Descriptive Statistics

(a) Population and sample. The population is the day's entire output, all $N=2400$ loaves; the sample is the $n=25$ loaves actually weighed. A parameter is a number computed from the whole population, fixed and unknown: the day's mean weight $\mu$ and variance ${\sigma }^2$ (or the proportion $p$ of underweight loaves). A statistic is computed from the sample and would change with a different sample: $\bar{x}=803$ g and $s=12$ g, so $s^2=144$ g${}^2$ (or $\hat{p}$).

(b) Random variable versus observed value. Upper-case $\bar{X}$ is the sample mean regarded as a random variable, before the sample is drawn; lower-case $\bar{x}=803$ is the one number it turned out to be. An expectation is a long-run average over all the samples that might have been drawn, so it is a statement about $\bar{X}$, and under the random-sample assumption the theorem that the sample mean is unbiased gives $E[\bar{X}]=\mu$ — an unknown constant, not $803$. Substituting one realisation into an expectation is a category error. The two correct sentences are $\bar{x}=803$ and $E[\bar{X}]=\mu$.

(c) Bias of the shifted estimator. By linearity of expectation and $E[\bar{X}]=\mu$,

$$\mathrm{Bias}(\hat{\theta })=E[\bar{X}+5]-\mu =\mu +5-\mu =5\text{g}.$$

The estimator is biased upward by $5$ g whatever the ovens do: if they run light, that is already built into $\mu$, and $\bar{X}$ tracks it without help. Unbiasedness of $\bar{X}$ promises only that, averaged over every random sample of $25$ loaves that could have been drawn, the sample means would centre exactly on $\mu$. It promises nothing about how far the single value $803$ is from $\mu$; that is governed by $\mathrm{SD}(\bar{X})=\sigma /\sqrt{n}$, which here is estimated by $12/\sqrt{25}=2.4$ g and is a separate quantity altogether.

(a) Mean and deviations. The sum is $11+6+12+3+8=40$, so $\bar{x}=40/5=8$. The deviations are $3,-2,4,-5,0$, and $3-2+4-5+0=0$, as the proposition that deviations from the mean sum to zero guarantees for every dataset. Their squares are $9,4,16,25,0$, so $S(\bar{x})=54$.

(b) Shift the centre with the identity. The theorem that the mean minimises total squared deviation gives the exact identity $S(c)=S(\bar{x})+n(c-\bar{x}{)}^2$. With $c=10$,

$$S(10)=54+5(10-8{)}^2=54+20=74.$$

Direct check: $(11-10{)}^2+(6-10{)}^2+(12-10{)}^2+(3-10{)}^2+(8-10{)}^2=1+16+4+49+4=74$. ✓ The excess $20$ is the cost of centring two degrees away from the minimiser.

(c) Linear transformation. $F=aC+b$ with $a=\frac{9}{5}$ and $b=32$, so by the proposition on the mean under a linear transformation,

$$\bar{y}=\frac{9}{5}\cdot 8+32=\frac{72}{5}+32=\frac{232}{5}=46.4{}^{\circ }\text{F}.$$

No individual reading was needed.

(d) Combined mean. By the proposition on the combined mean of two groups,

$$\bar{x}=\frac{n_A{\bar{x}}_A+n_B{\bar{x}}_B}{n_A+n_B}=\frac{5\cdot 8+15\cdot 12}{20}=\frac{40+180}{20}=11.$$

The pooled mean is a weighted mean of the two group means with the group sizes as weights, here $\frac{1}{4}\cdot 8+\frac{3}{4}\cdot 12=2+9=11$. The unweighted average $10$ treats the two rooms as equally represented, but the second room contributes three times as many readings, so the truth is pulled toward $12$. Averaging averages is correct only when $n_A=n_B$.

(a) Sort first. In order: $9,12,14,16,18,20,22,27,31,45$. With $n=10$ even the median is the average of the $5$th and $6$th values, $(18+20)/2=19$. The sum is $9+12+14+16+18=69$ and $20+22+27+31+45=145$, total $214$, so $\bar{x}=214/10=21.4$.

(b) One value changes. The $45$ was already the largest observation, so replacing it by $450$ leaves the order of the other nine untouched and $450$ is still last. The $5$th and $6$th values are still $18$ and $20$: the median is $19$, a move of $0$. The sum becomes $214-45+450=619$, so $\bar{x}=61.9$, a move of $40.5$ minutes — one observation out of ten has tripled the mean.

