Data Distributions and Statistics: Additional Practice (Set 1)

With ten values the halves are $4, 6, 7, 9, 10$ and $12, 13, 15, 16, 28$, each with five values, so each quartile is the single middle value of its half.

$Q_1 = 7, \qquad Q_3 = 15$

The interquartile range is $15 - 7 = 8$, so $1.5 \times \text{IQR} = 12$.

$\text{upper fence} = Q_3 + 12 = 15 + 12 = 27$

Adding the IQR itself rather than $1.5$ times it would give $23$, which is the common slip. Since $28 > 27$, that largest value is flagged as an outlier.

The mean is $\mu = \frac{3 + 7 + 11 + 15}{4} = \frac{36}{4} = 9$, so the deviations are $-6, -2, 2, 6$.

Square them and add.

$\sum (x_i - \mu)^2 = 36 + 4 + 4 + 36 = 80$

These are the whole population, so the variance divides that total by $N = 4$.

$\sigma^2 = \frac{80}{4} = 20$

Dividing by $n - 1 = 3$ would give the sample variance $\frac{80}{3} \approx 26.7$, and $\sqrt{20}$ is the population standard deviation, which is a different quantity in different units.

Total squared distance is a quadratic in $c$ that opens upward, and its vertex sits at the mean, so this is the mean's problem.

$\bar{x} = \frac{3 + 8 + 10 + 10 + 14}{5} = \frac{45}{5} = 9$

Check it against the median, $10$. At $c = 9$ the total squared distance is

$36 + 1 + 1 + 1 + 25 = 64,$

while at $c = 10$ it is $49 + 4 + 0 + 0 + 16 = 69$, which is worse.

Compare this with the previous question, where the same data made the median the winner. The data has not changed; the question has. Each center is the champion of exactly one of these two contests.

For a sample the divisor is $n - 1 = 9$, not $n = 10$.

$s^2 = \frac{90}{9} = 10$

That is the variance, so take the square root to return to the units of the data.

$s = \sqrt{10} \approx 3.16$

Dividing by $n = 10$ instead would give a variance of $9$ and a population standard deviation of exactly $3$, which is the tempting round answer and the wrong one for a sample. Stopping at $10$ reports the variance without the square root.

Use the identity that fell out of completing the square, which turns the sum of squared deviations into two sums you already have.

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

Substitute $n = 5$, $\mu = 6$, and $\sum x_i^2 = 200$.

$\sum (x_i - \mu)^2 = 200 - 5(36) = 200 - 180 = 20$

That is the total, so divide by $N = 5$ to get the average squared deviation.

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

Dividing the same total by $n - 1 = 4$ would give the sample variance, $5$, which answers a different question.

The bars are packed on the left and thin out into a long tail on the right, which is the picture of a right-skewed distribution. There is a single peak, so the data is not bimodal.

The median only counts how many values sit out in the tail, and there are not many. The mean feels how far out they are, so those few large values drag it to the right of the median.

That makes the mean the larger of the two here. Keep the hedge, though: "the mean is pulled toward the tail" is a reliable rule of thumb rather than a theorem, and constructed counterexamples exist, so the honest claim is that the mean is probably the larger one.

Adding a constant to every value slides the whole data set along the number line without reordering it, so each quartile simply moves by that same constant.

$Q_1 \to Q_1 + 12, \qquad Q_3 \to Q_3 + 12$

The IQR is a difference of the two, so the shift cancels.

$(Q_3 + 12) - (Q_1 + 12) = Q_3 - Q_1$

The spread is unchanged, which is exactly what you want from a measure of spread: moving the data does not stretch it. The same cancellation is why adding a constant leaves the range and the standard deviation alone too.

Convert the old mean into a total, since that is what the new value gets added to.

$\text{old total} = 8 \times 20 = 160$

Now add the new value and divide by the new count of $9$.

$\bar{x}_{\text{new}} = \frac{160 + 29}{9} = \frac{189}{9} = 21$

Averaging $20$ and $29$ directly gives $24.5$, which is the trap: it would treat the single new value as though it carried the same weight as the eight values already there. It gets one vote out of nine, so it lifts the mean by only $1$.

The five numbers of a summary must come in non-decreasing order, and that is forced by how they are defined, not by convention.

$\text{min} \le Q_1 \le \text{median} \le Q_3 \le \text{max}$

The median splits the sorted data in half, and $Q_1$ is the median of the lower half, so $Q_1$ is built entirely out of values at or below the median. It cannot possibly exceed it.

Here the report claims $Q_1 = 10$ while the median is $8$, which is impossible for any data set. A maximum sitting far above $Q_3$ is not an error at all; it just suggests a right-skewed data set, possibly with an outlier.

The deviation of the outlier is $70 - 28 = 42$, and the variance uses its square.

$(70 - 28)^2 = 42^2 = 1764$

Compare that with the total of $2300$ contributed by all twelve values.

$\frac{1764}{2300} \approx 0.77$

So one value out of twelve supplies more than three quarters of the entire sum of squared deviations. This is squaring doing what squaring does: it magnifies a large deviation out of all proportion, which is exactly why the standard deviation is sensitive to outliers while the IQR, which ignores this value completely, is not.

Repeated values change nothing about the method. With eight values the halves are $6, 6, 6, 6$ and $10, 10, 10, 10$, and each quartile is the median of its half.

$Q_1 = \frac{6 + 6}{2} = 6, \qquad Q_3 = \frac{10 + 10}{2} = 10$

Subtracting gives the width of the box.

$\text{IQR} = 10 - 6 = 4$

The answer $0$ is the trap, since each half really does have zero internal spread. But the IQR measures the distance between the quartiles, not the spread inside a half, and the two clusters sit $4$ apart. The median of the whole set, $8$, sits in the gap where no value lives.

Only the total changed, and it rose by exactly $100$. Since the mean is the total divided by the count, moving one value by $d$ moves the mean by $\frac{d}{n}$.

$\bar{x}_{\text{new}} = 40 + \frac{100}{25} = 40 + 4 = 44$

The damage is diluted across all $25$ values, so a $100$-point change to one of them lifts the mean by only $4$. That $\frac{d}{n}$ is also the precise statement of why the mean is sensitive: it can be dragged anywhere you like, but the more data you have, the harder a single value has to pull.