Data Distributions and Statistics: Practice

There are $12$ values, an even count, so cut the sorted list straight down the middle into two halves of six.

The lower half is $8, 10, 12, 15, 18, 20$ and the upper half is $22, 25, 27, 30, 33, 40$. The third quartile is the median of that upper half, and since the half has an even count, average its two middle values, $27$ and $30$.

$Q_3 = \frac{27 + 30}{2} = 28.5$

The median of the whole list is $\frac{20 + 22}{2} = 21$, which marks the boundary between the two halves rather than a quartile, and $25$ and $27$ are simply data values inside the upper half.

The count is even, so the halves are $11, 13, 14, 16$ and $19, 21, 24, 60$. Each quartile is the median of its half.

$Q_1 = \frac{13 + 14}{2} = 13.5, \qquad Q_3 = \frac{21 + 24}{2} = 22.5$

The interquartile range is $22.5 - 13.5 = 9$, so the fences sit $1.5 \times 9 = 13.5$ beyond each end of the box.

$Q_1 - 13.5 = 0 \qquad \text{and} \qquad Q_3 + 13.5 = 36$

Now test the extremes. The smallest value, $11$, is above the lower fence of $0$, so it is ordinary. The value $24$ sits below the upper fence of $36$, so it is ordinary too. Only $60$ lies beyond $36$, so it alone is flagged.

The mean pins down the total, so work out what the five values must add to.

$\text{sum} = 5 \times 20 = 100$

The four known values contribute $12 + 18 + 25 + 30 = 85$, so the missing value is the shortfall.

$100 - 85 = 15$

Check it: $\frac{12 + 18 + 25 + 30 + 15}{5} = \frac{100}{5} = 20$. The value $85$ is the sum of the four known numbers, not the answer.

Both formulas start from the same sum of squared deviations, $40$. Only the divisor differs.

$\sigma^2 = \frac{40}{n} = \frac{40}{5} = 8, \qquad s^2 = \frac{40}{n - 1} = \frac{40}{4} = 10$

Subtracting gives the gap.

$10 - 8 = 2$

The sample version is always the larger of the two, because dividing by the smaller number $n - 1$ inflates the answer. That inflation is deliberate: deviations measured from the sample's own mean are guaranteed to come out too small, since the sample mean is the value that makes them as small as they can be.

Both sets have the same mean, $20$, so compare their deviations directly.

For set A the deviations are $-2, 0, 2$ and for set B they are $-6, 0, 6$. Square and average each list.

$\sigma_A^2 = \frac{4 + 0 + 4}{3} = \frac{8}{3}, \qquad \sigma_B^2 = \frac{36 + 0 + 36}{3} = 24$

Set B's variance is nine times set A's, so its standard deviation is three times as large. Equal centers say nothing about spread, which is exactly why a mean on its own is never a summary of a data set.

You cannot average the two means directly, because they summarize different numbers of values. Recover the totals first, using the fact that a total is the count times the mean.

$\text{total}_A = 3 \times 12 = 36, \qquad \text{total}_B = 5 \times 20 = 100$

Now pool everything and divide by the combined count.

$\bar{x} = \frac{36 + 100}{3 + 5} = \frac{136}{8} = 17$

Averaging $12$ and $20$ naively gives $16$, which is the trap. The true mean is pulled toward $20$ because group B has more values, and the correct calculation weights each group by its size.

The sample mean is $\bar{x} = \frac{50}{5} = 10$, so the deviations are $-2, -2, 0, 2, 2$ and their squares add to

$4 + 4 + 0 + 4 + 4 = 16.$

The values are a sample, so divide by $n - 1 = 4$, then take the square root to return to the units of the data.

$s^2 = \frac{16}{4} = 4, \qquad s = \sqrt{4} = 2$

Dividing by $n = 5$ would give the population variance $3.2$ and a population standard deviation of $\sqrt{3.2} \approx 1.79$. Reporting $4$ stops at the variance without taking the root.

Pair the values from the outside in: $(1, 9)$ and $(4, 6)$. Any guess $c$ lying between a pair's two members contributes exactly that pair's gap, and any guess outside it contributes more.

So the outer pair wants $1 \le c \le 9$ and the inner pair wants $4 \le c \le 6$. Both are satisfied at once precisely on the inner interval, where the total is the sum of the gaps.

$(9 - 1) + (6 - 4) = 10$

Test it: at $c = 5$ the total is $4 + 1 + 1 + 4 = 10$, and at $c = 4$ it is $3 + 0 + 2 + 5 = 10$, the same. At $c = 3$ it is $2 + 1 + 3 + 6 = 12$, which is worse.

With an even count the minimizer is not unique. Every value between the two middle values ties, and the median, $5$, is simply the midpoint of that winning interval.

A box plot draws exactly five numbers: the minimum, $Q_1$, the median, $Q_3$, and the maximum. The line inside the box is the median, so it can be read straight off the picture.

The other three cannot. The mean and the standard deviation depend on the sizes of every individual value, and a box plot has thrown those away. The count is invisible too: the same box plot could summarize $20$ values or $20{,}000$.

This is the honest cost of a box plot. It shows you the order structure of the data, and nothing else.

Only one value changed, so only the total changed, and it changed by

$d = 70 - 30 = 40.$

The mean is the total divided by $n$, so an increase of $40$ in the total raises the mean by $\frac{d}{n}$.

$\bar{x}_{\text{new}} = 50 + \frac{40}{20} = 50 + 2 = 52$

Notice how the correction is diluted: a $40$-point error in one value moved the mean by only $2$, because $20$ values shared the damage. With $n = 2$ the same error would have moved the mean by $20$.

The range is $\text{max} - \text{min}$. A new value can only push the maximum up or leave it alone, and can only push the minimum down or leave it alone, so

$\text{new range} \ge \text{old range}$

always. The range grows or stays put, and never shrinks.

Each of the others can fall. Adding a value below everything else drags the mean down, and it drags the median down too.

Even the IQR can shrink. Take $10, 20, 60, 100, 110$ and use this lesson's convention for an odd count, leaving the middle value out of both halves. The halves are $10, 20$ and $100, 110$, so $Q_1 = 15$ and $Q_3 = 105$, giving an IQR of $90$. Now add one value, $55$.

$10, \; 20, \; 55, \; 60, \; 100, \; 110$

The halves are now $10, 20, 55$ and $60, 100, 110$, so $Q_1 = 20$ and $Q_3 = 100$, and the IQR has fallen to $80$.

The distribution is bimodal: it has two separate peaks, which almost always means two different groups have been mixed together, here the people who walk and the people who drive.

Both centers land at about $30$ minutes, in the empty valley between the peaks. That value is arithmetically correct and describes nobody's morning, since essentially no one commutes for $30$ minutes.

When a distribution has two peaks, the fix is not to hunt for a better center. It is to split the data and report each group, which is why looking at the shape has to come before computing any summary at all.