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

meansmediannormal distributionratiostandard deviation

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.

Best Answer

(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.

Related Solutions

[Math] Normal distribution involving $\Phi(z)$ and standard deviation

Hint:

Write,

$$ \tag{1}\textstyle P[\,X>31\,] =P\bigl[\,Z>{31-\mu\over\sigma}\,\bigr]=.2743\Rightarrow {31-\mu\over\sigma} = z_1 $$ $$\tag{2}\textstyle P[\,X<39\,] =P\bigl[\,Z<{39-\mu\over\sigma}\,\bigr]=.9192\Rightarrow {39-\mu\over\sigma} =z_2 , $$

where $Z$ is the standard normal random variable.

You can find the two values $z_1$ and $z_2$ from a cdf table for the standard normal distribution. Then you'll have two equations in two unknowns. Solve those for $\mu$ and $\sigma$.

For example, to find $z_1$ and $z_2$, you can use the calculator here. It gives the value $z$ such that $P[Z<z]=a$, where you input $a$.

To use the calculator for the first equation first write $$\textstyle P\bigl[\,Z<\underbrace{31-\mu\over\sigma}_{z_1}\,\bigr]=1-P\bigl[\,Z>{31-\mu\over\sigma}\,\bigr] =1-.2743=.7257.$$ You input $a=.7257$, and it returns $z_1\approx.59986$.

To use the calculator for the second equation, $$\textstyle P\bigl[\,Z<\underbrace{39-\mu\over\sigma}_{z_2}\,\bigr]= .9192,$$ input $a=.9192$, the calculator returns $z_2\approx1.3997$.

So, you have to solve the system of equations: $$ \eqalign{ {31-\mu\over\sigma}&=.59986\cr {39-\mu\over\sigma}&=1.3997\cr } $$ (The solution is $\sigma\approx 10$, $\mu\approx 25$.)

[Math] Is it possible to calculate the mean and standard deviation from a median and quartiles

It's mathematically impossible to deduce mean or standard deviation from median/quartiles, because medians and quartiles discard most of the data on which the mean and standard deviation are based.

Example:

data   frequency  
   0       50      
 1.4        4     
   2       50

That has a mean of 1.0 and standard deviation of 0.9. (I'm using 2 significant figures so I don't have to go into population versus sample standard deviation.)

data     frequency    
   0       30        
 1.4       44        
   2       30

That data also has the median and quartiles the same as in your example, but now the mean is 1.2 and the standard deviation is 0.8.

data     frequency        
   0       30        
 1.4        3        
   2       70        
10000000    1

Now I've changed my maximum without changing the median or quartiles, you can see even more clearly how the median and quartiles exclude extreme data, because the mean is now 96000 and the standard deviation is 98000 (still 2 sig.fig.).

Best Answer

Related Solutions

[Math] Normal distribution involving $\Phi(z)$ and standard deviation

[Math] Is it possible to calculate the mean and standard deviation from a median and quartiles

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