Chapter 6, Problem R6.3RE

(a)

To determine

Graphically represent the probability distribution.

Answer to Problem R6.3RE

Explanation of Solution

Given information:

Table showing possible outcomes of the game and their probabilities, along with the amount of money (payout) that a player wins for $1 bet:

From the table,

We have two payouts of $0.

By adding the corresponding probabilities of both payouts of $0,

We will get the probability of payout of $0 as 0.741.

For Probability Histogram:

The width of each bar has to be same.

And

The bars have to be centered at the amount of money collected.

Whereas,

The height has to be equal to probability.

(b)

To determine

Interpretation for expected value and standard deviation of X .

Answer to Problem R6.3RE

On an average, the expected payout is $0.70 and the payout varies about $6.58 from the expected payout of $0.70.

Explanation of Solution

Given information:

X : payout for a single $1 bet

Such that

For X :

Mean,

  ${\mu }_X=\$0.70$

Standard deviation,

  ${\sigma }_X=\$6.58$

Calculations:

X represents the payout for a single $1 bet.

For expected value (or mean):

We have

  ${\mu }_X=\$0.70$

This means

On an average, the expected payout is $0.70.

For Standard deviation:

We have

  ${\sigma }_X=\$6.58$

This means

On an average, the payout varies about $6.58 from the expected payout of $0.70.

(c)

To determine

Expected value and standard deviation of Jerry’s winnings.

Answer to Problem R6.3RE

For Jerry’s winnings:

Expected value (or mean),

  ${\mu }_Y=\$3.50$

Standard deviation,

  ${\sigma }_Y=\$32.9$

Explanation of Solution

Given information:

X : payout for a single $1 bet

Such that

For X :

Mean,

  ${\mu }_X=\$0.70$

Standard deviation,

  ${\sigma }_X=\$6.58$

Calculations:

X represents the payout for a single $1 bet.

Let

Y represents the payout for a single $5 bet.

If the bet is multiplied by 5, then the payout is also multiplied by 5.

Such that

  $Y=5X$

Then

Every data value in the distribution of X is multiplied by the same constant 5.

If every data value is multiplied by the same constant, the center of the distribution is also multiplied by the same constant.

We know that

The mean is the measure of the center.

Thus,

The mean of Y is the mean of X multiplied by 5.

  ${\mu }_Y=5{\mu }_X=5\left(0.70\right)=\$3.50$

If every data value is multiplied by the same constant, the spread of the distribution is also multiplied by the same constant.

We know that

The standard deviation is the measure of the spread.

Thus,

The standard deviation is multiplied by 5.

  ${\sigma }_Y=5{\sigma }_X=\frac{9}{5}\left(6.58\right)=\$32.9$

(d)

To determine

Expected value and standard deviation of Marla’s total winnings.

Answer to Problem R6.3RE

For Marla’s total winnings:

Expected value (or mean),

  ${\mu }_{{}_{X_1+X_2+\mathrm{...}+X_5}}=\$3.50$

Standard deviation,

  ${\sigma }_{{}_{X_1+X_2+\mathrm{...}+X_5}}\approx \$14.71$

Explanation of Solution

Given information:

X : payout for a single $1 bet

Such that

For X :

Mean,

  ${\mu }_X=\$0.70$

Standard deviation,

  ${\sigma }_X=\$6.58$

Calculations:

Property mean:

  ${\mu }_{{\displaystyle {\sum }_i{X}_i}}={\displaystyle \sum _i\mu X_i}$

Property variance:

  ${\sigma }_{{\displaystyle {\sum }_i{X}_i}}^2={\displaystyle \sum _i{\sigma }_{X_i}^2}$

We know that

X represents the payout for a single $1 bet.

Thus,

  $X_1+X_2+\mathrm{...}+X_5$ represents five single $1 bets.

For expected value (or mean),

  ${\mu }_{X_1+X_2+\mathrm{...}+X_5}=5{\mu }_X=5\left(0.70\right)=\$3.50$

For variance,

  ${\sigma }_{X_1+X_2+\mathrm{...}+X_5}^2=5{\sigma }_X^2=5{\left(6.58\right)}^2=\$216.482$

We also know

The standard deviation is the square root of the variance:

  ${\sigma }_{{}_{X_1+X_2+\mathrm{...}+X_5}}=\sqrt{{\sigma }_{{}_{X_1+X_2+\mathrm{...}+X_5}}^2}=\sqrt{216.482}\approx \$14.71$