Question

Exercise 4.4. Liz is standing on the real number line at position O. She rolls a die repeatedly. If the roll is 1 or 2, she takes one step to the right (in the positive direction). If the roll is 3, 4, 5 or 6, she takes two steps to the right. Let Xn be Liz's position after n flips of the coin. Estimate the probability that x90 is at least 160.

Answer 1

Probability that
`X_(90)` is at least 160 is 0.0127

Explanation:

Probability that
`X_(90)` is at least 160

`P(X_(90) \geq 160) = 1 - P(X_(90)\leq160)`

`P(X_(90)\leq160) = P((X_(90)- \mu )/(\sigma)\leq (160- \mu )/(\sigma))`

`Probability = (number of possible outcomes)/(number of total outcomes)`

Probability that she rolls 1 or 2 i.e. probability that she takes one step to the right:

P(X=1) = 2/6 = 1/3

Probability that she rolls 3,4,5,6 i.e. Probability that she takes two steps to the right:

P(X=2) = 4/6 = 2/3

`\mu = E(X) = \sum xP(X)\\`

when x = 1,2

`\mu = E(X) = (1*(1)/(3) )+(2*(2)/(3) )`

`\mu = (5)/(3)`

The mean value after n flips

`\mu_(90) = (5)/(3) * 90\\\mu_(90) = 150`

For the standard deviation:

`\sigma_(90) =\sqrt{ [E(x^(2)) -((E(x))^(2) ]*90} \\\sigma_(90) =\sqrt{ [(1^(2)*(1)/(3))+(2^(2)*(2)/(3) ) -((5)/(3)) ^(2) ]*90}`

`\sigma_(90) = \sqrt{(3-(25)/(3))*90 } \\\sigma_(90) = 4.47`

`P(X_(90)\leq160) = P(Z\leq (160- 150 )/(4.47))`

Where Z =
`(X_(90)- \mu )/(\sigma)`

`P(X_(90)\leq160) = P(Z\leq 2.24)` = 0.9873

`P(X_(90) \geq 160) = 1 - 0.9873\\P(X_(90) \geq 160) =0.0127`

Answer 2

The solution is attached in the pictures below

Explanation: