without $23$ .
3. Which summaries are robust to the outlier? Conclude with a recommendation.

**Solution of Exercise 17.10.**

*1.* With $23$: sum $50$, [mean](#def-g11-stat-mean) $5$; [median](#def-g11-stat-median) = [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $5$th and $6$th values $= 3$. Without $23$: sum $27$, [mean](#def-g11-stat-mean) $3$; [median](#def-g11-stat-median) (5th of 9 values) $= 3$.

*2.* With $23$ ($n = 10$): $Q_1$ is the $3$rd value ($2$), $Q_3$ the $8$th ($4$): [interquartile range](#def-g11-stat-quartiles) $2$. Without ($n = 9$): $Q_1$ is the $3$rd value ($2$), $Q_3$ the $7$th ($4$): still $2$.

*3.* The single outlier drags the [mean](#def-g11-stat-mean) from $3$ up to $5$ but leaves the [median](#def-g11-stat-median) and the [interquartile range](#def-g11-stat-quartiles) unchanged: those two are *robust*. When a dataset may contain aberrant values, report the [median](#def-g11-stat-median) and [interquartile range](#def-g11-stat-quartiles) rather than the [mean](#def-g11-stat-mean) and [standard deviation](#def-g11-stat-variance).

## 17.5 Problem: The z-score, a universal ruler

**Problem 17.1.**

Weekend problem — standard deviations compare apples with oranges, Chebyshev’s inequality turns $\sigma$ into a guarantee, and the mean wins its own medal

Is $15/20$ on a hard exam better than $13/20$ on an easy one? Is a $23$ in a list of small numbers an accident or an outlier? Raw numbers cannot say — but divided by the right [standard deviation](#def-g11-stat-variance) they all speak one language, the *[z-score](#pb-g11-stat-1)*. This problem masters the [variance](#def-g11-stat-variance) machinery ([Definition 17.6](#def-g11-stat-variance), [Proposition 17.7](#prop-g11-stat-shortcut)), builds the universal ruler, and proves the remarkable inequality that turns $\sigma$ from a descriptive number into a guarantee.

**Part I — [Variance](#def-g11-stat-variance) mechanics.**

1. For the dataset $4, 8, 6, 5, 2$ : compute the [mean](#def-g11-stat-mean) , then the [variance](#def-g11-stat-variance) straight from the definition, then $\sigma$ .
2. Recompute the [variance](#def-g11-stat-variance) with the shortcut formula ( [Proposition 17.7](#prop-g11-stat-shortcut) ) — [mean](#def-g11-stat-mean) of squares minus square of [mean](#def-g11-stat-mean) — and check the agreement.
3. With frequencies: marks $8$ , $10$ , $14$ obtained by $3$ , $5$ , $2$ students. Compute [mean](#def-g11-stat-mean) , [variance](#def-g11-stat-variance) and $\sigma$ (two decimals).
4. Temperatures have [mean](#def-g11-stat-mean) $20$ and $\sigma = 3$ (degrees Celsius). Give the [mean](#def-g11-stat-mean) and $\sigma$ in Fahrenheit ( $F = 1.8\,C + 32$ ; [Proposition 17.9](#prop-g11-stat-affine) ).
5. Prove the two rules just used: adding a constant $b$ to every value shifts the [mean](#def-g11-stat-mean) by $b$ and leaves $\sigma$ unchanged; multiplying by $a$ multiplies the [mean](#def-g11-stat-mean) by $a$ and $\sigma$ by $\abs a$ .

**Part II — The universal ruler.** The *z-score* of a value $x$ in a dataset of [mean](#def-g11-stat-mean) $\bar x$ and [standard deviation](#def-g11-stat-variance) $\sigma$ is $z = \dfrac{x - \bar x}{\sigma}$: the number of [standard deviations](#def-g11-stat-variance) separating $x$ from the [mean](#def-g11-stat-mean).

