Question

1) Suppose you start at a point and every minute you flip a coin. If the coin is head you move 1 foot north. If it is tails you stay in the same spot. A) At n minutes, what is the exact probability distribution of the number of feet north you have moved. B) What is the standard deviation of the number of feet you have moved. C) After 2000 minutes what is the approximate probability you have moved between 955 and 1045 feet north? (Hint, using the mean and the standard deviation, and assume that the random variable is approximately normal)

Answer 1

A) At every minute you flip a coin

n = number of times flip a coin

p = prob. of success = P ( Getting a head at every flip) =P ( Move 1 feet north) = 1/2 = 0.50

q =Prob. of failure =P( Getting a tail at every flip) = P ( Stay in the same spot) = 1/2 = 0.50

Let X denotes Number of feet north you have moved.

Hence X takes value 0,1, 2, ..........., n.

Since every flip is independent to each other and probability of getting head at every flip is constant.

The distribution of random variable X is binomial with parameter n and p = 0.50

X ~ Bin ( n, p = 0.5)

The p.m.f. of random variable X is

B) Since X ~ Bin ( n , p)

E(X) = n* p and SD(X) = sqrt (n* p *q)

C) n= number of times flip a coin = 2000

E(X) = 2000 *0.5 = 1000

SD(X) = sqrt ( 2000 *0.5 *0.5 ) = sqrt(500) = 22.3607

P ( Moved between 955 and 1045 feet north ) = P ( 955 < X < 1045)

Using Normal approximation to binomial distribution

Z = ( X - E(X)) / SD (X) ~ N(0,1).

= P ( -2.0125 < Z < 2.0125)

= P ( Z < 2.0125) - P ( Z < -2.0125)

from normal probability table

P ( Z< 2.0125) = 0.9779, P ( Z <-2.0125) = 0.0221

P ( 955 < X < 1045) = 0.9779 - 0.0221 = 0.9558.

P ( Moved between 955 and 1045 feet north ) = 0.9558.