St According to an almanac, $$ 80 \% $$ 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 before 18 in 200 trials of the probability experiment. (b) Interpret the mean. (c) Would it be unusual to observe 170 smokers who started smoking before turning 18 years old in a random sample of 200 adult smokers? Why? 1 $$ \sigma_{x}=\square $$ (Round to the nearest tenth as needed.) (a) $$ \mu_{\mathrm{x}}=160 $$

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To solve the problem, we will use the properties of the binomial distribution since we are dealing with a fixed number of trials (200 smokers) and a constant probability of success (80% of smokers starting before age 18).

### (a) Compute the mean and standard deviation of the random variable $$ X $$.

1. **Mean ($$ \mu_X $$)**:
   The mean of a binomial distribution is given by the formula:
   $$
   \mu_X = n \cdot p
   $$
   where $$ n $$ is the number of trials and $$ p $$ is the probability of success.

Here, $$ n = 200 $$ and $$ p = 0.80 $$:
   $$
   \mu_X = 200 \cdot 0.80 = 160
   $$

2. **Standard Deviation ($$ \sigma_X $$)**:
   The standard deviation of a binomial distribution is given by the formula:
   $$
   \sigma_X = \sqrt{n \cdot p \cdot (1 - p)}
   $$
   Using the same values for $$ n $$ and $$ p $$:
   $$
   \sigma_X = \sqrt{200 \cdot 0.80 \cdot (1 - 0.80)} = \sqrt{200 \cdot 0.80 \cdot 0.20}
   $$
   $$
   = \sqrt{200 \cdot 0.16} = \sqrt{32} \approx 5.7
   $$

Thus, the answers for part (a) are:
- Mean ($$ \mu_X $$): $$ 160 $$
- Standard Deviation ($$ \sigma_X $$): $$ 5.7 $$ (rounded to the nearest tenth)

### (b) Interpret the mean.

The mean $$ \mu_X = 160 $$ indicates that, on average, in a random sample of 200 adult smokers, we would expect about 160 of them to have started smoking before turning 18 years old. This value represents the expected number of smokers who began smoking at a young age in such a sample.

### (c) Would it be unusual to observe 170 smokers who started smoking before turning 18 years old in a random sample of 200 adult smokers? Why?

To determine if observing 170 smokers is unusual, we can calculate how many standard deviations away from the mean this value is. We can use the z-score formula:
$$
z = \frac{X - \mu_X}{\sigma_X}
$$
where $$ X = 170 $$.

Substituting the values:
$$
z = \frac{170 - 160}{5.7} \approx \frac{10}{5.7} \approx 1.75
$$

A z-score of 1.75 indicates that 170 is 1.75 standard deviations above the mean. In general, a z-score greater than 2 or less than -2 is often considered unusual. A z-score of 1.75 is not typically considered unusual, as it falls within the range of common variation.

### Conclusion:
It would not be unusual to observe 170 smokers who started smoking before turning 18 years old in a random sample of 200 adult smokers, as it is within 2 standard deviations of the mean.
(a) Mean ($$ \mu_X $$): 160, Standard Deviation ($$ \sigma_X $$): 5.7 (b) The mean indicates an average of 160 smokers started smoking before 18 in a sample of 200. (c) Observing 170 smokers is not unusual as it is within 2 standard deviations of the mean.

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