Spread, Outliers and Standard Deviation, SAT math lesson

Two archers both average a score of 10 on five arrows. One lands every arrow between 8 and 12. The other scatters arrows from 2 to 18. Their means are identical, yet nobody would call them equally reliable. Center tells you where data sits, and spread tells you how tightly it clusters there. On the SAT you will rarely compute a standard deviation by hand, but you will often be asked which data set has the larger one, or how it changes when the data change.

Key terms

Spread

Definition The spread, or variability, of a data set describes how far its values tend to lie from one another and from the center. Two data sets can have the same center and very different spreads. Common measures of spread are the range, the interquartile range and the standard deviation. Each is a nonnegative number. The range and the standard deviation are $0$ exactly when all the values are equal, while the interquartile range can be $0$ even when they are not, as for $1, 5, 5, 5, 5, 5, 9$ .

In plain words Spread answers the question of how much the values differ. Tightly bunched values mean small spread.

Example The data sets $A = \{8, 9, 10, 11, 12\}$ and $B = \{2, 6, 10, 14, 18\}$ both have mean $10$ and median $10$ . On dot plots drawn with the same scale, the dots of $B$ stretch four times as far from the middle as the dots of $A$ . Comparing displays only makes sense when their scales match.

Range

Definition The range of a data set is

\text{range} = \text{maximum} - \text{minimum}.

It depends only on the two most extreme values.

In plain words The range is the distance from the smallest value to the largest.

Example The range of $A$ is $12 - 8 = 4$ and the range of $B$ is $18 - 2 = 16$ . Because it uses only two values, the range can jump when a single value changes, as the outlier section below shows.

Mean and deviations

Definition The mean of $n$ values $x_1, \dots, x_n$ is

\bar{x} = \frac{x_1 + x_2 + \dots + x_n}{n}.

The deviation of a value $x_i$ from the mean is $x_i - \bar{x}$ . Deviations are positive for values above the mean and negative for values below it, and the deviations of any data set always add up to $0$ .

In plain words A deviation measures how far a value sits from the mean, and on which side. Since the positive and negative deviations cancel, their plain average is always zero and says nothing about spread.

Example For $A$ the deviations from the mean $10$ are $-2, -1, 0, 1, 2$ , which add to $0$ . For $B$ they are $-8, -4, 0, 4, 8$ , each four times as large.

Standard deviation

Definition For a whole population of $n$ values with mean $\mu$ , the population standard deviation is

\sigma = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \mu)^2}{n}}.

For a sample of $n \ge 2$ values with mean $\bar{x}$ , the sample standard deviation divides by $n - 1$ instead,

s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n - 1}}.

Either way, standard deviation squares the deviations, averages them (dividing by $n$ or by $n - 1$ ) and takes the square root. It is not the same as the mean absolute deviation, which averages $\lvert x_i - \bar{x} \rvert$ without squaring, and the two usually give different numbers. In the Desmos graphing calculator, stdev applied to a list gives the sample version and stdevp gives the population version.

In plain words Standard deviation is a typical distance from the mean, with large distances counting extra because they are squared. Bigger means more spread out.

Example For $A$ the squared deviations are $4, 1, 0, 1, 4$ , with sum $10$ . Treating $A$ as a whole population,

\sigma_A = \sqrt{\frac{10}{5}} = \sqrt{2} \approx 1.41.

For $B$ the squared deviations add to $160$ , so $\sigma_B = \sqrt{32} \approx 5.66$ , exactly $4$ times $\sigma_A$ . The mean absolute deviation of $A$ is $6/5 = 1.2$ , which differs from $1.41$ . The sample standard deviation of $A$ would be $\sqrt{10/4} \approx 1.58$ .

Outliers and spread

Definition An outlier is a value that lies unusually far from the rest of the data. One common rule flags values below $Q_1 - 1.5 \cdot \text{IQR}$ or above $Q_3 + 1.5 \cdot \text{IQR}$ . Moving one value much farther from the others increases the range whenever that value becomes the new maximum or minimum, and it increases the standard deviation, since its squared deviation grows quickly. The median is much less affected, and in larger data sets so is the interquartile range.

In plain words One far away value stretches the range and inflates the standard deviation, while the median stays put.

