Use the following distribution to complete parts (a) through (d) below. $$ 3,5,9,12,11 $$ p a) Compute the mean and standard deviation of the distribution. The mean is 8 . The standard deviation is 3.87 . (Round to the nearest hundredth as needed.) b) Multiply each number in the distribution by 2 and compute the mean and the standard deviation of this new distribution. The mean is 16 . The standard deviation is 7.75 . (Round to the nearest hundredth as needed.) c) Multiply each number in the original distrhation by 4 and compute the mean and the standard deviation of this new distribution. The mean is $$ \square $$

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$$
\textbf{Step 1. Original Distribution Calculations:}
$$

The original data are: $$ 3,\, 5,\, 9,\, 12,\, 11 $$.

$$
\textbf{Mean ($$\bar{x}$$):} \quad \bar{x} = \frac{3 + 5 + 9 + 12 + 11}{5} = \frac{40}{5} = 8.
$$

$$
\textbf{Sample Standard Deviation ($$s$$):}
$$

Since we use the sample standard deviation formula,
$$
s = \sqrt{\frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n-1}},
$$
first compute the squared deviations:
$$
(3-8)^2 = 25,\quad (5-8)^2 = 9,\quad (9-8)^2 = 1,\quad (12-8)^2 = 16,\quad (11-8)^2 = 9.
$$
Sum of squared deviations:
$$
25 + 9 + 1 + 16 + 9 = 60.
$$
Then,
$$
s = \sqrt{\frac{60}{5-1}} = \sqrt{\frac{60}{4}} = \sqrt{15} \approx 3.87.
$$

$$
\textbf{Step 2. Multiplying by a Constant:}
$$

When each number in a distribution is multiplied by a constant $$k$$, the new mean and standard deviation transform as follows:
$$
\text{New mean} = k \cdot \bar{x}, \quad \text{New standard deviation} = |k| \cdot s.
$$

$$
\textbf{Part (b): Multiplying by } k = 2.
$$

$$
\text{New mean} = 2 \cdot 8 = 16.
$$
$$
\text{New standard deviation} = 2 \cdot 3.87 \approx 7.75.
$$

$$
\textbf{Part (c): Multiplying by } k = 4.
$$

$$
\text{New mean} = 4 \cdot 8 = 32.
$$
$$
\text{New standard deviation} = 4 \cdot 3.87 \approx 15.48.
$$

Thus, for part (c), the mean of the new distribution is $$ \boxed{32} $$ and the standard deviation is approximately $$ \boxed{15.48} $$.
For part (c), the mean is 32 and the standard deviation is approximately 15.48.

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