(c) Trimmed mean. With $\alpha =0.1$ and $n=10$ we discard the lowest $1$ and the highest $1$ value, leaving $12,14,16,18,20,22,27,31$, whose sum is $160$. The trimmed mean is $160/8=20$. It sits between the median $19$ and the mean $61.9$, but right next to the median: trimming removed the single value that was driving the mean, which is exactly what the trimmed mean is for.

(d) Why the median cannot move. The theorem that the median minimises total absolute deviation shows that $A(c)=\sum |x_i-c|$ has, between data values, slope $L(c)-R(c)$: the number of observations to the left of $c$ minus the number to the right. That slope depends only on how many points lie on each side, never on how far away they are. Replacing $45$ by $450$ leaves it on the right of every $c<45$, so for $c$ in $(18,20)$ the counts are still $L=R=5$, the slope is still zero there, and the minimising set is still $[18,20]$ with midpoint $19$.

The argument holds for any replacement value that is at least $20$, since then $450$'s replacement still lies to the right of the middle interval and the sorted positions $5$ and $6$ are unchanged. A replacement below $20$ would change the ordering, and the median could then move.

(a) From the definition. The sum is $7+7+11+11+14=50$, so $\bar{x}=10$. The deviations are $-3,-3,1,1,4$ (they sum to $0$, as they must), with squares $9,9,1,1,16$ summing to $36$. Dividing by $n-1=4$,

$$s^2=\frac{36}{4}=9{\text{cm}}^2,s=3\text{cm}.$$

If the five parts were the whole population the divisor would be $N=5$: ${\sigma }^2=36/5=7.2$ cm${}^2$ and $\sigma =6/\sqrt{5}$ cm.

(b) Computational formula. $\sum x_i^2=49+49+121+121+196=536$ and $n{\bar{x}}^2=5\cdot 100=500$, so $\sum (x_i-\bar{x}{)}^2=536-500=36$. ✓

(c) Linear transformation. Here $a=10$ and $b=-100$. By the proposition on the mean under a linear transformation, $\bar{y}=10\cdot 10-100=0$. By the proposition on the variance under a linear transformation, $s_y^2=a^2{s}_x^2=100\cdot 9=900$ mm${}^2$ and $s_y=|a|s_x=30$ mm. The shift by $-100$ moved the mean and left the spread alone; the factor $10$ scaled the standard deviation by $10$ and the variance by $100$.

(d) Coefficient of variation. ${\mathrm{CV}}_x=s/\bar{x}=3/10=0.3$, that is $30\%$. For the $y$ data the formula would divide by $\bar{y}=0$, but the failure is deeper than a zero denominator: the CV measures spread relative to level, which only means something when zero is a genuine zero of the quantity. The shift $-100$ moved the origin to an arbitrary target, so “level” no longer has meaning and the ratio is uninterpretable. A pure rescaling with $a>0$ and $b=0$ would have left the CV unchanged, since $|a|s/(a\bar{x})=s/\bar{x}$; it is the shift that destroys it, which is why the CV is a ratio-scale tool.

(a) The two-point sample. By definition $\bar{X}=\frac{1}{2}(X_1+X_2)$. The deviations are

$$X_1-\bar{X}=\frac{1}{2}(X_1-X_2),X_2-\bar{X}=\frac{1}{2}(X_2-X_1)=-\frac{1}{2}(X_1-X_2),$$

which sum to zero as deviations from the mean must. Each squares to $\frac{1}{4}(X_1-X_2{)}^2$, so the sum of squares is $\frac{1}{2}(X_1-X_2{)}^2$, and dividing by $n-1=1$ gives $s^2=\frac{1}{2}(X_1-X_2{)}^2$.

