91影视

To interpret the values of ${渭}_X$and ${蟽}_X$ in context.

Since, the given values are:
${渭}_X=\$0.70$
${蟽}_X=\$6.58$
Determine the mean and standard deviation as:
${渭}_{aX+b}=a{渭}_X+b$
${蟽}_{aX+b}=a{蟽}_X$

The mean of $X$indicates that the predicted pay out is $\$0.70$on average, and that the payout will vary by $\$6.58$on average from the payout of $\$0.70$.

To find the expected value and the standard deviation of his winnings. Let, Jerry places a single $\$5$ bet on $4$-Spot Keno.

Determine the expected value and standard deviation as follows:

${渭}_{5X}=5脳{渭}_X$

$=5(0.70)$

$=\$3.50$

${蟽}_{5X}=5脳{蟽}_X$

$=5(6.58)$

$=\$32.90$

As a result, the mean and standard deviation are $\$3.50$and$\$32.90$respectively.

To find the expected value and the standard deviation of the total winnings. Let, Marla plays five games of $4-$Spot Keno, betting $\$1$each time.

Determine the expected value and standard deviation as follows:
${渭}_{X_1+X_2+鈥�+X_5}=5脳{渭}_X$

$=5(0.70)$

$=\$3.50$

$\sqrt{{蟽}^2{X}_1+X_2+鈥�+X_5}=\sqrt{5脳{蟽}^{2}x}$

$=\sqrt{5(6.58{)}^2}$

$=\$14.71$

As a result, the mean and standard deviation are $\$3.50$and $\$14.71$respectively.

To justify the answer that based on the answers to (b) and (c), which player would the casino prefer.

Since the result from part (b) as for Jerry:
${渭}_{5X}=\$3.50$
${蟽}_{5X}=\$32.90$
And the result from part (c) as for Marla:
$渭=\$3.50$
$蟽=\$14.71$
Because both players have the same mean, the standard deviation will determine the outcome.
A casino wants a player who can best forecast their wins and losses, and hence the person with the lowest standard deviation, which corresponds to the player Marla.