According to an almanac, $$ 60 \% $$ 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 300 trials of the probability experiment. (b) Interpret the mean. (c) Would it be unusual to observe 240 smokers who started smoking before turning 18 years old in a random sample of 300 adult smokers? Why? (a) $$ \mu_{x}=\square $$

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To 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, we can use the given information that 60% of adult smokers started smoking before turning 18 years old.

Given:
- $$ P(X = 1) = 0.6 $$ (60% of adult smokers started smoking before turning 18 years old)

(a) To find the mean and standard deviation of $$ X $$, we can use the formula for the mean and standard deviation of a binomial distribution:
$$ \mu = np $$
$$ \sigma = \sqrt{np(1-p)} $$

where:
- $$ n $$ is the number of trials (300 in this case)
- $$ p $$ is the probability of success (0.6 in this case)

Let's calculate the mean and standard deviation of $$ X $$ using the given values.
Calculate the value by following steps:
- step0: Calculate:
  $$300\times 0.6$$
- step1: Multiply the numbers:
  $$180$$
Calculate or simplify the expression $$ \sqrt{300*0.6*(1-0.6)} $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\sqrt{300\times 0.6\left(1-0.6\right)}$$
- step1: Subtract the numbers:
  $$\sqrt{300\times 0.6\times 0.4}$$
- step2: Multiply the terms:
  $$\sqrt{72}$$
- step3: Write as a product:
  $$\sqrt{36\times 2}$$
- step4: Write in exponential form:
  $$\sqrt{6^{2}\times 2}$$
- step5: Use the properties of radicals:
  $$\sqrt{6^{2}}\times \sqrt{2}$$
- step6: Simplify the root:
  $$6\sqrt{2}$$
The mean of the random variable $$ X $$ is 180, and the standard deviation is approximately 8.485281.

(b) The mean of 180 represents the average number of smokers who started smoking before turning 18 years old in 300 trials of the probability experiment.

(c) To determine if it would be unusual to observe 240 smokers who started smoking before turning 18 years old in a random sample of 300 adult smokers, we can compare the observed value to the mean and standard deviation calculated.

Since the observed value (240) is less than the mean (180), it would be unusual to observe 240 smokers who started smoking before turning 18 years old in a random sample of 300 adult smokers.
(a) $$ \mu_{x}=180 $$ (b) The mean of 180 represents the average number of smokers who started smoking before turning 18 years old in 300 trials. (c) It would be unusual to observe 240 smokers who started smoking before turning 18 years old in a random sample of 300 adult smokers.

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