Data Distributions and Statistics: Practice

The minimizer is the mean, so find it first.

$\bar{x} = \frac{5 + 7 + 8 + 10 + 10}{5} = \frac{40}{5} = 8$

Now evaluate the total squared distance at $c = 8$. The deviations are $-3, -1, 0, 2, 2$.

$S(8) = 9 + 1 + 0 + 4 + 4 = 18$

The shortcut $\sum x_i^2 - n\bar{x}^2$ gives the same thing: $\sum x_i^2 = 25 + 49 + 64 + 100 + 100 = 338$, and $338 - 5(64) = 18$. The value $8$ is the mean itself, not the minimum of $S$.

With eight values the halves are $2, 30, 33, 35$ and $37, 39, 41, 44$, and each quartile is the median of its half.

$Q_1 = \frac{30 + 33}{2} = 31.5, \qquad Q_3 = \frac{39 + 41}{2} = 40$

So $\text{IQR} = 40 - 31.5 = 8.5$, and the fences sit $1.5 \times 8.5 = 12.75$ beyond each end of the box.

$Q_1 - 12.75 = 18.75 \qquad \text{and} \qquad Q_3 + 12.75 = 52.75$

The largest value, $44$, is comfortably below the upper fence, so it is ordinary. The smallest value, $2$, falls below $18.75$, so it is flagged. Outliers can sit at either end, and here the low one is the anomaly.

Delete the $28$ and recompute both summaries on the nine remaining values, $4, 6, 7, 9, 10, 12, 13, 15, 16$.

The new total is $92$, so the mean falls from $12$ to

$\bar{x} = \frac{92}{9} \approx 10.22,$

a drop of about $1.78$. The new median is the fifth of the nine sorted values, which is $10$, a drop of exactly $1$ from the old median of $11$.

Both summaries fall, because the value removed was above each of them, but the mean falls by nearly twice as much. The median only re-indexed by one position, while the mean gave up the full size of the deleted value.

The least-squares theorem says $\sum (x_i - c)^2$ is smallest at $c = \bar{x}$. Feeding the population's true mean $\mu$ into that same sum can therefore only give something at least as big.

$\sum (x_i - \bar{x})^2 \le \sum (x_i - \mu)^2$

Equality happens in exactly one case, the coincidence $\bar{x} = \mu$, which a sample essentially never delivers.

The spread you want is the one measured from $\mu$, on the right. The spread you can actually compute is the one on the left, and it can never come out too big: it is too small in every sample but that lucky one, and not by bad luck but by construction, because the sample fitted its center to itself. An estimate that misses the same way nearly every time is biased, and this one is biased downward.

Dividing by $n - 1$ rather than $n$ inflates the result to compensate, so it makes the answer larger, not smaller. Proving that $n - 1$ is the exactly right correction needs expected values, which is a debt this lesson leaves open.

The interquartile range is $Q_3 - Q_1$, and both quartiles are medians of halves, so they depend only on the order of the values. In any data set big enough that the extremes sit outside the box, pushing the largest value further out leaves it as the largest value, so it changes no quartile and the IQR does not move. (With only three or four values the largest value is itself part of $Q_3$, and then even the IQR shifts, which is why that caveat belongs on the claim.)

The other three feel the size of the extreme values in every data set. The range is $\text{max} - \text{min}$, which is built from precisely the two most outlier-prone points. The variance and the standard deviation square the deviations, which makes a far-off value dominate the sum.

The rule to carry away is that the median pairs with the IQR, and the mean pairs with the standard deviation. Never mix a resistant center with a sensitive spread.

The two means cannot simply be averaged, because they summarize different numbers of students. Turn each mean back into a total.

$20 \times 70 = 1400, \qquad 30 \times 80 = 2400$

Now pool the scores and divide by the total number of students.

$\bar{x} = \frac{1400 + 2400}{50} = \frac{3800}{50} = 76$

The naive average of $70$ and $80$ is $75$, which is the trap. The true mean is pulled toward $80$ because more students sit in the higher-scoring class, and $76$ reflects that weighting.

Read the five numbers off the picture. The whiskers reach the minimum, $10$, and the maximum, $90$. The box runs from $Q_1 = 25$ to $Q_3 = 55$, and the line drawn inside it is the median, $40$.

The interquartile range is the width of the box.

$\text{IQR} = Q_3 - Q_1 = 55 - 25 = 30$

The value $80$ is the range, $90 - 10$, which is the distance from whisker tip to whisker tip and a far less trustworthy measure of spread. The box holds the middle half of the data, so its width ignores both extremes.

The minimum of $S$ sits at $c = \mu$, so the smallest possible total squared distance is

$S(\mu) = \sum (x_i - \mu)^2 .$

Now read the definition of the population variance backwards. It is the average squared deviation, $\sigma^2 = \frac{1}{n}\sum (x_i - \mu)^2$, so multiplying both sides by $n$ recovers the sum.

$\sum (x_i - \mu)^2 = n\sigma^2$

So the variance is exactly the minimum total squared distance, shared out one part per value. That is not a coincidence of notation: the variance is defined to be the best score any center could achieve, divided by $n$. It is zero only when every value equals the mean.

Compute the IQR and then the fences.

$\text{IQR} = 24 - 12 = 12, \qquad 1.5 \times 12 = 18$

$Q_1 - 18 = -6 \qquad \text{and} \qquad Q_3 + 18 = 42$

Now test the two extremes. The minimum, $3$, sits above the lower fence of $-6$, so it is an ordinary value even though it looks far from the box. The maximum, $60$, sits beyond the upper fence of $42$, so it is flagged.

The lopsided result is itself informative. A box whose median sits closer to $Q_3$ than to $Q_1$, with a long reach to the maximum, is the fingerprint of a right-skewed data set.

Compute the total absolute distance at each candidate. At the mean, $c = 5$,

$\lvert 2 - 5\rvert + \lvert 4 - 5\rvert + \lvert 9 - 5\rvert = 3 + 1 + 4 = 8.$

At the median, $c = 4$,

$\lvert 2 - 4\rvert + \lvert 4 - 4\rvert + \lvert 9 - 4\rvert = 2 + 0 + 5 = 7.$

The median wins, and by the pairing theorem it is the unique winner for this odd-sized data set. Run the same contest with squared distance and the result flips: the mean scores $9 + 1 + 16 = 26$ while the median scores $4 + 0 + 25 = 29$. Each center wins its own game and loses the other one.

The divisor is a statement about what you are claiming, not about how many numbers you happen to have.

Here the twelve players are the whole group of interest. There is no larger population being estimated, so the mean you computed is the true mean $\mu$, not a stand-in for one. Nothing is being fitted, so nothing needs correcting.

$\sigma^2 = \frac{1}{12}\sum (x_i - \mu)^2$

The $n - 1$ correction exists only to compensate for measuring deviations from a sample mean that was fitted to the sample itself. If you had chosen $12$ players out of a whole league in order to estimate the league, you would divide by $11$.

Adding a constant to every value shifts the mean by that same constant, so every deviation survives untouched.

$(x_i + 7) - (\bar{x} + 7) = x_i - \bar{x}$

The standard deviation is built only from those deviations, so it does not move: the data $11, 13, 15, 17, 19$ still has variance $8$.

Each of the other three changes it. Multiplying by $7$ multiplies every deviation by $7$ and so multiplies the variance by $49$. Adding a value equal to the mean contributes a zero deviation while raising the count, dropping the variance to $\frac{40}{6} \approx 6.67$. Removing the $12$ leaves $4, 6, 8, 10$ with mean $7$ and variance $\frac{20}{4} = 5$.