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According to an almanac, \( 70 \% \) 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 smoking before 18 based on a random sample of 200 adults: (b) Interpret the mean. (a) \( \mu_{x}=\square \)

Ask by Robbins Hall. in the United States
Mar 16,2025

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(a) Mean \( \mu_X = 140 \) and standard deviation \( \sigma_X \approx 6.48 \). (b) On average, 140 out of 200 adult smokers started smoking before turning 18 years old.

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To compute the mean and standard deviation of the random variable \( X \), we can model this scenario using a binomial distribution. Given that 70% of adult smokers started smoking before age 18, we have: - Number of trials \( n = 200 \) - Probability of success (starting smoking before 18) \( p = 0.70 \) (a) The mean \( \mu_{X} \) is calculated as: \[ \mu_{X} = n \cdot p = 200 \cdot 0.70 = 140 \] The standard deviation \( \sigma_{X} \) can be calculated using the formula: \[ \sigma_{X} = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{200 \cdot 0.70 \cdot 0.30} = \sqrt{42} \approx 6.48 \] (a) \( \mu_{x} = 140 \) (b) The mean of 140 indicates that, in a random sample of 200 adults, you can expect about 140 of them to have started smoking before turning 18 years old. This represents a significant portion of the sample, reflecting the tendency for many smokers to pick up the habit at a young age.

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