Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
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Chapter 5.1, Problem 15E
To determine

(a)

The Expected value of Mike’s bet.

Expert Solution
Check Mark

Answer to Problem 15E

Solution:

The expected value of Mike’s bet is -4.517.

Explanation of Solution

Given Information:

Mike’s older brother, Jeff, bets him that he can’t roll two dice and get doubles three times in a row. If Mike does it, Jeff will give him $100.00. Otherwise, Mike has to give Jeff $5.00.

Formula Used:

The expected value for a discrete random variable X is equal to the mean of the probability distribution of X and is given by,

E(X)=[xi.P(X=xi)]

Where xi is the ith value of the random variable X.

Calculation:

There are 6 faces in a die. The total number of outcomes of rolling two dice is 36. There are six different rolls that are doubles (double 1, double 2, double 3, double 4, double 5, double 6) out of a total of 36 possible outcomes when rolling two dice

The probability of rolling a double is

P(X=x)=636=16

Each roll is independent of the others. So the probability of rolling two dice and getting doubles three times in a row is

P(X=x)=16×16×16=1216=0.0046

Each roll is independent of the others. So the probability of rolling two dice and not getting doubles three times in a row is

P(X=x)=116×16×16=11216=10.0046=0.9954

X $100 $-5
P(X = x) 0.0046 0.9954

The expected value is given by

E(X)=[xi.P(X=xi)]

The expected value

E(X)=(100×0.0046+(5)×0.9954)=4.517

Conclusion:

Thus the expected value of Mike’s bet is -4.517.

To determine

(b)

The expected value of Jeff’s bet.

Expert Solution
Check Mark

Answer to Problem 15E

Solution:

The expected value of Jeff’s bet is 4.517

Explanation of Solution

Given Information:

Mike’s older brother, Jeff, bets him that he can’t roll two dice and get doubles three times in a row. If Mike does it, Jeff will give him $100.00. Otherwise, Mike has to give Jeff $5.00.

Formula Used:

The expected value for a discrete random variable X is equal to the mean of the probability distribution of X and is given by

E(X)=[xi.P(X=xi)]

Where xi is the ith value of the random variable X.

Calculation:

There are 6 faces in a die. The total number of outcomes of rolling two dice is 36. There are six different rolls that are doubles (double 1, double 2, double 3, double 4, double 5, double 6) out of a total of 36 possible outcomes when rolling two dice

The probability of rolling a double is

P(X=x)=636=16

Each roll is independent of the others. So the probability of rolling two dice and getting doubles three times in a row is

P(X=x)=16×16×16=1216=0.0046

Each roll is independent of the others. So the probability of rolling two dice and not getting doubles three times in a row is

P(X=x)=116×16×16=11216=10.0046=0.9954

X $-100 $5
P(X = x) 0.0046 0.9954

The expected value is given by

E(X)=[xi.P(X=xi)]

The expected value

E(X)=((100)×0.0046+5×0.9954)=4.517

Conclusion:

Thus the expected value of Jeff’s bet is 4.517.

To determine

(c)

The expected value of Mike’s bet.

Expert Solution
Check Mark

Answer to Problem 15E

Solution:

If Mike and Jeff make the same bet 30 times, Mike is expected to lose $135.51

Explanation of Solution

Given:

Mike’s older brother, Jeff, bets him that he can’t roll two dice and get doubles three times in a row. If Mike does it, Jeff will give him $100.00. Otherwise, Mike has to give Jeff $5.00.

Formula Used:

The expected value for a discrete random variable X is equal to the mean of the probability distribution of X and is given by

E(X)=[xi.P(X=xi)]

Where xi is the ith value of the random variable X.

Calculation:

There are 6 faces in a die. The total number of outcomes of rolling two dice is 36. There are six different rolls that are doubles (double 1, double 2, double 3, double 4, double 5, double 6) out of a total of 36 possible outcomes when rolling two dice

The probability of rolling a double is

P(X=x)=636=16

Each roll is independent of the others. So the probability of rolling two dice and getting doubles three times in a row is

P(X=x)=16×16×16=1216=0.0046

Each roll is independent of the others. So the probability of rolling two dice and not getting doubles three times in a row is

P(X=x)=116×16×16=11216=10.0046=0.9954

X $100 $-5
P(X = x) 0.0046 0.9954

The expected value is given by

E(X)=[xi.P(X=xi)]

The expected value

E(X)=(100×0.0046+(5)×0.9954)=4.517

If Mike and Jeff make the same bet 30 times, then the expected value of Mike is

E(x)=30×(4.517)=135.51

Conclusion:

Thus, if Mike and Jeff make the same bet 30 times, Mike is expected to lose $135.51.