(b) Independent case. Put $D=X_1-X_2$. By linearity, $E[D]=\mu -\mu =0$, so by the computational formula for the variance, $E[D^2]=\mathrm{Var}(D)+(E[D]{)}^2=\mathrm{Var}(D)$. For independent summands the variance of a sum is the sum of the variances, and $\mathrm{Var}(-X_2)=(-1{)}^2\mathrm{Var}(X_2)={\sigma }^2$, so

$$\mathrm{Var}(D)=\mathrm{Var}(X_1)+\mathrm{Var}(X_2)=2{\sigma }^2,\text{hence}E[s^2]=\frac{1}{2}E[D^2]={\sigma }^2.■$$

The $n$-divisor estimator is ${\hat{\sigma }}^2=\frac{1}{2}\cdot \frac{1}{2}{D}^2=\frac{1}{2}{s}^2$, so $E[{\hat{\sigma }}^2]=\frac{1}{2}{\sigma }^2$ and its bias is $-\frac{1}{2}{\sigma }^2=-{\sigma }^2/n$ with $n=2$. This is the corollary that the $n$-divisor estimator is biased downward, in the case where it is off by half.

(c) Correlated case. Expand $D-E[D]=(X_1-\mu )-(X_2-\mu )$ and take expectations of the square:

$$\mathrm{Var}(D)=E[(X_1-\mu {)}^2]+E[(X_2-\mu {)}^2]-2E[(X_1-\mu )(X_2-\mu )]={\sigma }^2+{\sigma }^2-2\rho {\sigma }^2=2(1-\rho ){\sigma }^2.$$

Still $E[D]=0$, so $E[s^2]=\frac{1}{2}\mathrm{Var}(D)=(1-\rho ){\sigma }^2$. $■$

This equals ${\sigma }^2$ only when $\rho =0$. The indispensable hypothesis is independence of the observations (uncorrelatedness is what the algebra actually uses); it enters the general proof through $\mathrm{Var}(\bar{X})={\sigma }^2/n$, which fails when the cross terms $2{\sum }_{i<j}\mathrm{Cov}(X_i,X_j)$ do not vanish. Normality was never needed and still is not. When $\rho >0$ the bias $-\rho {\sigma }^2$ is negative: positively correlated observations sit closer to each other than independent ones would, so their spread understates the population spread — two measurements on the same subject look less variable than two subjects do.

(a) The square root is not linear. The invalid step is $E\left[\sqrt{s^2}\right]=\sqrt{E[s^2]}$: the expectation has been moved inside the square root. Expectation commutes with linear operations only. The square root is strictly concave, so Jensen's inequality gives $E[\sqrt{V}]\le \sqrt{E[V]}$ for any non-negative $V$, with strict inequality whenever $V$ is genuinely random. Hence $E[s]<\sigma$: the sample standard deviation underestimates $\sigma$ on average even though $s^2$ is unbiased for ${\sigma }^2$.

Example: $E[V]=\frac{1}{2}(1+9)=5$, so $\sqrt{E[V]}=\sqrt{5}$, while $E[\sqrt{V}]=\frac{1}{2}(1+3)=2$. Since $2^2=4<5$, we have $2<\sqrt{5}$ and the two sides differ. The size of the gap depends on the distribution of $V$, which is why no universal correction to $s$ exists.

(b) Two false claims. First, “for large $n$ dividing by $n$ is also unbiased.” By the corollary that the $n$-divisor estimator is biased downward, $\mathrm{Bias}({\hat{\sigma }}^2)=-{\sigma }^2/n$, which is nonzero for every finite $n$. It tends to zero, but unbiasedness is an exact property and this estimator never has it. For $n=1000$ and ${\sigma }^2=50$,

$$E[{\hat{\sigma }}^2]=\frac{999}{1000}\cdot 50=49.95,\mathrm{Bias}=-\frac{50}{1000}=-\mathrm{0.05.}$$

Small, but not zero, and “negligible” is a different claim from “unbiased”.

Second, “the reason is that it makes the answer bigger.” The divisor $n-1$ is chosen because the lemma on the expected sum of squares about the sample mean gives $E[\sum (X_i-\bar{X}{)}^2]=(n-1){\sigma }^2$, so $n-1$ is the unique divisor making $E[s^2]={\sigma }^2$. Being larger than the $n$-divisor version is a consequence, not a justification: dividing by $n-2$ would be larger still, and biased upward.

(c) The correct statement. “For a random sample — independent, identically distributed observations with finite variance, with no assumption of normality — dividing the sum of squared deviations about $\bar{X}$ by $n-1$ produces an estimator $s^2$ with $E[s^2]={\sigma }^2$ exactly; the same is not true of $s$, for which $E[s]<\sigma$.”

