If $(Q_3-Q_2)=\frac34(Q_2-Q_1)$, then

Question

If $(Q_3-Q_2)=\frac34(Q_2-Q_1)$, then

1. There are more data which are less than the median value
2. There are more data which are less than the modal value
3. There are less data greater than the mean value
4. There are more data more than the median value
5. There are less data greater than the median value

My Try

I tried to find median $Q_2$ in terms of $Q_1$ & $Q_3$ $$Q_2=\frac{3Q_1+4Q_3}{7}$$ Also I know that in general, $$Q_2=\frac{Q_1+Q_3}{2}$$

All I can understand is that this has something to do with the mean. I'm completely lost in this problem. Can you please help me.

Answer 1

Now ,if the graph was symmetric then $Q_2 = \frac{Q_1+Q_3}{2}$ but for the given set $Q_2 = \frac{3Q_1+4Q_3}{7}$

Determining quartile skewness; $$B_1=\frac{\frac{Q_1+Q_3}{2} - \frac{3Q_1+4Q_3}{7}}{\frac{Q_3-Q_1}{2}}$$

then we have $$ B_1= \frac{\frac{Q_1-Q_3}{14}}{\frac{Q_3-Q_1}{2}}$$ $$B_1=\frac{-1}{7}$$

Therefore the graph is negatively skewed,Hence there's a majority of data less than mode of the distribution

Option 2# would be the answer.