For a normal distribution, what is the ratio between the standard deviation and the interquartile range?

Question

Textbook question:

For a normal distribution, find the following ratios:

a)$\dfrac{\text{median}}{\text{mean}}$

b $\dfrac{\text{standard deviation}}{\text{interquartile range}}$

Answer 1

In a normal distribution the median, mode and mean are all equal. Assuming that they aren't equal to $0$ then this doesn't evaluate to anything.

$$\frac{\text{median}}{\text{mean}} =1 $$

The interquartile range covers the middle 50%, which is $-.67 \sigma $ to $ .67 \sigma$ $$ \frac{\sigma}{ \text{interquartile range }}$$ $$ \frac{\sigma}{ 1.34 \sigma }$$ $$ \frac{1}{1.34} $$