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Chapter 5 Solutions

Beginning Statistics, 2nd Edition

Ch. 5.1 - Prob. 11ECh. 5.1 - Prob. 12ECh. 5.1 - Prob. 13ECh. 5.1 - Prob. 14ECh. 5.1 - Prob. 15ECh. 5.1 - Prob. 16ECh. 5.1 - Prob. 17ECh. 5.1 - Prob. 18ECh. 5.1 - Prob. 19ECh. 5.1 - Prob. 20ECh. 5.1 - Prob. 21ECh. 5.1 - Prob. 22ECh. 5.1 - Prob. 23ECh. 5.2 - Prob. 1ECh. 5.2 - Prob. 2ECh. 5.2 - Prob. 3ECh. 5.2 - Prob. 4ECh. 5.2 - Prob. 5ECh. 5.2 - Prob. 6ECh. 5.2 - Prob. 7ECh. 5.2 - Prob. 8ECh. 5.2 - Prob. 9ECh. 5.2 - Prob. 10ECh. 5.2 - Prob. 11ECh. 5.2 - Prob. 12ECh. 5.2 - Prob. 13ECh. 5.2 - Prob. 14ECh. 5.2 - Prob. 15ECh. 5.2 - Prob. 16ECh. 5.2 - Prob. 17ECh. 5.2 - Prob. 18ECh. 5.2 - Prob. 19ECh. 5.2 - Prob. 20ECh. 5.2 - Prob. 21ECh. 5.2 - Prob. 22ECh. 5.2 - Prob. 23ECh. 5.2 - Prob. 24ECh. 5.2 - Prob. 25ECh. 5.3 - Prob. 1ECh. 5.3 - Prob. 2ECh. 5.3 - Prob. 3ECh. 5.3 - Prob. 4ECh. 5.3 - Prob. 5ECh. 5.3 - Prob. 6ECh. 5.3 - Prob. 7ECh. 5.3 - Prob. 8ECh. 5.3 - Prob. 9ECh. 5.3 - Prob. 10ECh. 5.3 - Prob. 11ECh. 5.3 - Prob. 12ECh. 5.3 - Prob. 13ECh. 5.3 - Prob. 14ECh. 5.3 - Prob. 15ECh. 5.3 - Prob. 16ECh. 5.3 - Prob. 17ECh. 5.3 - Prob. 18ECh. 5.3 - Prob. 19ECh. 5.3 - Prob. 20ECh. 5.3 - Prob. 21ECh. 5.3 - Prob. 22ECh. 5.3 - Prob. 23ECh. 5.3 - Prob. 24ECh. 5.4 - Prob. 1ECh. 5.4 - Prob. 2ECh. 5.4 - Prob. 3ECh. 5.4 - Prob. 4ECh. 5.4 - Prob. 5ECh. 5.4 - Prob. 6ECh. 5.4 - Prob. 7ECh. 5.4 - Prob. 8ECh. 5.4 - Prob. 9ECh. 5.4 - Prob. 10ECh. 5.4 - Prob. 11ECh. 5.4 - Prob. 12ECh. 5.4 - Prob. 13ECh. 5.4 - Prob. 14ECh. 5.4 - Prob. 15ECh. 5.4 - Prob. 16ECh. 5.4 - Prob. 17ECh. 5.4 - Prob. 18ECh. 5.4 - Prob. 19ECh. 5.CR - Prob. 1CRCh. 5.CR - Prob. 2CRCh. 5.CR - Prob. 3CRCh. 5.CR - Prob. 4CRCh. 5.CR - Prob. 5CRCh. 5.CR - Prob. 6CRCh. 5.CR - Prob. 7CRCh. 5.CR - Prob. 8CRCh. 5.CR - Prob. 9CRCh. 5.CR - Prob. 10CRCh. 5.CR - Prob. 11CRCh. 5.CR - Prob. 12CRCh. 5.CR - Prob. 13CRCh. 5.P - Prob. 1PCh. 5.P - Prob. 2PCh. 5.P - Prob. 3PCh. 5.P - Prob. 4PCh. 5.P - Prob. 5PCh. 5.P - Prob. 6PCh. 5.P - Prob. 7PCh. 5.P - Prob. 8PCh. 5.P - Prob. 9PCh. 5.P - Prob. 10PCh. 5.P - Prob. 11PCh. 5.P - Prob. 12PCh. 5.P - Prob. 13PCh. 5.P - Prob. 14PCh. 5.P - Prob. 15PCh. 5.P - Prob. 16PCh. 5.P - Prob. 17P
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