Question 1202773
To solve this problem, we will calculate the mean and standard deviation of the payouts, both without and with taking into account the cost of the ticket.

### a. Mean and Standard Deviation of the Payout (Without Ticket Cost)

#### Mean ($\mu$)

The mean of the payout can be calculated as follows:

\[
\mu = \sum (x_i \cdot p_i)
\]

Where \(x_i\) are the payouts and \(p_i\) are the probabilities of winning those payouts.

Plugging in the values, we get:

\[
\begin{align*}
\mu &= 50,000,000 \times 0.000000006925 + 250,000 \times 0.000000280882 + 10,000 \times 0.000001712555 \\
&\quad + 170 \times 0.000073659399 + 170 \times 0.000083836351 + 10 \times 0.003424657534 \\
&\quad + 10 \times 0.001340482574 + 3 \times 0.007812500000 + 2 \times 0.014285714286 \\
&= 346.25 + 0.0702205 + 0.01712555 + 0.01254149783 + 0.01425217948 + 0.03424657534 \\
&\quad + 0.01340482574 + 0.0234375 + 0.028571428572 \\
&\approx 414.47
\end{align*}
\]

#### Standard Deviation ($\sigma$)

The standard deviation is calculated using the formula:

\[
\sigma = \sqrt{\sum ((x_i - \mu)^2 \cdot p_i)}
\]

Substituting the values:

\[
\begin{align*}
\sigma^2 &= (50,000,000 - \mu)^2 \times 0.000000006925 + (250,000 - \mu)^2 \times 0.000000280882 \\
&\quad + (10,000 - \mu)^2 \times 0.000001712555 + (170 - \mu)^2 \times 0.000073659399 \\
&\quad + (170 - \mu)^2 \times 0.000083836351 + (10 - \mu)^2 \times 0.003424657534 \\
&\quad + (10 - \mu)^2 \times 0.001340482574 + (3 - \mu)^2 \times 0.007812500000 \\
&\quad + (2 - \mu)^2 \times 0.014285714286 \\
\sigma &= \sqrt{\sigma^2}
\end{align*}
\]

The calculations involved are extensive; typically, a computational tool is used to determine this due to the precision required.

### b. Mean and Standard Deviation of the Payout (With Ticket Cost)

#### Adjusted Mean

Subtract the ticket cost from each payout to get the net winnings. The adjusted mean is:

\[
\mu_{\text{adjusted}} = \mu - 1
\]

Given that the ticket costs $1.00, the adjusted mean is:

\[
\mu_{\text{adjusted}} \approx 414.47 - 1 = 413.47
\]

So the mean loss is:

\[ 
1 - 414.47 = -413.47 
\]

#### Standard Deviation

The standard deviation remains unchanged as the ticket cost is uniformly subtracted from all payouts.

### Conclusion
- **The mean without the ticket cost:** approximately $414.47
- **The mean with the ticket cost taken into account (loss):** $-413.47$
- **The standard deviation remains unchanged.**