Solved – QR ever be larger than standard deviation

Question

As I understand it the IQR specifies the dispersion of the data by taking the difference between Q3 and Q1 (i.e., the range that the middle 50% of the data lies), while std specifies dispersion using all the data, to where 68% of the data belong. Does this imply that the std is always larger than (or equal to?) iqr?

Answer 1

Your argument about $68\%$ of the density contained within $1$ standard deviation (of the mean) is true for the Normal distribution, but not in general. There are examples where the standard deviation exceeds the IQR and examples in the other direction as well.

Let $X$ have a t-distribution with $v$ degrees of freedom.

The standard deviation of $X$ is $$\sigma = \sqrt{\text{Var}(X)} = \sqrt{\frac{v}{v-2}}$$

The IQR can be found using the quantile function of the t-distribution. In R, we have

qt(.75, v) - qt(.25, v)

We have that $\sigma = IQR$ for $v\approx 3.61$. When $v < 3.61$, we have $\sigma > IQR$ and we have $\sigma < IQR$ otherwise. Since a normal distribution can be achieved in the limit as $v\rightarrow \infty$, this demonstrates that $\sigma < IQR$ for a normal distribution.

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Sal considers the interesting question, can IQR be larger than $2\sigma$?. Indeed, this will occur for the symmetric Beta distribution when the shape parameter is small.

Let $X \sim Beta(\alpha, \alpha)$. Then $IQR = 2\sigma$ when $\alpha = 0.5$, $IQR > 2\sigma$ when $\alpha < 0.5$ and $IQR < 2\sigma$ otherwise.

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We have established examples, where $IQR > \sigma$ and examples where $IQR > 2\sigma$. It is not possible, however, for $IQR$ to exceed $4\sigma$ when the distribution of $X$ is symmetric about its mean $\mu$. Applying Chebyshev's Inequality, we have $$P(|X-\mu| \leq 2\sigma) \geq \frac{3}{4}.$$

If the density function of $X$ is symmetric about $\mu$, we have $$P(|X-\mu| \leq \text{IQR}/2) = \frac{3}{4}.$$

This implies that $2\sigma \geq \text{IQR}/2$, or equivalently $$\text{IQR} \leq 4\sigma$$