(a) Sort, then split. In order: $4,7,9,10,11,12,13,15,18,24,40$. With $n=11$ odd the median is the $6$th value, $Q_2=12$. The rule excludes the median itself from both halves: the lower half is $\{4,7,9,10,11\}$ with median $Q_1=9$, and the upper half is $\{13,15,18,24,40\}$ with median $Q_3=18$. The five-number summary is $\{4,9,12,18,40\}$ and $\mathrm{IQR}=18-9=9$.

(b) Fences. $1.5\mathrm{IQR}=13.5$, so the mild fences are $9-13.5=-4.5$ and $18+13.5=31.5$; $3\mathrm{IQR}=27$, so the extreme fences are $-18$ and $45$. The only value outside a fence is $40$: it exceeds $31.5$ but not $45$, so it is a mild outlier and not an extreme one. Nothing lies below $-4.5$. The lower whisker runs from $Q_1=9$ down to the minimum $4$; the upper whisker runs from $Q_3=18$ to the largest observation inside the fence, $24$, with $40$ plotted separately.

(c) Mean versus median. The sum is $4+7+9+10+11=41$ and $12+13+15+18+24+40=122$, total $163$, so

$$\bar{x}=\frac{163}{11}=14\frac{9}{11}\approx \mathrm{14.8.}$$

The mean exceeds the median $12$ by $\frac{31}{11}\approx 2.8$ ms. By the proposition that skew separates the mean from the median, this suggests right skew — a long upper tail, visible here as $24$ and $40$. It is a rule of thumb, to be confirmed by a plot.

(d) What a distant value can and cannot touch. The $40$ was already the largest observation, so replacing it by $400$ changes no sorted position. Minimum, $Q_1$, $Q_2$, $Q_3$, the IQR, both sets of fences and both whisker ends are all unchanged; in the five-number summary only the maximum becomes $400$. The classification does change: $400>45$, so the point is now an extreme outlier. The mean rises by $\frac{400-40}{11}=\frac{360}{11}\approx 32.7$ ms to $\frac{523}{11}\approx 47.5$ ms, which now exceeds ten of the eleven observations — the skew diagnosis in (c) is sharpened, and the IQR's indifference to the tail is the robustness claim made concrete.

(a) Standardise. $z_{\text{mid}}=\frac{74-62}{8}=\frac{12}{8}=1.5$ and $z_{\text{fin}}=\frac{79-70}{12}=\frac{9}{12}=0.75$. The midterm was the relatively better performance: $1.5$ standard deviations above its mean against $0.75$, even though the raw final score is higher. The z-score is the common ruler that makes the two comparable.

(b) Invariance under rescaling. On the curved scale the mean is $\frac{5}{4}\cdot 62+5=77.5+5=82.5$, the standard deviation is $\frac{5}{4}\cdot 8=10$ (by the proposition on the variance under a linear transformation, $s_y=|a|s_x$), and the student's score is $\frac{5}{4}\cdot 74+5=92.5+5=97.5$. So $z=\frac{97.5-82.5}{10}=1.5$, unchanged.

In general, with $a>0$: $\bar{y}=a\bar{x}+b$ and $s_y=|a|s_x=a{s}_x$, hence

$$z_y=\frac{y-\bar{y}}{s_y}=\frac{(ax+b)-(a\bar{x}+b)}{a{s}_x}=\frac{a(x-\bar{x})}{a{s}_x}=z_x.■$$

If $a<0$ then $s_y=|a|s_x=-a{s}_x$, and the same computation gives $z_y=\frac{a(x-\bar{x})}{-a{s}_x}=-z_x$: a reflection swaps which side of the mean each score is on, so every z-score changes sign but not size.

(c) Chebyshev with $k=2$. $[46,94]=70\pm 24=\mu \pm 2\sigma$. Chebyshev's inequality, which needs only a finite variance, gives $P(|X-\mu |\ge 2\sigma )\le \frac{1}{2^2}=\frac{1}{4}$, so at most $25\%$ of scores lie outside the interval. Under approximate normality the empirical rule puts about $95\%$ inside $\mu \pm 2\sigma$, so only about $5\%$ outside. The universal guarantee is far weaker than the normal figure, as it must be — it has to hold for the worst-case shape.

