Question

Keno In a game of 4-Spot Keno, the player picks 4 numbers from 1 to 80 . The casino randomly selects 20 winning numbers from 1 to 80 . The table shows the possible outcomes of the game and their probabilities, along with the amount of money (Payout) that the player wins for a 1\( bet. If X= the payout for a single 1 bet, you can check that ${\mu }_X=$$0.70$and ${\sigma }_X=$$6.58$.

a. Make a graph of the probability distribution. Describe what you see.

b. Interpret the values of ${\mu }_X$and ${\sigma }_X$.

c. Jerry places a single 5\) bet on 4-Spot Keno. Find the expected value and the standard deviation of his winnings.

d. Marla plays five games of 4-Spot Keno, betting 1$ each time. Find the expected value and the standard deviation of her total winnings.

Short answer

(a)The width of each bar has to be same.AndThe bars have to be centered at the amount of money collected.

(b)On an average, the payout varies about $6.58 from the expected payout of $0.70

(c)The standard deviation is multiplied by $5 .

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

(d)The standard deviation is the square root of the variance:

${\sigma }_{x_1+x_2+\dots +x_5}=\sqrt{{\sigma }_{x_1+x_2+\dots +x_5}^2}=\sqrt{216.482}\approx \$14.71$

Step-by-step solution

Part (a) Step 1: 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.

Part (a) Step 2: Simplification

X : payout for a single 1$ bet

Such that

For X:

Mean,

${\mu }_X=\$0.70$

Standard deviation,

${\sigma }_X=\$6.58$

Part (b) Step 1: Given Information

X : payout for a single 1$ bet

Such that

For X:

Mean,

$m{u}_X=\$0.70$

Standard deviation,

${\sigma }_X=\$6.58$

Part (b) Step 2 : Simplification

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 $0.70\$.$

Part (c) Step 1: Given Information

X : payout for a single 1$ bet

Such that

For X:

Mean,

$\mu x=\$0.70$

Standard deviation,

${\sigma }_x=\$6.58$

Part (c) Step 2: Simplification

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=5 X

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 .

$\mu Y=5\mu x=5(0.70)=\$3.50$

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 .

$\sigma \gamma =5\sigma x=\frac{9}{5}(6.58)=\$32.9$

Part (d) Step 1: Given Information

X : payout for a single 1$ bet

Such that

For X :

Mean,

${\mu }_X=\$0.70$

Standard deviation,

${\sigma }_X=\$6.58$

Part (d) Step 2: Simplification

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:

$\mu \sum _i{x}_i=\sum _i\mu X_i$

Property variance:

${\sigma }_i^2{x}_i=\sum _i{\sigma }_{x_i}^2$

We know that

X represents the payout for a single 1$ bet.

Thus,

$X_1+X_2+\dots +X_5$represents five single 1$ bets.

For expected value (or mean),

${\mu }_{X_1+X_2+\dots +X_5}=5{\mu }_X=5(0.70)=\$3.50$

For variance,

${\sigma }_{x_1+x_2+\dots +x_5}^2=5{\sigma }_X^2=5(6.58{)}^2=\$216.482$

We also know

The standard deviation is the square root of the variance:

${\sigma }_{x_1+x_2++1x_5}=\sqrt{{\sigma }_{x_1+x_2++x_5}^2}=\sqrt{216.482}\approx \$14.71$