6. Exam A has [mean](#def-g11-stat-mean) $12$ and $\sigma = 2$ ; exam B has [mean](#def-g11-stat-mean) $8$ and $\sigma = 4$ . Alice scored $15$ on A, Bob $13$ on B. Compute both [z-scores](#pb-g11-stat-1) : whose performance stands further above the crowd?
7. Compute the five [z-scores](#pb-g11-stat-1) of the dataset of question 1, and verify that they have [mean](#def-g11-stat-mean) $0$ and [standard deviation](#def-g11-stat-variance) $1$ . Explain from question 5 why standardization *always* produces [mean](#def-g11-stat-mean) $0$ and $\sigma = 1$ .
8. An aptitude test has [mean](#def-g11-stat-mean) $100$ and $\sigma = 15$ . What raw score has $z = 2$ ? What is the [z-score](#pb-g11-stat-1) of $76$ ?
9. A pediatrician records a baby’s weight as “ $z = -2.5$ ”. Say what this means, and why growth is tracked in [z-scores](#pb-g11-stat-1) rather than kilograms as the child ages.
10. The outlier of [Exercise 17.10](#exo-g11-stat-10) (dataset $1, 2, 2, 3, 3, 3, 4, 4, 5, 23$ ): compute the [mean](#def-g11-stat-mean) , $\sigma$ , and the [z-score](#pb-g11-stat-1) of $23$ . The classic alarm threshold is $\abs z \geq 3$ : verdict?
11. Recompute $\sigma$ *without* the value $23$ , and the [z-score](#pb-g11-stat-1) that $23$ would have against the clean data’s [mean](#def-g11-stat-mean) and $\sigma$ . What did the outlier do to the very ruler meant to measure it, and what does [Exercise 17.10](#exo-g11-stat-10) suggest as a countermeasure?

**Part III — Chebyshev’s guarantee.**

12. Prove the Bienaymé–Chebyshev inequality for a dataset: if a proportion $p$ of the values satisfy $\abs{x - \bar x} \geq k\sigma$, then, keeping only those terms in the [variance](#def-g11-stat-variance) sum, $$\sigma^2 \geq p \cdot (k\sigma)^2,  \qquad\text{hence}\qquad  p \leq \frac{1}{k^2} .$$
13. What does the inequality guarantee about the proportion of any dataset beyond $2\sigma$ of its [mean](#def-g11-stat-mean) ? Beyond $3\sigma$ ? Check the $2\sigma$ statement on the dataset of question 1.
14. A factory produces bolts with [mean](#def-g11-stat-mean) $50$ mm and $\sigma = 0.1$ mm. What does Chebyshev guarantee about the proportion of bolts outside $\intcc{49.7}{50.3}$ ? (Real production does far better — the bell curve of grade 12 turns this cautious $\frac19$ into a fraction of a percent.)
15. Two funds both return $5\,\%$ per year on average, with $\sigma = 2\,\%$ for fund A and $\sigma = 12\,\%$ for fund B. For each fund, what does Chebyshev say about the probability-like proportion of years with returns below $-19\,\%$ ? What is $\sigma$ ’s financial name?

**Part IV — The [mean](#def-g11-stat-mean)’s medal, and a warning.**

16. Let $f(c)$ be the [mean](#def-g11-stat-mean) of the squared deviations $(x_i - c)^2$ from an arbitrary center $c$. Prove the identity $$f(c) = \sigma^2 + (\bar x - c)^2 ,$$ and conclude: the [mean](#def-g11-stat-mean) is the unique minimizer of [mean](#def-g11-stat-mean) squared deviation — its medal, matching the [median](#def-g11-stat-median)’s minimization of absolute deviations in [Problem 8.1](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#pb-g10-stats-1).
17. Verify the identity on the dataset of question 1 with $c = 6$ : compute $f(6)$ directly and via the formula.
18. Same summaries, different data: check that $P = \{3, 5, 7, 9, 11\}$ and $Q = \{1, 7, 7, 7, 13\}$ share the same [mean](#def-g11-stat-mean) but not the same [variance](#def-g11-stat-variance) ; then construct a dataset of the form $\{7 - a, 7, 7, 7, 7 + a\}$ with exactly the [mean](#def-g11-stat-mean) and [variance](#def-g11-stat-variance) of $P$ — yet a shape all its own.
19. Prove that $\sigma = 0$ forces all values equal. (What must every term of a zero sum of squares be?)
20. Finale — the statistician’s ladder, one line per rung: centers ( [mean](#def-g11-stat-mean) , [median](#def-g11-stat-median) ), spreads ( $\sigma$ , IQR), the universal ruler ( $z$ ), the guarantee (Chebyshev), and the warning of question 18: summaries compress, and compression loses — with the bell curve of grade 12 waiting to turn [z-scores](#pb-g11-stat-1) into exact probabilities.