(d) Solve for $k$. We need $1-\frac{1}{k^2}\ge 0.91$, i.e. $\frac{1}{k^2}\le 0.09=\frac{9}{100}$, i.e. $k^2\ge \frac{100}{9}$, so $k\ge \frac{10}{3}$. Then $k\sigma =\frac{10}{3}\cdot 12=40$ and the interval is $[70-40,70+40]=[30,110]$. Its upper end exceeds any score the exam could award, which is a reminder that Chebyshev uses only two numbers and knows nothing about the scale of the data; the bound is a guarantee, not a description.

(a) Two sums. Since $s$ does not depend on $i$,

$$\sum _{i=1}^n{z}_i=\frac{1}{s}\sum _{i=1}^n(x_i-\bar{x})=\frac{1}{s}(\sum _{i=1}^n{x}_i-n\bar{x})=\frac{1}{s}(n\bar{x}-n\bar{x})=0,$$

using $\sum x_i=n\bar{x}$, the definition of the mean rearranged. For the squares, the definition $s^2=\frac{1}{n-1}\sum (x_i-\bar{x}{)}^2$ says $\sum (x_i-\bar{x}{)}^2=(n-1)s^2$, so

$$\sum _{i=1}^n{z}_i^2=\frac{1}{s^2}\sum _{i=1}^n(x_i-\bar{x}{)}^2=\frac{(n-1)s^2}{s^2}=n-1.■$$

(These are the statements $\bar{z}=0$ and $s_z^2=1$ of the proposition on standardised data, proved here without the linear-transformation propositions.)

(b) A bound on every z-score. Every term $z_i^2$ is non-negative, so for any fixed $j$,

$$z_j^2\le \sum _{i=1}^n{z}_i^2=n-1,\text{hence}|z_j|\le \sqrt{n-1}.■$$

A value with $|z|=4$ requires $\sqrt{n-1}\ge 4$, i.e. $n\ge 17$; for $n\le 16$ it is impossible. In a sample of $10$ no z-score can exceed $3$. Equality in the bound would need every other $z_i$ to be $0$, i.e. every other $x_i=\bar{x}$, but then the deviations summing to zero would force $x_j=\bar{x}$ as well and $s=0$; so the bound is never attained for $n\ge 2$, though it is the right order of magnitude. The practical moral: an “outlier” rule of the form $|z|>3$ cannot fire in a small sample, however extreme the point, which is one reason Tukey's fences are preferred.

(c) A Chebyshev inequality for data. Discard the non-negative terms that do not satisfy the condition, then bound each remaining term below:

$$(n-1)s^2=\sum _{i=1}^n(x_i-\bar{x}{)}^2\ge \sum _{i:|x_i-\bar{x}|\ge ks}(x_i-\bar{x}{)}^2\ge m\cdot k^2{s}^2.$$

Dividing by $n{k}^2{s}^2>0$ gives $\frac{m}{n}\le \frac{n-1}{n{k}^2}$. $■$

Chebyshev's inequality says $P(|X-\mu |\ge k\sigma )\le 1/k^2$ for a random variable with finite variance; this is the same statement for the empirical distribution of a dataset, and slightly stronger, by the factor $\frac{n-1}{n}<1$ that Bessel's divisor introduces. It rests on nothing probabilistic: it is an algebraic inequality about $n$ fixed numbers, so there is no randomness, no independence and no distribution to assume. (Chebyshev's own hypothesis is almost as light — only a finite variance — which is why both bounds are loose: they must survive the worst possible shape.)

(a) Counts and shape. Binning the sorted list:

$[0,5)$: $3,4$ — count $2$. $[5,10)$: $6,6,7,8,8,9,9$ — count $7$. $[10,15)$: $11,12,12,13,14$ — count $5$. $[15,20)$: $16,17,19$ — count $3$. $[20,25)$: $22$ — count $1$. $[25,30)$: $27$ — count $1$. $[30,35)$: none — count $0$. $[35,40)$: $38$ — count $1$. Total $2+7+5+3+1+1+0+1=20$. ✓

The histogram has bars of heights $2,7,5,3,1,1,0,1$ over those intervals, with the horizontal axis labelled “waiting time (minutes)” and the vertical “count”. It is unimodal, peaking in $[5,10)$, and right-skewed: the bars fall off steeply to the left (nothing below $3$) and slowly to the right, with a tail reaching $38$, an empty bin at $[30,35)$ and an isolated bar beyond it.

