According to an almanac, $$ 80 \% $$ 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 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? (a) $$ \mathrm{p}_{\mathrm{x}}=160 $$ $$ \sigma_{\mathrm{x}}=5.7 $$ (Round to the nearest tenth às needed.) (b) What is the correct interpretation of the mean? A. It is expected that in a random sample of 200 adult smokers, 160 will have started smoking before furning 18 . B. It is expected that in $$ 50 \% $$ of random samples of 200 adult smokers, 160 will have started smoking before turning 18 . C. It is expected that in a random sample of 200 adult smokers, 160 will have started smoking after turning 18 .

Question

UpStudy AI · Accepted Answer

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 started before turning 18).

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

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

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

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

Here, $$ n = 200 $$, $$ p = 0.80 $$, and $$ (1 - p) = 0.20 $$:
   $$
   \sigma = \sqrt{200 \cdot 0.80 \cdot 0.20} = \sqrt{32} \approx 5.7
   $$

Thus, the results are:
- Mean ($$ \mu $$) = 160
- Standard Deviation ($$ \sigma $$) ≈ 5.7 (rounded to the nearest tenth)

### (b) Interpret the mean.

The correct interpretation of the mean is:
**A. It is expected that in a random sample of 200 adult smokers, 160 will have started smoking before turning 18.**

### (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 who started before turning 18 is unusual, we can calculate how many standard deviations away from the mean this value is. We can use the z-score formula:

$$
z = \frac{X - \mu}{\sigma}
$$

Where:
- $$ X = 170 $$
- $$ \mu = 160 $$
- $$ \sigma \approx 5.7 $$

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

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. Since 1.75 is not greater than 2, it would not be considered unusual to observe 170 smokers who started smoking before turning 18 in a random sample of 200 adult smokers.

### Summary:
- (a) Mean = 160, Standard Deviation ≈ 5.7
- (b) Correct interpretation: A
- (c) No, it would not be unusual to observe 170 smokers who started smoking before turning 18.
(a) Mean = 160, Standard Deviation ≈ 5.7 (b) A. It is expected that in a random sample of 200 adult smokers, 160 will have started smoking before turning 18. (c) No, it would not be unusual to observe 170 smokers who started smoking before turning 18.

Upstudy AI Solution

Answer

Solution

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

(b) Interpret the mean.

© 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?

Summary:

Beyond the Answer

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