Standard deviation vs IQR: which spread to use

By Jude Wallis · Published July 28, 2026

The IQR is the width of the middle half of the data, the third quartile minus the first, and it resists outliers. The standard deviation uses every value's distance from the mean, so one extreme point can inflate it. Pair the IQR with the median, the standard deviation with the mean.

AP Statistics: Unit 1 (topics 1.7). Range, IQR, and standard deviation are the three measures of variability in topic 1.7 of Unit 1 in the Fall 2026 AP Statistics course. The IQR is defined as Q3 minus Q1, and this site uses the median-excluded (TI-84) quartile convention.

Standard deviation vs IQR: the short answer

Both are measures of spread, but they differ in exactly the way the median and the mean differ. The interquartile range, or IQR, is resistant: it depends only on the quartiles that bracket the middle half of the data, so values far out in a tail barely move it. The standard deviation is not resistant: it is built from every value's distance to the mean, so a single outlier can pull it up sharply.

That one contrast decides which to use. When a distribution is skewed or has outliers you report the median for center and the IQR for spread, because both ignore the extremes. When a distribution is roughly symmetric with no outliers you report the mean for center and the standard deviation for spread, because they use all the information the data carry.

What the standard deviation measures

The standard deviation is the typical distance of a value from the mean, in the data's own units. For a sample of size $n$ with mean $\bar{x}$ (read "x-bar"), it is

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

where $x_i$ is one data value and $\sum$ means add across all the values. Every value enters the sum, so every value affects the answer.

That is its strength and its weakness. Using all the data makes the standard deviation efficient for symmetric distributions, but it also means a lone extreme value, whose squared distance from the mean is huge, can dominate the sum and blow the standard deviation up. You can build one by hand in the standard deviation guide.

What the IQR measures

The IQR is the width of the middle 50% of the data. You find it by locating the first quartile $Q_1$ , the value a quarter of the way through the sorted data, and the third quartile $Q_3$ , the value three quarters of the way through, then subtracting:

\text{IQR} = Q_3 - Q_1

This site uses the median-excluded (TI-84) convention for quartiles: split the sorted data at the median, then take $Q_1$ as the median of the lower half and $Q_3$ as the median of the upper half.

Because the IQR reads off only two positions in the sorted list, changing the largest or smallest values does not move it, exactly the resistance the median has. The IQR is also the spread shown by the box of a boxplot and the engine behind the 1.5 times IQR rule for outliers.

The differences side by side

Feature	Standard deviation	IQR
Measures	Typical distance from the mean	Width of the middle 50%
Uses	Every value	Only the two quartiles
Resistant to outliers	No	Yes
Pairs with	The mean	The median
Best for	Symmetric data, no outliers	Skewed data or outliers
Shown by	Spread around the mean	The box of a boxplot

The resistant-versus-not row drives every other row, just as it does for the median versus the mean.

When to use which

Check the shape first, ideally from a graph. If the distribution is skewed or shows outliers, report the median and the IQR, because both stay put when a few values are extreme and so describe a typical case honestly. Incomes, home prices, and reaction times usually call for this pair.

If the distribution is roughly symmetric with no outliers, report the mean and the standard deviation, because they use every value and feed the z-scores, confidence intervals, and tests that come later in the course. On the AP exam, name your reason: point to skew or an outlier when you pick the IQR, and to a symmetric shape when you pick the standard deviation. Keep the pairs together, since a mean with an IQR, or a median with a standard deviation, mixes a resistant measure with a non-resistant one.

The classic mix-up and how to avoid it

The common error is reaching for the standard deviation out of habit when the data are skewed or contain an outlier, then reporting a spread that the outlier alone inflated. A commute that is usually 5 minutes but once hit 45 minutes has a standard deviation dragged upward by that one day, while the IQR of the ordinary days is unchanged; quoting the large standard deviation makes the typical commute look far more variable than it is.

Avoid it with a two-step habit. First look at the shape: skew or a visible outlier is a signal to switch to the median and IQR. Second, keep center and spread matched, so median with IQR and mean with standard deviation. When you are unsure, compute both spreads: a standard deviation much larger than the IQR would suggest is itself a sign that outliers are stretching it, which you can confirm with the five-number summary calculator.

One outlier inflates the standard deviation but not the IQR

A commuter records travel time, in minutes, on seven ordinary days: 4, 5, 6, 7, 8, 9, 10. Find the IQR and the standard deviation. Then one snowy day the 10 becomes 45. Recompute both and compare what happened.

The data are already sorted, and $n = 7$ . The median is the 4th value, 7.
Use the median-excluded convention. Lower half (below the median) is 4, 5, 6, so $Q_1 = 5$ . Upper half is 8, 9, 10, so $Q_3 = 9$ . Then $\text{IQR} = 9 - 5 = 4$ minutes.
For the standard deviation, find the mean: sum $= 4 + 5 + 6 + 7 + 8 + 9 + 10 = 49$ , so $\bar{x} = \frac{49}{7} = 7$ .
Squared deviations from 7: $9, 4, 1, 0, 1, 4, 9$ , which add to $28$ . Then $s = \sqrt{\frac{28}{7-1}} = \sqrt{\frac{28}{6}} = \sqrt{4.6667} \approx 2.16$ minutes.
Now replace 10 with 45: the data are 4, 5, 6, 7, 8, 9, 45. The median is still the 4th value, 7, so the positions did not shift.
Recompute the IQR. Lower half is still 4, 5, 6, so $Q_1 = 5$ . Upper half is now 8, 9, 45, whose median is 9, so $Q_3 = 9$ and $\text{IQR} = 9 - 5 = 4$ minutes, unchanged.
Recompute the standard deviation. New sum $= 4 + 5 + 6 + 7 + 8 + 9 + 45 = 84$ , so $\bar{x} = \frac{84}{7} = 12$ . Squared deviations from 12: $64, 49, 36, 25, 16, 9, 1089$ , which add to $1288$ . Then $s = \sqrt{\frac{1288}{6}} = \sqrt{214.67} \approx 14.65$ minutes.

Before the outlier: IQR = 4 minutes, standard deviation about 2.16 minutes. After the 45-minute day: IQR = 4 minutes, unchanged, while the standard deviation jumps to about 14.65 minutes, more than six times larger. The IQR ignored the extreme value; the standard deviation was inflated by it, which is why skewed or outlier-prone data are summarized with the median and IQR.

Frequently asked questions

Which is better, the standard deviation or the IQR?

Neither is always better. Use the IQR for skewed data or data with outliers, because it is resistant, and use the standard deviation for roughly symmetric data with no outliers, because it uses every value and drives the inference procedures later in the course.

Do the standard deviation and IQR pair with different measures of center?

Yes. The IQR pairs with the median, since both are resistant, and the standard deviation pairs with the mean, since both use every value. Reporting a median with a standard deviation, or a mean with an IQR, mixes a resistant measure with a non-resistant one.

Why does an outlier change the standard deviation so much more than the IQR?

The standard deviation squares each value's distance from the mean, so a far-out point contributes an enormous squared distance and dominates the total. The IQR reads only the two quartile positions, so a value beyond them does not change it at all.