walker starts at \(0\),

At given unit time he moves to the right or left randomly. The probability of moving to the right or left is \(\frac{1}{2}\).

Let value to the left be \(-1\) and value to the right be \(1\). Hence at any given time walker is moving towards \(0\) or away from \(0\) on either side.

Hence mean \(\mu = 0\). And variance \(\sigma = \sum p(x_i - \mu)^2\).

\(\sigma^2 = \frac{1}{2}(-1-0)^2 + \frac{1}{2}(0-0)^2 + \frac{1}{2}(1-0)^2 = \frac{1}{2} + 0 + \frac{1}{2} = 1\)

After \(100\) steps, \(n = 100\)

\(E(S_{100}) = 100*\mu = 100*0 = 0\).

\(V(S_{100}) = 100*\sigma^2 = 100 * 1 = 100\)

Standard deviation \(\sigma = \sqrt V(S_{100}) = \sqrt 100 = 10\)

If walker is 10 steps away to left, the value of \(x = -10\)

\(z = \frac{x - \mu}{\sigma} = \frac{-10 - 0}{10} = -1\)

If walker is 10 steps away to right, the value of \(x = 10\)

\(z = \frac{x - \mu}{\sigma} = \frac{10 - 0}{10} = 1\)

The probability of walker is 10 steps away in either direction, can be defined as the area between tails under the curve.

\(P(S_{100} \ge 10) = 1 - P(S_{100} \le 10)\), \(1 - area~ of~ the~ tails\)

\(P(S_{100} \le 10)\)

\(= P(-1 \le z \le 1)\)

\(= P(-1 \le z \le 1)\)

\(= P(z\le 1) - P(z \le -1)\)

\(= P(z\le 1) - P(z \le -1)\)

As we are calculating for steps > 10, \(x > 10\), using normal distribution table values for \(1~ and~ -1\)

\(= 0.8413 - 0.1587 = 0.6826\)

Hence \(P(S_{100} \le 10) = 0.6826\),

\(P(S_{100} \ge 10) = 1 - P(S_{100} \le 10)\)

\(P(S_{100} \ge 10) = 1 - 0.6826 = 0.3174\).

Therefore the probability that walker will be 10 steps away from \(0 = 0.3174\). It means there is \(31.74 \%\) chances that walker will be away from \(0\).