**Solution of Problem 17.1.**

**1.** [Mean](#def-g11-stat-mean) $\bar x = \frac{25}{5} = 5$. Deviations squared: $1, 9, 1, 0, 9$: [variance](#def-g11-stat-variance) $\frac{20}{5} = 4$, so $\sigma = 2$.

**2.** [Mean](#def-g11-stat-mean) of squares: $\frac{16 + 64 + 36 + 25 + 4}{5} = 29$; minus $\bar x^2 = 25$: [variance](#def-g11-stat-variance) $4$. Same answer, half the work.

**3.** [Mean](#def-g11-stat-mean) $= \frac{3 \times 8 + 5 \times 10 + 2 \times
14}{10} = 10.2$; [variance](#def-g11-stat-variance) $= \frac{3 \times 64 + 5 \times 100 +
2 \times 196}{10} - 10.2^2 = 108.4 - 104.04 = 4.36$; $\sigma \approx 2.09$.

**4.** [Mean](#def-g11-stat-mean): $1.8 \times 20 + 32 = 68\,^\circ$F; $\sigma$: $1.8 \times 3 = 5.4\,^\circ$F — the shift $+32$ spreads nothing.

**5.** Adding $b$: every deviation $x_i + b - (\bar x + b)
= x_i - \bar x$ is unchanged, so the [variance](#def-g11-stat-variance) is. Multiplying by $a$: deviations are multiplied by $a$, their squares by $a^2$, the [variance](#def-g11-stat-variance) by $a^2$, and $\sigma$ by $\sqrt{a^2} = \abs a$.

**6.** $z_A = \frac{15 - 12}{2} = 1.5$; $z_B = \frac{13 - 8}{4} = 1.25$. Alice stands further above her crowd: the harder exam’s $15$ beats the easier exam’s $13$.

**7.** $z$-scores: $-0.5,\ 1.5,\ 0.5,\ 0,\ -1.5$: [mean](#def-g11-stat-mean) $0$, [standard deviation](#def-g11-stat-variance) $1$. In general, standardizing subtracts $\bar x$ (making the [mean](#def-g11-stat-mean) $0$, $\sigma$ untouched) then divides by $\sigma$ (making the new $\sigma$ equal to $1$): question 5 twice.

**8.** $z = 2$: $100 + 2 \times 15 = 130$. And $z(76) = \frac{76 - 100}{15} = -1.6$.

**9.** The baby is $2.5$ [standard deviations](#def-g11-stat-variance) below the [mean](#def-g11-stat-mean) weight *of babies its age* — alarmingly light for its cohort. Raw kilograms cannot be compared across ages (all babies gain weight); [z-scores](#pb-g11-stat-1) measure each child against its own age group, one ruler for all ages.

**10.** [Mean](#def-g11-stat-mean) $5$; [mean](#def-g11-stat-mean) of squares $\frac{622}{10} = 62.2$, [variance](#def-g11-stat-variance) $37.2$, $\sigma \approx 6.10$; $z(23) = \frac{18}{6.10} \approx 2.95$: just under the alarm threshold — the test barely blinks.

**11.** Without $23$: [mean](#def-g11-stat-mean) $3$, [variance](#def-g11-stat-variance) $\frac{93}{9} - 9 \approx 1.33$, $\sigma \approx 1.15$; against the clean ruler, $23$ scores $z \approx \frac{20}{1.15} \approx 17$: a monster. The outlier inflated both the [mean](#def-g11-stat-mean) and $\sigma$, camouflaging itself — which is why [Exercise 17.10](#exo-g11-stat-10) recommends robust summaries ([median](#def-g11-stat-median), IQR) when contamination threatens.

