Answer :
Assume that a step to the north is in the positive direction and a step in the south is in the negative direction.
Then,
[tex]E(X_i)=50\left(\frac{1}{2}\right)-50\left(\frac{1}{2}\right)\\ \\=25-25=0[/tex]
and
[tex]\sigma^2(X_i)=E(X_i^2)-E(X_i)^2 \\ \\ =50^2\left(\frac{1}{2}\right)+50^2\left(\frac{1}{2}\right)-0^2=2500\left(\frac{1}{2}\right)+2500\left(\frac{1}{2}\right)-0 \\ \\ =2500 \\ \\ \Rightarrow\sigma(X_i)=50[/tex]
After 1 hour of independent steps,
[tex]E(X)=\sum_{i=1}^{60}E(X_i)=60(0)=0[/tex]
and
[tex]\sigma^2(X)=\sum_{i=1}^{60}\sigma^2(X_i) \\ \\ =60(50)=300 \\ \\ \Rightarrow\sigma(X)=\sqrt{300}=17.32[/tex]
Therefore, by Central Limit Theorem, the distribution of the random walk is normal with a mean of 0 cm and a standard deviation of 17.32 cm. He is expected to be somewhere around his starting point after 1 hour.
Then,
[tex]E(X_i)=50\left(\frac{1}{2}\right)-50\left(\frac{1}{2}\right)\\ \\=25-25=0[/tex]
and
[tex]\sigma^2(X_i)=E(X_i^2)-E(X_i)^2 \\ \\ =50^2\left(\frac{1}{2}\right)+50^2\left(\frac{1}{2}\right)-0^2=2500\left(\frac{1}{2}\right)+2500\left(\frac{1}{2}\right)-0 \\ \\ =2500 \\ \\ \Rightarrow\sigma(X_i)=50[/tex]
After 1 hour of independent steps,
[tex]E(X)=\sum_{i=1}^{60}E(X_i)=60(0)=0[/tex]
and
[tex]\sigma^2(X)=\sum_{i=1}^{60}\sigma^2(X_i) \\ \\ =60(50)=300 \\ \\ \Rightarrow\sigma(X)=\sqrt{300}=17.32[/tex]
Therefore, by Central Limit Theorem, the distribution of the random walk is normal with a mean of 0 cm and a standard deviation of 17.32 cm. He is expected to be somewhere around his starting point after 1 hour.
You should get the mean through:
Mean= 50 (1/2)- 50(1/2) which will give you zero.
Second, you will get the standard deviation.
Then multiply it 50 by 60
So:
50 (60) = 300
Square root of 300 = 17.32.
Through this, you can say that he is just near to the place where he is walking after 1 hour.