Example Replace $12$ in $A$ with $30$ , a value far from the others, giving $8, 9, 10, 11, 30$ . The range grows from $4$ to $22$ , the mean rises from $10$ to $13.6$ and the population standard deviation jumps from about $1.41$ to about $8.26$ . The median is still $10$ . With so few values the quartiles move too. Using the quartile convention from Reading Data Displays, the new set has $Q_1 = 8.5$ and $Q_3 = 20.5$ , so the upper cutoff is $20.5 + 1.5 \cdot 12 = 38.5$ and the common rule does not flag $30$ as an outlier, even though it sits far from the other values.

Shift and shifted distributions

Definition Adding the same constant $k$ to every value of a data set shifts the distribution. The mean, the median and the quartiles all increase by $k$ , while the range, the interquartile range and the standard deviation do not change, because every deviation from the mean stays the same. Two distributions that differ only by such a shift have equal spread and different centers.

In plain words Sliding all the dots along the number line moves the center but keeps the shape and width exactly the same.

Example Add $5$ to every value of $A$ to get $13, 14, 15, 16, 17$ . The mean goes from $10$ to $15$ , but the deviations are still $-2, -1, 0, 1, 2$ , so the range is still $4$ and the standard deviation is still $\sqrt{2}$ . If an SAT question shows two dot plots with the same shape in different places, their standard deviations are equal.

Scale

Definition Multiplying every value of a data set by a constant $c$ scales the distribution. The mean and the median are multiplied by $c$ , and every deviation is multiplied by $c$ , so the range, the interquartile range and the standard deviation are multiplied by $\lvert c \rvert$ . If $\sigma$ is the original standard deviation, the new one is

\sigma_{\text{new}} = \lvert c \rvert \, \sigma.

Combining both moves, the values $cx_i + k$ have mean $c\bar{x} + k$ and standard deviation $\lvert c \rvert \sigma$ .

In plain words Stretching the number line stretches the spread by the same factor. A negative factor also flips the data, but a distance is never negative, so the spread uses the absolute value.

Example Multiply every value of $A$ by $3$ to get $24, 27, 30, 33, 36$ . The mean becomes $30$ , the range becomes $12$ and the standard deviation becomes $3\sqrt{2} \approx 4.24$ . Multiplying by $-2$ instead gives a mean of $-20$ and a standard deviation of $2\sqrt{2} \approx 2.83$ , not a negative number.

Check yourself

Data set $X$ is $4, 5, 5, 5, 6$ and data set $Y$ is $1, 3, 5, 7, 9$ . Which statement is true? (A) The standard deviation of $X$ is greater than that of $Y$ . (B) The standard deviation of $Y$ is greater than that of $X$ . (C) The standard deviations are equal because the means are equal. (D) The standard deviations are equal because each set has five values.

Show the answer

The correct choice is (B). Both sets have mean $5$ , but the values of $Y$ lie much farther from $5$ , with deviations up to $4$ compared with at most $1$ for $X$ . Choice (A) reverses the comparison. Choices (C) and (D) fail because neither the mean nor the number of values decides the spread.

A data set has mean $50$ and standard deviation $6$ . Every value is multiplied by $2$ , and then $10$ is added to each result. What is the standard deviation of the new data set? (A) $6$ (B) $12$ (C) $16$ (D) $22$

Show the answer

The correct choice is (B). Multiplying by $2$ doubles every deviation, so the standard deviation becomes $12$ , and adding $10$ shifts the data without changing the spread. Choice (D) wrongly adds $10$ to the spread, (C) adds $10$ to the original $6$ , and (A) ignores the doubling. The new mean is $2 \cdot 50 + 10 = 110$ .

The data set $10, 12, 14, 16, 18$ is changed by replacing $18$ with $40$ . Which of the following must be true for the new data set? (A) The mean decreases. (B) The range stays the same. (C) The standard deviation increases. (D) The median increases.

Show the answer

The correct choice is (C). The value $40$ lies far from the rest, so the deviations grow and the population standard deviation rises from about $2.83$ to about $10.98$ . The mean increases from $14$ to $18.4$ , so (A) fails. The range grows from $8$ to $30$ , so (B) fails. The middle value is still $14$ , so (D) fails.

Spread, Outliers and Standard Deviation

Theory, part 79

Practice, part 80

Key terms

Spread

Range

Mean and deviations

Standard deviation

Outliers and spread

Shift and shifted distributions

Scale

Check yourself

How this lesson connects

Builds on