Question

Suppose 200 people are lined up side-by-side, each one holding a fair coin. Each person flips their coin 64 times; every time it lands heads they step 1 meter forward, each time it lands tails they step 1 meter backward. Use a normal approximation to answer the following question: after everyone finishes their 64 steps, approximately how many people will be standing between 4 and 8 meters behind the starting line?

Answer 1

Solution:

Given,

Xi = +1 for head where p = 1/2

= - 1 for tail where p = 1/2

Mean E(Xi) = xi*p(x)

= 1*(1/2) - 1(1/2)

= 1/2 - 1/2

= 0

E(X^2) = x^2*P(x)

substitute values then

= 1^2*(1/2) + 1^2*(1/2)

= 1*1/2 + 1*1/2

= 1/2 + 1/2

= 1

Difference = E(x^2) - E(x)^2

substituting the values then we get

= 1 - 0

= 1

Mean = 0

variance = n*xi

= 64*1

= 64

Standard deviation = sqrt(variance)

= sqrt(64)

= 8

Behind the beginning time

P(- 8 < X < - 4) = P((- 8-0)/8 < (x - u)/s < (- 4-0)/1)

using central limit theorem

= P(- 1 < z < - 0.5)

= P(z < - 0.5) - P(z < - 1)

= 0.3085375 - 0.1586553 [since from z table]

= 0.1499

= 0.15

Expected individuals = np

substitute values

= 200*0.15

= 30 people