Data Distributions and Statistics: Practice

The median is the middle value of the sorted data, so sort first.

$5, \; 7, \; 9, \; 12, \; 20$

There are five values, an odd count, so a single value sits in the middle: the third one.

$\text{median} = 9$

The value $10.6$ is the mean of this data, which is a different summary and a different question.

Sort the values before touching the middle.

$2, \; 6, \; 9, \; 15$

The count is even, so no single value is in the middle. Average the two middle values, $6$ and $9$.

$\text{median} = \frac{6 + 9}{2} = 7.5$

Averaging the middle two of the unsorted list would give $\frac{2 + 15}{2} = 8.5$, which is the trap here. The median is defined on the sorted order.

The range is the distance from the smallest value to the largest, so identify both ends. The maximum is $31$ and the minimum is $9$.

$\text{range} = \text{max} - \text{min} = 31 - 9 = 22$

The range measures spread, not center. Note that it is built from only the two most extreme values, which is exactly what makes it fragile.

The count is even, so cut the list into two halves of four. The lower half is $2, 4, 5, 9$.

$Q_1$ is the median of that lower half, and the half has an even count, so average its two middle values, $4$ and $5$.

$Q_1 = \frac{4 + 5}{2} = 4.5$

The value $10$ is the median of the whole list and $13$ is $Q_3$, so neither answers this question.

The interquartile range is the width of the box, which runs from $Q_1$ to $Q_3$.

$\text{IQR} = Q_3 - Q_1 = 31 - 18 = 13$

The value $28$ is the range, $40 - 12$, which stretches from whisker tip to whisker tip. The IQR describes only the middle half of the data, which is what makes it resistant to outliers.

The sample mean is $\bar{x} = \frac{1 + 3 + 4 + 5 + 7}{5} = \frac{20}{5} = 4$, so the deviations are $-3, -1, 0, 1, 3$ and their squares add to

$\sum (x_i - \bar{x})^2 = 9 + 1 + 0 + 1 + 9 = 20.$

These values are a sample, not a whole population, so divide by $n - 1 = 4$.

$s^2 = \frac{20}{4} = 5$

Dividing by $n = 5$ would give $4$, the population variance. The larger sample answer is the correction doing its job, compensating for deviations that were measured from the sample's own mean.

Expanding the total squared distance shows it is a quadratic in the guess $c$.

$S(c) = \sum (x_i - c)^2 = nc^2 - 2c\sum x_i + \sum x_i^2$

The leading coefficient $n$ is positive, so this parabola opens upward and its minimum sits at the vertex.

$c = \frac{2\sum x_i}{2n} = \frac{1}{n}\sum x_i = \bar{x}$

So the mean is the unique minimizer of the total squared distance. The total absolute distance is minimized by the median instead, which is a different optimization problem with a different champion.

Pair the sorted values from the outside in, the smallest with the largest and so on. For a pair $a \le b$, any guess $c$ lying between them gives

$\lvert a - c\rvert + \lvert b - c\rvert = (c - a) + (b - c) = b - a,$

the fixed gap between the partners, while a guess outside the pair gives strictly more. So each pair is happiest when $c$ sits between its two members, and the only point inside every pair at once is the middle of the data.

That makes the median the minimizer of the total absolute distance. The total squared distance is the mean's problem, not the median's.

The mean fixes the total, so find the total the four numbers must have.

$\text{sum} = 4 \times 15 = 60$

The three known values contribute $10 + 14 + 20 = 44$, so the missing value is whatever is left over.

$60 - 44 = 16$

Check it: $\frac{10 + 14 + 20 + 16}{4} = \frac{60}{4} = 15$, as required.

The median cuts the sorted data in half. $Q_1$ is the median of the lower half, so roughly a quarter of the data sits below it, and $Q_3$ is the median of the upper half, so roughly a quarter sits above it.

That leaves the middle for the box.

$1 - \tfrac{1}{4} - \tfrac{1}{4} = \tfrac{1}{2}$

The box holds the middle half of the data, which is why its width, the IQR, is a measure of spread that ignores the extremes on both ends.

Find the IQR, then the fences.

$\text{IQR} = 32 - 20 = 12, \qquad 1.5 \times 12 = 18$

The fences sit that far beyond each end of the box.

$Q_1 - 18 = 2 \qquad \text{and} \qquad Q_3 + 18 = 50$

A value is flagged only if it falls outside those fences. Now test the candidates: $48$ and $20$ sit inside the upper fence, and $4$ sits above the lower fence of $2$, so none of them is flagged. Only $52$ lies beyond $50$, so it is the outlier.

The count is $8$, so the quartiles come from halves of four. The lower half is $10, 12, 14, 16$ and the upper half is $18, 20, 22, 80$.

$Q_1 = \frac{12 + 14}{2} = 13, \qquad Q_3 = \frac{20 + 22}{2} = 21$

Changing $80$ into $800$ leaves it the largest value, so the upper half is still four values whose middle two are $20$ and $22$. Both quartiles are untouched, and so is their difference.

$\text{IQR} = 21 - 13 = 8 \quad \text{before and after}$

Everything else moves: the range jumps from $70$ to $790$, the mean climbs from $24$ to $114$, and the standard deviation explodes along with it.