Data Distributions and Statistics: Additional Practice (Set 2)

The mean comes straight from the first sum.

$\mu = \frac{48}{6} = 8$

Completing the square in the earlier proof left the identity $\sum (x_i - \mu)^2 = \sum x_i^2 - n\mu^2$, which lets you get the sum of squared deviations without listing the data.

$\sum (x_i - \mu)^2 = 408 - 6(8^2) = 408 - 384 = 24$

These are the whole population, so divide by $N = 6$.

$\sigma^2 = \frac{24}{6} = 4$

Dividing by $n - 1 = 5$ would give the sample variance $4.8$, and stopping at $24$ reports the sum of squared deviations rather than its average.

Check the two centers. The total is $180$, so the mean is $\frac{180}{6} = 30$, and the median averages the third and fourth sorted values.

$\text{median} = \frac{12 + 48}{2} = 30$

They agree perfectly, and yet the nearest actual data values are $12$ and $48$. The distribution is bimodal: two tight clusters with a wide empty gap between them, and both centers land in that gap.

The data is not skewed at all, in fact it is symmetric about $30$, and its standard deviation is far from zero since the values are spread widely. This is the case where no single center is honest, and the right response is to report the two groups separately.

For six values to have a mean of $10$, their total must be

$6 \times 10 = 60.$

The five existing values already total $4 + 8 + 10 + 12 + 16 = 50$, so the new value must supply the difference.

$60 - 50 = 10$

The value that leaves the mean alone is the mean itself, which makes sense from the balance argument: a point sitting exactly at the balance point adds no turning force to either side. Notice, though, that it does change the population variance, since it adds a zero deviation to the pile while increasing the count.

The mean fixes the total, so start there.

$\text{sum} = 6 \times 15 = 90$

The five known values contribute $11 + 13 + 14 + 18 + 20 = 76$, so the missing value must make up the difference.

$90 - 76 = 14$

It is tempting to guess $15$, the mean itself, but that would only be right if the five known values already averaged $15$, and they do not: they average $\frac{76}{5} = 15.2$, slightly high, so the sixth value has to pull the mean back down.

The variance is an average of squared deviations, and a square is never negative. For that average to be $0$, every single term must be $0$.

$(x_i - \bar{x})^2 = 0 \quad \text{for every } i \;\Longrightarrow\; x_i = \bar{x} \quad \text{for every } i$

So every value equals the mean, which means all the values are identical. The data set $7, 7, 7, 7$ has standard deviation $0$, and its mean and median are both $7$, not $0$.

That same example rules out the claim that the data must contain exactly one value: a lone value does have zero spread, but it is not the only kind of data set that does.

Total absolute distance is the median's problem, not the mean's. The data is already sorted and has an odd count, so the median is the single middle value, the third one.

$\text{median} = 10$

Check it against its nearest rival, the mean, which is $\frac{3 + 8 + 10 + 10 + 14}{5} = \frac{45}{5} = 9$. At $c = 10$ the total absolute distance is

$7 + 2 + 0 + 0 + 4 = 13,$

while at $c = 9$ it is $6 + 1 + 1 + 1 + 5 = 14$, which is worse. With an odd count the median is the unique minimizer, so no other value can tie it.

The sample mean is $\bar{x} = \frac{90}{9} = 10$, so four values sit $3$ below it, four sit $3$ above it, and one sits on it.

$\sum (x_i - \bar{x})^2 = 4(9) + 0 + 4(9) = 72$

The values are a sample, so divide by $n - 1 = 8$ and then take the square root.

$s^2 = \frac{72}{8} = 9, \qquad s = \sqrt{9} = 3$

Dividing by $n = 9$ instead would give a population variance of $8$ and a population standard deviation of $2\sqrt{2} \approx 2.83$, which is the smaller, different answer to a different question.

Both start from the same sum of squared deviations. Call it $Q$, so that the two variances differ only in their divisor.

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

Because the values are not all equal, $Q > 0$, so dividing one variance by the other is legitimate. The sum $Q$ cancels and only the divisors survive.

$\frac{s^2}{\sigma^2} = \frac{n}{n-1} = \frac{5}{4} = 1.25$

The sample variance is always the larger, and the ratio $\frac{n}{n-1}$ shrinks toward $1$ as $n$ grows. With $5$ values the correction inflates the answer by a full $25$ percent, while with $500$ values it changes almost nothing.

The "not all equal" condition is doing real work. If all five values were identical then $Q = 0$, both variances would be $0$, and they really would be equal, with no ratio to speak of.

The mean feels the size of every value in a long tail, while the median only counts how many values are out there, so a right tail usually drags the mean above the median. That signal is reliable enough to be a good skew detector, which is why a gap between the two is worth investigating.

But it is not a theorem, and the word "always" is what breaks the other options. Statisticians have constructed data sets with a long right tail whose mean nonetheless falls below the median.

The claim that the mean and median agree only for symmetric data fails too, because equality does not force symmetry. The data $0, 2, 4, 4, 10$ has

$\bar{x} = \frac{20}{5} = 4 \qquad \text{and} \qquad \text{median} = 4,$

yet it is plainly not symmetric about $4$.

Doubling and shifting is an increasing transformation, so it does not reorder the data. Whichever value was in the middle before is still in the middle, and it simply gets transformed.

$\text{new median} = 2(12) + 3 = 27$

The quartiles transform the same way, so

$\text{new IQR} = (2Q_3 + 3) - (2Q_1 + 3) = 2(Q_3 - Q_1) = 2(5) = 10.$

The $+3$ cancels in the subtraction, which is the point: adding a constant slides the data without stretching it, so it moves the center but leaves every measure of spread alone. Only the factor of $2$ affects the IQR.

The sample variance divides the sum of squared deviations by $n - 1$, so run that definition backwards.

$s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1} \;\Longrightarrow\; \sum (x_i - \bar{x})^2 = s^2 \times (n - 1)$

With $n = 4$ the divisor is $3$, not $4$.

$\sum (x_i - \bar{x})^2 = 10 \times 3 = 30$

Multiplying by $n = 4$ gives $40$, which is what you would get if this were a population variance. The divisor is the only thing separating the two, so it has to be read off the wording of the question.

The mean sits above the median, and the one extreme value is far out on the high side, so the long tail runs to the right.

$\bar{x} = 12 > 11 = \text{median}$

That is the standard signature of right skew. The outlier at $28$ is exactly what is doing it: it contributes its full size to the mean, but to the median it is just one more value on the high side.

So the median is the safer summary of a typical value here, and the mean is the one being dragged. Reporting the mean would not be wrong arithmetic, it would be a misleading description, which is the distinction the whole lesson turns on.