(b) Five-number summary. $n=20$ is even, so the median is $\frac{1}{2}(x_{(10)}+x_{(11)})=\frac{1}{2}(11+12)=11.5$. The lower half is the first ten values, whose median is $\frac{1}{2}(7+8)=7.5=Q_1$; the upper half is $12,12,13,14,16,17,19,22,27,38$, whose median is $\frac{1}{2}(16+17)=16.5=Q_3$. The summary is $\{3,7.5,11.5,16.5,38\}$ and $\mathrm{IQR}=9$. The mild fences are $7.5-13.5=-6$ and $16.5+13.5=30$. The single outlier is $38>30$; it is mild, not extreme, since the extreme fence is $16.5+27=43.5$.

(c) The box plot beneath. On the same axis draw a box from $7.5$ to $16.5$ with a line at $11.5$; a lower whisker from $7.5$ down to the minimum $3$ (inside the fence); an upper whisker from $16.5$ to $27$, the largest value not exceeding $30$; and $38$ as a separate point.

Preserved: the centre (the median line sits under the tall left-of-centre bars), the spread of the middle half (the box), the skew (the median is left of the box's centre, $4$ from $Q_1$ but $5$ from $Q_3$, and the upper whisker, of length $10.5$, is more than twice the lower one, of length $4.5$), and the outlier. Lost: modality — the box cannot show that the data pile up in $[5,10)$ rather than being spread evenly across $[7.5,16.5]$ — and the gap at $[30,35)$. Five numbers do not determine a shape.

(d) Mean against median. The first ten values sum to $3+4+6+6+7+8+8+9+9+11=71$ and the last ten to $12+12+13+14+16+17+19+22+27+38=190$, so $\bar{x}=261/20=13.05$. Mark the median at $11.5$ and the mean at $13.05$, both inside the $[10,15)$ bar, with the mean to the right.

The order $\text{median}<\text{mean}$ is what right skew predicts, and the two minimisation theorems say why. The mean minimises $\sum (x_i-c{)}^2$, so the far-out values $27$ and $38$ contribute squared distances that grow fast, and the minimiser is dragged toward them. The median minimises $\sum |x_i-c|$, whose slope between data points is the count on the left minus the count on the right — so $38$ counts exactly once, like every other observation, and its distance is irrelevant.

(a) Averaging averages. The error is treating the two branch means as equally weighted. By the proposition on the combined mean of two groups, the pooled mean is weighted by group size:

$$\bar{x}=\frac{40\cdot 6+10\cdot 16}{40+10}=\frac{240+160}{50}=8\text{minutes}.$$

The figure $11$ would be right only if the branches had served equally many customers; with A four times busier, the truth sits much closer to A's $6$.

(b) Wrong centre for a skewed distribution. A mean of $16$ against a median of $9$ says, by the proposition that skew separates the mean from the median, that the waits at B are strongly right-skewed: a few very long waits. “Uses every observation” is the mean's weakness here, not its strength — the mean minimises total squared deviation, so it chases the tail, while the median minimises total absolute deviation and only counts observations, so it stays with the bulk. The recipe for choosing a summary says: skewed data or unremovable outliers, report the median and IQR. Corrected statement: “A typical wait at B is about $9$ minutes (median, $n=10$); the mean of $16$ is inflated by a few very long waits and should be reported with the IQR, not on its own.”

(c) A flag is not a licence to delete. Tukey's fences are calibrated on normal data, where they flag about $0.7\%$ of values. On right-skewed waiting times the long tail is the distribution, not a contamination of it, and the rule flags legitimate observations wholesale — $3$ out of $10$ is itself evidence of skew, not of three errors. Deleting them removes the longest waits from the record, biases $s$ downward, and produces a summary of a dataset that does not exist. Corrected procedure: investigate each flagged wait and delete only if a recording error is found; otherwise keep all $10$, report the five-number summary and IQR (which the tail cannot distort) alongside $s$ computed on all the data, and state $n$.

(d) A z-score is not a percentile. The computation $z=(18-8)/5=2$ is fine. The conversion of $z=2$ into the $97.7$th percentile assumes a normal distribution, and the data are not normal: they are right-skewed by (b), and waits cannot be negative while a normal with $\mu =8$, $\sigma =5$ extends well below zero. Without a shape, only Chebyshev's inequality applies: $P(|X-8|\ge 10)\le \frac{1}{4}$. Since $X\ge 0$, the event $|X-8|\ge 10$ is exactly $X\ge 18$ — the lower tail $X\le -2$ is empty — so the honest statement is “at most $25\%$ of customers waited $18$ minutes or longer; nothing sharper can be said from $\bar{x}$ and $s$ alone.” The $97.7$ figure is unsupported, and on skewed data could be badly wrong in either direction.

