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 100 trials of the probability experiment. (b) Interpret the mean. (c) Would it be unusual to observe 85 smokers who started smoking before turning 18 years old in a random sample of 100 adult smokers? Why? (a) $$ \mu_{\mathrm{x}}= $$

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To solve this problem, we can model the situation using a binomial distribution since we are dealing with a fixed number of trials (100 smokers) and two possible outcomes (started smoking before 18 or did not).

### (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 (in this case, the probability that a smoker started smoking before 18).

Here, $$ n = 100 $$ and $$ p = 0.80 $$.

$$
   \mu_X = 100 \cdot 0.80 = 80
   $$

2. **Standard Deviation ($$ \sigma_X $$)**:
   The standard deviation of a binomial distribution is given by the formula:
   $$
   \sigma_X = \sqrt{n \cdot p \cdot (1 - p)}
   $$
   where $$ (1 - p) $$ is the probability of failure.

$$
   \sigma_X = \sqrt{100 \cdot 0.80 \cdot (1 - 0.80)} = \sqrt{100 \cdot 0.80 \cdot 0.20}
   $$
   $$
   \sigma_X = \sqrt{100 \cdot 0.16} = \sqrt{16} = 4
   $$

Thus, the mean and standard deviation are:
$$
\mu_X = 80, \quad \sigma_X = 4
$$

### (b) Interpret the mean.

The mean $$ \mu_X = 80 $$ indicates that, on average, we expect 80 out of 100 adult smokers to have started smoking before turning 18 years old. This is a measure of the central tendency of the distribution of smokers who began smoking at a young age.

### (c) Would it be unusual to observe 85 smokers who started smoking before turning 18 years old in a random sample of 100 adult smokers? Why?

To determine if observing 85 smokers is unusual, we can calculate how many standard deviations away from the mean this observation is. We can use the z-score formula:

$$
z = \frac{X - \mu_X}{\sigma_X}
$$

where $$ X = 85 $$.

$$
z = \frac{85 - 80}{4} = \frac{5}{4} = 1.25
$$

A z-score of 1.25 indicates that 85 is 1.25 standard deviations above the mean.

In general, a z-score greater than 2 or less than -2 is often considered unusual in a normal distribution. A z-score of 1.25 is not considered unusual; it falls within the range of typical variation.

Thus, it would not be unusual to observe 85 smokers who started smoking before turning 18 years old in a random sample of 100 adult smokers.
(a) $$ \mu_X = 80 $$, $$ \sigma_X = 4 $$

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Solution

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

(b) Interpret the mean.

© Would it be unusual to observe 85 smokers who started smoking before turning 18 years old in a random sample of 100 adult smokers? Why?

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