Exercise 5.1.4

• When the number $n$ is $10,20,\dot{\text{,}}100,\dots ,900,1000$, we have its last digit as 0. There are 9 numbers ending with 0 from $199$, there are 10 numbers ending with 0 for each hundred, e.g. $100199$ and we also have 1000. So $p_0=\frac{1}{1000}(9+10\ast 9+1)=\frac{100}{1000}$
• There are 10 numbers ending with 1 between $[1,100]$, and $p_1=\frac{100}{1000}$ So we have $p_i=\frac{100}{1000}=\frac{1}{10}$ for $i=0,1,\dots ,9$.

The mean of that last digit is thus $m=0\times \frac{1}{10}+1\times \frac{1}{10}+\cdots +9\times \frac{1}{10}=4.5$

The variance is ${\sigma }^2=0^2\times \frac{1}{10}+1^2\times \frac{1}{10}+\cdots +9^2\times \frac{1}{10}=13.5$