(a) Decompose each group about its own mean. Within group $A$, write $x_i-\bar{x}=(x_i-{\bar{x}}_A)+({\bar{x}}_A-\bar{x})$ and expand the square:

$$\sum _{i\in A}(x_i-\bar{x}{)}^2=\sum _{i\in A}(x_i-{\bar{x}}_A{)}^2+2({\bar{x}}_A-\bar{x})\sum _{i\in A}(x_i-{\bar{x}}_A)+n_A({\bar{x}}_A-\bar{x}{)}^2.$$

The middle sum is the total deviation of group $A$ from its own mean, which is zero by the proposition that deviations from the mean sum to zero. What remains is ${\mathrm{SS}}_A+n_A({\bar{x}}_A-\bar{x}{)}^2$ — this is exactly the identity $S(c)=S({\bar{x}}_A)+n_A(c-{\bar{x}}_A{)}^2$ from the theorem that the mean minimises total squared deviation, applied to group $A$ with $c=\bar{x}$. The same computation for group $B$ gives ${\mathrm{SS}}_B+n_B({\bar{x}}_B-\bar{x}{)}^2$, and since every observation belongs to exactly one group, adding the two gives the claim. $■$

(b) The between-groups term. Since $\bar{x}=\frac{n_A{\bar{x}}_A+n_B{\bar{x}}_B}{N}$,

$${\bar{x}}_A-\bar{x}=\frac{N{\bar{x}}_A-n_A{\bar{x}}_A-n_B{\bar{x}}_B}{N}=\frac{n_B({\bar{x}}_A-{\bar{x}}_B)}{N},{\bar{x}}_B-\bar{x}=\frac{n_A({\bar{x}}_B-{\bar{x}}_A)}{N}.$$

Writing $d={\bar{x}}_A-{\bar{x}}_B$,

$$n_A({\bar{x}}_A-\bar{x}{)}^2+n_B({\bar{x}}_B-\bar{x}{)}^2=\frac{n_A{n}_B^2{d}^2+n_B{n}_A^2{d}^2}{N^2}=\frac{n_A{n}_B(n_B+n_A)}{N^2}{d}^2=\frac{n_A{n}_B}{N}{d}^2.■$$

The between-groups contribution depends only on the two sizes and the gap between the two means.

(c) Numbers. The pooled mean is $\bar{x}=\frac{4\cdot 10+6\cdot 15}{10}=\frac{40+90}{10}=13$. Recover each sum of squares from its variance via $\mathrm{SS}=(n-1)s^2$: ${\mathrm{SS}}_A=3\cdot 4=12$ and ${\mathrm{SS}}_B=5\cdot 9=45$. The between term is $\frac{4\cdot 6}{10}(10-15{)}^2=\frac{24}{10}\cdot 25=60$; as a check, $4(10-13{)}^2+6(15-13{)}^2=36+24=60$. ✓ So the pooled sum of squares is $12+45+60=117$ and

$$s^2=\frac{117}{10-1}=13,s=\sqrt{13}.$$

(d) Why the pooled variance can exceed both. Each group's variance measures spread about that group's own centre. The pooled variance measures spread about $13$, a point that group $A$ sits $3$ below and group $B$ sits $2$ above; that displacement is what the between term $60$ records, and here it outweighs the two within terms combined ($57$). Sums of squares add; variances about different centres do not. Naively averaging $4$ and $9$ would be wrong twice over — wrong weights, and no between term at all.

The between term vanishes exactly when ${\bar{x}}_A={\bar{x}}_B$. Then $\mathrm{SS}={\mathrm{SS}}_A+{\mathrm{SS}}_B$ and

$$s^2=\frac{(n_A-1)s_A^2+(n_B-1)s_B^2}{N-1},$$

a weighted average of the two variances with weights $n_A-1$ and $n_B-1$ — the degrees of freedom of each group. This is the pooled variance that a two-sample comparison of means uses, and the condition of equal group means is precisely what makes it legitimate as a description of a single spread.