Data Distributions and Statistics: Free Response

The mean is the total divided by the count:

$\bar{x} = \frac{6 + 10 + 20}{3} = \frac{36}{3} = 12.$

The data is already sorted, and with three values the median is the single middle one, $10$.

Score the median first:

$S(10) = (6-10)^2 + (10-10)^2 + (20-10)^2 = 16 + 0 + 100 = 116.$

Now score the mean:

$S(12) = (6-12)^2 + (10-12)^2 + (20-12)^2 = 36 + 4 + 64 = 104.$

Since $104 < 116$, the mean scores lower, exactly as the theorem guarantees: no other guess can score lower than the mean does on total squared distance.

Score the median:

$T(10) = |6-10| + |10-10| + |20-10| = 4 + 0 + 10 = 14.$

Score the mean:

$T(12) = |6-12| + |10-12| + |20-12| = 6 + 2 + 8 = 16.$

Here $14 < 16$, so the median wins this contest even though the mean won part B. That is not a contradiction: $S$ and $T$ are two different scoring rules. The mean is always the unique minimizer of total squared distance. The median minimizes total absolute distance, and because this data set has an odd count of values ($n=3$), that minimizer is unique here too. A guess built to win one contest has no reason to also win the other.

The mean is

$\bar{x} = \frac{6+7+9+13+15}{5} = \frac{50}{5} = 10.$

The deviations are $6-10=-4$, $7-10=-3$, $9-10=-1$, $13-10=3$, and $15-10=5$. Adding them: $-4-3-1+3+5 = 0$, as the mean always guarantees.

Square the deviations from part A and add them:

$(-4)^2+(-3)^2+(-1)^2+3^2+5^2 = 16+9+1+9+25 = 60.$

Since this is the whole population, divide by $N = 5$:

$\sigma^2 = \frac{60}{5} = 12, \qquad \sigma = \sqrt{12} = 2\sqrt{3}.$

The sum of squared deviations does not change; only the divisor does. As a sample,

$s^2 = \frac{60}{n-1} = \frac{60}{4} = 15, \qquad s = \sqrt{15}.$

This is bigger than $\sigma^2 = 12$ because the same numerator, $60$, is divided by a smaller number. That is not an accident: the sample mean is, by definition, the value that makes the sum of squared deviations from itself as small as possible, no larger than the sum of squared deviations from the true population mean would be. So on average, across repeated samples, deviations taken from a sample's own mean understate the spread of the larger population the sample came from. Dividing by $n-1$ instead of $n$ corrects for that average understatement.

With $n=10$, an even count, split the sorted list exactly in half. The median is the average of the $5$th and $6$th values:

$\text{median} = \frac{7+8}{2} = 7.5.$

The lower half is $3,4,5,6,7$, whose median is the middle value, $Q_1 = 5$. The upper half is $8,9,10,11,30$, whose median is $Q_3 = 10$. The minimum is $3$ and the maximum is $30$.

$\text{IQR} = Q_3 - Q_1 = 10 - 5 = 5.$

The fences sit $1.5$ IQRs beyond each end of the box:

$Q_1 - 1.5(5) = 5 - 7.5 = -2.5, \qquad Q_3 + 1.5(5) = 10 + 7.5 = 17.5.$

Every value from $3$ through $11$ lies between $-2.5$ and $17.5$, but $30 > 17.5$, so the $30$-hour value is the one outlier by Tukey's convention.

The mean is the total divided by the count:

$\bar{x} = \frac{3+4+5+6+7+8+9+10+11+30}{10} = \frac{93}{10} = 9.3.$

That sits above the median of $7.5$ found in part A. Nine of the ten students studied at most $11$ hours, and the tenth studied $30$, which part B flagged as an outlier by Tukey's convention. That single student's hours are large enough in size to drag the mean up past almost everyone else's total, while the median only reflects rank and is unmoved by how far out that one value sits. So the median, together with the IQR from part B, is the honest description of a typical student's week; the mean overstates it.

The mean is

$\bar{x} = \frac{10+15+20+25+300}{5} = \frac{370}{5} = 74.$

The data is already sorted, and with five values the median is the middle one, $20$.

The claim treats the mean as though it describes most donors, but no individual donation actually sits close to it. Line the donations up against the mean:

$10, \; 15, \; 20, \; 25 \; < \; 74 \; < \; 300.$

Four of the five donations sit far below $74$ dollars, and the fifth, $300$ dollars, sits far above it in the other direction. The mean is sensitive to size, so that single large gift is enough to drag it far above where four-fifths of the donors actually gave. The median, $20$ dollars, is what a genuinely typical donor gave here, and it is unmoved by exactly how large that one big donation happens to be.

Only one value changed, by $d = 3000 - 300 = 2700$, so the mean shifts by $d/n$:

$\text{new mean} = 74 + \frac{2700}{5} = 74 + 540 = 614.$

The median does not move at all. The $300$-dollar donation was already the largest value, and raising it further keeps it the largest; the sorted order of the other four donations, and therefore which one sits in the middle, is completely untouched, so the median stays at $20$ dollars. The mean and the median responded so differently because one is built from the sizes of the values and the other only from their order.

The original mean is $\bar{x} = S/n$. Only one term of the sum changed, from $x_n$ to $x_n + d$, so the new total is

$S' = S + d.$

Dividing by the same count $n$,

$\bar{x}' = \frac{S'}{n} = \frac{S+d}{n} = \frac{S}{n} + \frac{d}{n} = \bar{x} + \frac{d}{n}.$

This holds for any $n$ and any $d$, since nothing about the argument used the value of either.

The median is a function of ORDER, not size: it is defined as whichever value (or average of two values) sits at a fixed rank once the data is sorted. Before the change, $x_n$ already held the largest rank, position $n$, meaning

$x_1 \le x_2 \le \cdots \le x_{n-1} \le x_n.$

Adding a positive amount $d$ to $x_n$ can only make it larger still, so $x_n + d \ge x_n \ge x_{n-1} \ge \cdots \ge x_1$, and it continues to hold rank $n$: no other value can leapfrog past it, since none of them changed. With $x_n$'s rank fixed at the top both before and after, the ranks of every other value are unchanged too, and in particular whichever value or pair of values defines the median sits at exactly the same rank it always did. The size of $d$ never enters the argument, so this holds no matter how large the change is.

With $n=3$ sorted values $x_1 \le x_2 \le x_3$, the convention leaves the middle value out of both halves. The lower half is the single value $x_1$, so $Q_1$, its median, is $x_1$ itself. The upper half is the single value $x_3$, so $Q_3$, the median of a one-element set, is that element:

$Q_3 = x_3.$

When $x_3$ becomes $x_3+d$, this equation says $Q_3$ becomes $x_3+d$ too, an increase of exactly $d$. Since $Q_1 = x_1$ never moved, the IQR increases by that same $d$:

$\text{IQR}' = (x_3+d) - x_1 = \text{IQR} + d.$

This is the opposite of part B's conclusion. The median survived the change untouched for any $n \ge 3$, but the IQR here moves in lockstep with $d$, because at $n=3$ the extreme value $x_3$ is not merely nearby the quartile calculation, it IS the quartile. The IQR's resistance depends on having enough data for the extremes to sit outside the quartiles, and $n=3$ is one such small-sample case where they do not.