A five digit integer is always greater than $7000.$
The number of such integers are ${\mathrm{}}^5{P}_5=5!=120.$
For a four digit integer to be greater than $7000,$ it must begin with $\mathrm{7,8}$ or $9.$
The number of such integers are $\left(3\right){\mathrm{}}^4{P}_3=72.$
Hence, the required number of such integers are $120+72=192.$
$\Rightarrow m=192$
$\Rightarrow \frac{m}{100}=1.92.$