Answer :
To solve this problem, we begin by clearly outlining the known information and utilizing the concepts of probability and statistics related to a binomial distribution.
### Part (a) Compute the mean and standard deviation of the random variable [tex]\( X \)[/tex].
Given:
- Probability of success (starting smoking before 18) [tex]\( p = 0.80 \)[/tex]
- Number of trials [tex]\( n = 200 \)[/tex]
#### Mean ([tex]\(\mu_X\)[/tex]):
The mean of a binomial distribution can be calculated using the formula:
[tex]\[ \mu_X = n \cdot p \][/tex]
So,
[tex]\[ \mu_X = 200 \cdot 0.80 = 160 \][/tex]
#### Standard Deviation ([tex]\(\sigma_X\)[/tex]):
The standard deviation of a binomial distribution is given by:
[tex]\[ \sigma_X = \sqrt{n \cdot p \cdot (1 - p)} \][/tex]
So,
[tex]\[ \sigma_X = \sqrt{200 \cdot 0.80 \cdot (1 - 0.80)} = \sqrt{200 \cdot 0.80 \cdot 0.20} \][/tex]
[tex]\[ \sigma_X = \sqrt{32} \approx 5.65685424949238 \][/tex]
Thus, the mean ([tex]\(\mu_X\)[/tex]) is 160 and the standard deviation ([tex]\(\sigma_X\)[/tex]) is approximately 5.65685424949238.
### Part (b) Interpret the mean.
The mean, or expected value, [tex]\( \mu_X = 160 \)[/tex], means that on average, out of 200 adult smokers, we would expect 160 of them to have started smoking before turning 18 years old. This provides a central value around which the number of smokers starting before 18 is centered.
### Part (c) Determine if it would be unusual to observe 170 smokers who started smoking before turning 18 years old in a random sample of 200 adult smokers.
To determine if observing 170 smokers is unusual, we calculate the z-score and then find the corresponding probability.
#### Z-score Calculation:
The z-score for 170 smokers is calculated using the formula:
[tex]\[ z = \frac{x - \mu_X}{\sigma_X} \][/tex]
Given:
- [tex]\( x = 170 \)[/tex]
- [tex]\( \mu_X = 160 \)[/tex]
- [tex]\( \sigma_X = 5.65685424949238 \)[/tex]
So,
[tex]\[ z = \frac{170 - 160}{5.65685424949238} \][/tex]
[tex]\[ z \approx 1.7677669529663689 \][/tex]
#### Probability Calculation:
To find the probability associated with this z-score, we look up the z-score in a standard normal distribution table or use a computational tool. The probability of obtaining a z-score less than 1.7677669529663689 corresponds to the cumulative distribution function (CDF) value at that point.
For [tex]\( z \approx 1.7677669529663689 \)[/tex]:
[tex]\[ P(Z < 1.7677669529663689) \approx 0.9614500641282292 \][/tex]
Thus, the probability of observing more than 170 smokers is:
[tex]\[ P(X \geq 170) \approx 1 - 0.9614500641282292 \approx 0.0385499358717708 \][/tex]
### Conclusion:
Since this probability (approximately 0.0385) is less than the common significance level of 0.05 (or 5%), it would be considered unusual to observe 170 smokers who started smoking before turning 18 years old in a random sample of 200 adult smokers.
Thus, we can summarize:
(a) [tex]\(\mu_X = 160\)[/tex]
(b) On average, out of 200 adult smokers, we expect 160 to have started smoking before turning 18.
(c) Observing 170 smokers who started smoking before 18 would be unusual, as there is roughly a 3.85% chance of this happening, which is less than the typical threshold of 5%.
### Part (a) Compute the mean and standard deviation of the random variable [tex]\( X \)[/tex].
Given:
- Probability of success (starting smoking before 18) [tex]\( p = 0.80 \)[/tex]
- Number of trials [tex]\( n = 200 \)[/tex]
#### Mean ([tex]\(\mu_X\)[/tex]):
The mean of a binomial distribution can be calculated using the formula:
[tex]\[ \mu_X = n \cdot p \][/tex]
So,
[tex]\[ \mu_X = 200 \cdot 0.80 = 160 \][/tex]
#### Standard Deviation ([tex]\(\sigma_X\)[/tex]):
The standard deviation of a binomial distribution is given by:
[tex]\[ \sigma_X = \sqrt{n \cdot p \cdot (1 - p)} \][/tex]
So,
[tex]\[ \sigma_X = \sqrt{200 \cdot 0.80 \cdot (1 - 0.80)} = \sqrt{200 \cdot 0.80 \cdot 0.20} \][/tex]
[tex]\[ \sigma_X = \sqrt{32} \approx 5.65685424949238 \][/tex]
Thus, the mean ([tex]\(\mu_X\)[/tex]) is 160 and the standard deviation ([tex]\(\sigma_X\)[/tex]) is approximately 5.65685424949238.
### Part (b) Interpret the mean.
The mean, or expected value, [tex]\( \mu_X = 160 \)[/tex], means that on average, out of 200 adult smokers, we would expect 160 of them to have started smoking before turning 18 years old. This provides a central value around which the number of smokers starting before 18 is centered.
### Part (c) Determine if it would be unusual to observe 170 smokers who started smoking before turning 18 years old in a random sample of 200 adult smokers.
To determine if observing 170 smokers is unusual, we calculate the z-score and then find the corresponding probability.
#### Z-score Calculation:
The z-score for 170 smokers is calculated using the formula:
[tex]\[ z = \frac{x - \mu_X}{\sigma_X} \][/tex]
Given:
- [tex]\( x = 170 \)[/tex]
- [tex]\( \mu_X = 160 \)[/tex]
- [tex]\( \sigma_X = 5.65685424949238 \)[/tex]
So,
[tex]\[ z = \frac{170 - 160}{5.65685424949238} \][/tex]
[tex]\[ z \approx 1.7677669529663689 \][/tex]
#### Probability Calculation:
To find the probability associated with this z-score, we look up the z-score in a standard normal distribution table or use a computational tool. The probability of obtaining a z-score less than 1.7677669529663689 corresponds to the cumulative distribution function (CDF) value at that point.
For [tex]\( z \approx 1.7677669529663689 \)[/tex]:
[tex]\[ P(Z < 1.7677669529663689) \approx 0.9614500641282292 \][/tex]
Thus, the probability of observing more than 170 smokers is:
[tex]\[ P(X \geq 170) \approx 1 - 0.9614500641282292 \approx 0.0385499358717708 \][/tex]
### Conclusion:
Since this probability (approximately 0.0385) is less than the common significance level of 0.05 (or 5%), it would be considered unusual to observe 170 smokers who started smoking before turning 18 years old in a random sample of 200 adult smokers.
Thus, we can summarize:
(a) [tex]\(\mu_X = 160\)[/tex]
(b) On average, out of 200 adult smokers, we expect 160 to have started smoking before turning 18.
(c) Observing 170 smokers who started smoking before 18 would be unusual, as there is roughly a 3.85% chance of this happening, which is less than the typical threshold of 5%.
