According to an almanac, $$ 60 \% $$ of adull 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 300 trials of the probability experiment. (b) Interpret the mean. (c) Would it be unusual to observe 270 smokers who started smoking before turning 18 years old in a random sample of 300 adult smokers? Why? (a) $$ \mu_{\mathrm{x}}=\square $$

Question

UpStudy AI · Accepted Answer

**(a) Calculation of the mean and standard deviation**

Since the number of smokers who started before 18 follows a binomial distribution with parameters $$ n = 300 $$ and $$ p = 0.60 $$, the mean $$ \mu_X $$ and standard deviation $$ \sigma_X $$ are given by:

$$
\mu_X = np = 300 \times 0.60 = 180
$$

$$
\sigma_X = \sqrt{np(1-p)} = \sqrt{300 \times 0.60 \times 0.40} = \sqrt{72} \approx 8.49
$$

**(b) Interpretation of the mean**

The mean $$ \mu_X = 180 $$ indicates that, on average, out of 300 adult smokers, 180 are expected to have started smoking before turning 18.

**(c) Assessment of whether observing 270 smokers is unusual**

To determine if observing 270 smokers is unusual, we calculate the $$ z $$-score for $$ X = 270 $$:

$$
z = \frac{270 - \mu_X}{\sigma_X} = \frac{270 - 180}{\sqrt{72}} = \frac{90}{8.49} \approx 10.59
$$

A $$ z $$-score of approximately 10.59 is extremely high, as values beyond 2 or 3 standard deviations from the mean are typically considered unusual. Thus, observing 270 smokers who started smoking before turning 18 in a random sample of 300 is highly unusual.

**Final Answer for (a):**

$$
\mu_X = 180
$$
**(a) Mean and Standard Deviation** - **Mean ($$ \mu_X $$)**: 180 - **Standard Deviation ($$ \sigma_X $$)**: Approximately 8.49 **(b) Interpretation of the Mean** On average, out of 300 adult smokers, 180 are expected to have started smoking before turning 18. **(c) Unusual Observation** Observing 270 smokers who started smoking before turning 18 in a sample of 300 is highly unusual because the $$ z $$-score is approximately 10.59, which is far beyond typical thresholds for unusual events.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Statistics Questions