**12.** In the [variance](#def-g11-stat-variance) sum $\sigma^2 = \frac1N \sum (x_i - \bar x)^2$, keep only the terms with $\abs{x_i - \bar x} \geq k\sigma$: there are $pN$ of them, each at least $(k\sigma)^2$, so $\sigma^2 \geq \frac{1}{N} \cdot pN \cdot k^2\sigma^2 =
p\,k^2\sigma^2$: divide by $k^2 \sigma^2$ (nonzero spread): $p \leq \frac{1}{k^2}$.

**13.** Beyond $2\sigma$: at most $\frac14 = 25\,\%$ of any dataset whatsoever; beyond $3\sigma$: at most $\frac19 \approx 11\,\%$. Question 1’s data: beyond $2\sigma = 4$ from the [mean](#def-g11-stat-mean) $5$ means $\leq 1$ or $\geq 9$: none of the five values — $0\,\% \leq 25\,\%$, as promised.

**14.** Outside $\intcc{49.7}{50.3}$ means beyond $3\sigma$: at most $\frac19$ of production, *whatever* the shape of the distribution — Chebyshev asks nothing but a [mean](#def-g11-stat-mean) and a $\sigma$. Bell-shaped production concentrates far better (about $0.3\,\%$ beyond $3\sigma$): the normal law, grade 12.

**15.** $-19\,\%$ sits $24$ points below the [mean](#def-g11-stat-mean) $5$. Fund B: $24 = 2\sigma_B$: Chebyshev allows up to $\frac14$ of years that bad. Fund A: $24 = 12\sigma_A$: at most $\frac{1}{144}$. Same average, incomparable dangers — $\sigma$ is what finance calls *risk* (or volatility).

**16.** Expand around $\bar x$: $(x_i - c)^2 = \left((x_i - \bar x) + (\bar x - c)\right)^2
= (x_i - \bar x)^2 + 2(\bar x - c)(x_i - \bar x) +
(\bar x - c)^2$. Averaging: the middle term dies (deviations from the [mean](#def-g11-stat-mean) sum to zero), leaving $f(c) = \sigma^2 + (\bar x - c)^2$ — minimal exactly at $c = \bar x$, with [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $\sigma^2$. The [mean](#def-g11-stat-mean) is the least-squares center, as the [median](#def-g11-stat-median) is the least-absolute center ([Problem 8.1](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#pb-g10-stats-1)).

**17.** Directly: deviations from $6$ are $-2, 2, 0, -1, -4$, squares $4, 4, 0, 1, 16$: $f(6) = \frac{25}{5} = 5$. Formula: $\sigma^2 + (5 - 6)^2 = 4 + 1 = 5$. Agreement.

**18.** Both have [mean](#def-g11-stat-mean) $7$; [variances](#def-g11-stat-variance): $P$: $\frac{16 + 4 + 0 + 4 + 16}{5} = 8$; $Q$: $\frac{36 + 0 + 0 + 0 + 36}{5} = 14.4$: different spreads behind one [mean](#def-g11-stat-mean). For $\{7 - a, 7, 7, 7, 7 + a\}$: [variance](#def-g11-stat-variance) $\frac{2a^2}{5} = 8$ gives $a = \sqrt{20} \approx 4.47$: the dataset $\{2.53, 7, 7, 7, 11.47\}$ carries the exact [mean](#def-g11-stat-mean) and [variance](#def-g11-stat-variance) of $P$ with a completely different profile — summaries compress, and compression loses.

**19.** $\sigma^2 = 0$ means a sum of squares $(x_i - \bar x)^2$ vanishes; squares are $\geq 0$, so a zero sum forces every term to be zero: $x_i = \bar x$ for all $i$ — all values equal.

**20.** Centers say where the data lives; spreads say how loosely; the [z-score](#pb-g11-stat-1) lays one ruler across all datasets, exams, ages and currencies; Chebyshev converts $\sigma$ into an unconditional guarantee; and question 18 keeps us honest — no handful of numbers ever contains the whole story. Next year the bell curve sharpens the ruler into exact probabilities.