According to an almanac, $$ 70 \% $$ of adult smokers started smoking before turning 18 years old. (a) Compute the mean and standard deviation of the random variable $$ X $$, the number of smokers who started smoking before 18 based on a random sample of 200 adults: (b) Interpret the mean. (a) $$ \mu_{x}=\square $$

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To solve this problem, we will use the properties of a binomial distribution since we are dealing with a fixed number of trials (200 adults) and a constant probability of success (70% of smokers started smoking before turning 18).

### Step 1: Identify the known conditions
- The probability of success (starting smoking before 18) is $$ p = 0.70 $$.
- The number of trials (sample size) is $$ n = 200 $$.

### Step 2: Calculate the mean and standard deviation
For a binomial distribution, the mean $$ \mu_X $$ and standard deviation $$ \sigma_X $$ can be calculated using the following formulas:

1. Mean: 
   $$
   \mu_X = n \cdot p
   $$

2. Standard Deviation:
   $$
   \sigma_X = \sqrt{n \cdot p \cdot (1 - p)}
   $$

Now, let's calculate these values.

### Mean Calculation
$$
\mu_X = 200 \cdot 0.70
$$

### Standard Deviation Calculation
$$
\sigma_X = \sqrt{200 \cdot 0.70 \cdot (1 - 0.70)}
$$

Now, I will perform these calculations.
Calculate the value by following steps:
- step0: Calculate:
  $$200\times 0.7$$
- step1: Multiply the numbers:
  $$140$$
Calculate or simplify the expression $$ \sqrt(200 * 0.70 * (1 - 0.70)) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\sqrt{200\times 0.7\left(1-0.7\right)}$$
- step1: Subtract the numbers:
  $$\sqrt{200\times 0.7\times 0.3}$$
- step2: Multiply the terms:
  $$\sqrt{42}$$
### Step 3: Results of the calculations
- The mean $$ \mu_X $$ is calculated as:
  $$
  \mu_X = 140
  $$

- The standard deviation $$ \sigma_X $$ is approximately:
  $$
  \sigma_X \approx 6.48
  $$

### Step 4: Interpretation of the mean
The mean $$ \mu_X = 140 $$ indicates that, on average, out of a random sample of 200 adult smokers, we expect that 140 of them started smoking before turning 18 years old.

### Final Answers
(a) 
- Mean: $$ \mu_X = 140 $$
- Standard Deviation: $$ \sigma_X \approx 6.48 $$

(b) The mean indicates that in a sample of 200 adult smokers, we expect about 140 of them to have started smoking before the age of 18.
(a) Mean $$ \mu_X = 140 $$ and standard deviation $$ \sigma_X \approx 6.48 $$. (b) On average, 140 out of 200 adult smokers started smoking before turning 18 years old.

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