Sure, let's break down the solution step-by-step.
### Part (a) Compute the mean and standard deviation of [tex]\( X \)[/tex]
Given:
- The probability of a smoker starting before 18, [tex]\( p = 0.60 \)[/tex].
- The number of trials (the sample size), [tex]\( n = 100 \)[/tex].
To compute the mean ([tex]\( \mu \)[/tex]) of the binomial distribution:
[tex]\[ \mu = n \cdot p \][/tex]
So,
[tex]\[ \mu = 100 \cdot 0.60 = 60 \][/tex]
To compute the standard deviation ([tex]\( \sigma \)[/tex]) of the binomial distribution:
[tex]\[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} \][/tex]
So,
[tex]\[ \sigma = \sqrt{100 \cdot 0.60 \cdot (1 - 0.60)} = \sqrt{100 \cdot 0.60 \cdot 0.40} = \sqrt{24} \][/tex]
Rounding the standard deviation to the nearest tenth:
[tex]\[ \sigma \approx 4.9 \][/tex]
So,
[tex]\[ \mu_x = 60 \][/tex]
[tex]\[ \sigma_x = 4.9 \][/tex]
### Part (b) Interpret the mean
The mean ([tex]\( \mu \)[/tex]) represents the expected number of smokers out of 100 who started smoking before turning 18 years old.
The correct interpretation is:
C. It is expected that in a random sample of 100 adult smokers, 60 will have started smoking before turning 18.
### Part (c) Would it be unusual to observe 85 smokers who started smoking before turning 18 years old?
To determine if observing 85 smokers is unusual, we should consider the range defined by the mean [tex]\(\mu\)[/tex] plus or minus two standard deviations ([tex]\(\sigma\)[/tex]):
[tex]\[ \text{Lower bound} = \mu - 2 \sigma = 60 - 2 \cdot 4.9 = 60 - 9.8 \approx 50.2 \][/tex]
[tex]\[ \text{Upper bound} = \mu + 2 \sigma = 60 + 2 \cdot 4.9 = 60 + 9.8 \approx 69.8 \][/tex]
We compare the observed value (85):
- If 85 is outside this interval [50.2, 69.8], it is considered unusual.
- If 85 is within this interval, it is not considered unusual.
Since 85 is greater than 69.8, it falls outside the expected range of values within two standard deviations from the mean.
Thus:
Yes, it would be unusual to observe 85 smokers who started smoking before turning 18 years old.
The correct answer is:
A. Yes, because 85 is greater than [tex]\(\mu + 2 \sigma\)[/tex].
