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

Answer:

Probability that $X_{90}$ is at least 160 is 0.0127

Step-by-step explanation:

Probability that $X_{90}$ is at least 160

$P(X_{90} \geq 160) = 1 - P(X_{90}\leq160)$

$P(X_{90}\leq160) = P(\frac{X_{90}- \mu }{\sigma}\leq \frac{160- \mu }{\sigma})$

$Probability = \frac{number of possible outcomes}{number of total outcomes}$

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

$\mu = E(X) = (1*\frac{1}{3} )+(2*\frac{2}{3} )$

$\mu = \frac{5}{3}$

The mean value after n flips

$\mu_{90} = \frac{5}{3} * 90\\\mu_{90} = 150$

For the standard deviation:

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

$\sigma_{90} = \sqrt{(3-\frac{25}{3})*90 } \\\sigma_{90} = 4.47$

$P(X_{90}\leq160) = P(Z\leq \frac{160- 150 }{4.47})$

Where Z = $\frac{X_{90}- \mu }{\sigma}$

$P(X_{90}\leq160) = P(Z\leq 2.24)$ = 0.9873

$P(X_{90} \geq 160) = 1 - 0.9873\\P(X_{90} \geq 160) =0.0127$

Answer 2

Answer:

The solution is attached in the pictures below

Step-by-step explanation: