Ratios of median/mean and standard deviation/IQR in a normal distribution

Question

I have some queries on the following question

For a normal distribution, find the ratios of:

(a) $\frac {\mbox{median}}{\mbox{mean}}$

(b) $\frac {\mbox{standard deviation}}{\mbox{interquartile range}}$

For (a), I know that the ratio is 1, as the median and mean are the same and I can see that if I graph the normal distribution curve out. However, is there a mathematical way of doing it not involving a graphical method?

For (b), what I tried to do was do $\frac {\sigma}{X3-X1}$ and then using the Z values, I calculated that $X1=-0.675\sigma + \mu$

and

$X3= 0.675\sigma + \mu$

So $X3-X1=1.35 \mu$

which cancels out with the $\mu$ on the numerator, and making the ratio 1/1.35, however the answer states that the ratio should be 1.48. Please advise where I went wrong. Also, sorry in advance for any wrong tags and wrong title name.

Answer 1

(a) Yes. If $X \sim \operatorname{Normal}(\mu, \sigma^2)$, then the PDF of $X$ is given by $$f_X(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-(x-\mu)^2/(2\sigma^2)}, \quad -\infty < x < \infty.$$ We can also readily observe that $X$ is a location-scale transformation of a standard normal random variable $Z \sim \operatorname{Normal}(0,1)$, namely $$X = \sigma Z + \mu,$$ or equivalently, $$Z = \frac{X - \mu}{\sigma},$$ and the density of $Z$ is simply $$f_Z(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}, \quad -\infty < z < \infty.$$ Therefore, if $m$ is the median of $X$, then the median of $Z$ is $m^* = (m - \mu)/\sigma$. But we also know that $m^*$ satisfies $$F_Z(m^*) = \int_{z=-\infty}^{m^*} f_Z(z) \, dz = \Phi(m^*) = \frac{1}{2}.$$ But since $f_Z(z) = f_Z(-z)$ for all $z$, the substitution $$u = -z, \quad du = -dz$$ readily yields $$F_Z(m^*) = -\int_{u=\infty}^{-m^*} f_Z(-u) \, du = \int_{u=-m^*}^\infty f_Z(u) \, du = 1 - F_Z(-m^*),$$ and since both of these must equal $1/2$, we conclude $F_Z(m^*) = F_Z(-m^*)$, or $m^* = -m^*$, or $m^* = 0$. From this, we recover the median of $X$: $m = \sigma m^* + \mu = \mu$.

(b) The interquartile range is equal to $q_3 - q_1$, where $q_3$ satisfies $F_X(q_3) = \frac{3}{4}$ and $F_X(q_1) = \frac{1}{4}$. Again, using the location-scale relationship to $Z$, we first find the IQR of $Z$, then transform back to get the IQR of $X$. The conditions $$\Phi(q_1^*) = \frac{1}{4}, \quad \Phi(q_3^*) = \frac{3}{4}$$ are clearly symmetric (see part a). We can look up in a normal distribution table that $\Phi(-0.67449) \approx 0.25$, or to more precision with a computer, $$q_1^* \approx -0.67448975019608174320.$$ It follows that the IQR of $Z$ is $$q_3^* - q_1^* \approx 1.3489795003921634864,$$ hence the IQR of $X$ is $$q_3 - q_1 = (\sigma q_3^* + \mu) - (\sigma q_1^* + \mu) \approx 1.3489795 \sigma,$$ and so the desired ratio is simply approximately $$0.74130110925280093027.$$ Note this quantity does not depend on the parameters. Your error is that you performed the subtraction incorrectly.

Answer 2

(a) You are right about mean and median. Because the probability density function of a normal distribution takes an input parameter x, and its first steps are to subtract the mean, divide by the SD, and square it, it is symmetrical about the mean. For a symmetrical distribution, the mean is the same as the median.

(b) You are right about the interquartile range and SD. The 25th and 75th percentiles are at 0.674 SDs below and above the mean, respectively, so the interquartile range is 1.35 SDs, not 1.48 SDs.

In short, your answers are correct. Perhaps you can email the author of the book? I am sure they will be pleased to know of a correction sooner rather than leter.