Exercise 5.1.6

For $i=2,3,\dots ,9$, there are 1 $i$ between $[1,9]$, there are 10 $i$ between $[10,99]$, there are 100 $i$ between $[i00,i99]$, e.g. $[200,299]$. So in the total there are $100+10+1=111$ for each $i$. There are one extra number starting with 1, which is 1000, so for 1, we have 112 numbers starting with 1.

Then we have $p_0=0,p_1=\frac{112}{1000}$ and $p_i=\frac{111}{1000}$ for $i=2,\dots ,9$.

The mean $m=1\times \frac{112}{1000}+(2+3+\cdots +9)\times \frac{111}{1000}=4.996$, the variance ${\sigma }^2=1^2\times \frac{112}{1000}+(2^2+3^2+\cdots +9^2)\times \frac{111}{1000}=31.636$ #### Problem V.1.7 If we have $N=4$ samples $157,312,696,602$, the first digits $x_1=2,x_2=9,x_3=4,x_4=3$ of the squares $24649,97344,484416,362404$. The sample mean $\mu =\frac{2+9+4_3}{4}=4.5$, the sample variance $S^2=\frac{{(2-4.5)}^2+{(9-4.5)}^2+{(4-4.5)}^2+{(3-4.5)}^2}{N-1}=\frac{29}{3}$