Respuesta :
Answer:
There is a 68% chance that between 17% and 30% are smokers.
There is a 95% chance that between 10% and 37% are smokers.
There is a 99.7% chance that between 4% and 44% are smokers.
Step-by-step explanation:
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of the sampling distribution of sample proportion is:
[tex]\mu_{\hat p}=p\\[/tex]
The standard deviation of the sampling distribution of sample proportion is:
[tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}[/tex]
Given:
n = 40
p = 0.236
Compute the mean and standard deviation of this sampling distribution of sample proportion as follows:
[tex]\mu_{\hat p}=p=0.236[/tex]
[tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.236(1-0.236)}{40}}=0.067[/tex]
The Empirical Rule states that in a normal distribution with mean µ and standard deviation σ, nearly all the data will fall within 3 standard deviations of the mean. The empirical rule can be divided into three parts:
- 68% data falls within 1 standard-deviation of the mean.
That is P (µ - σ ≤ X ≤ µ + σ) = 0.68.
- 95% data falls within 2 standard-deviations of the mean.
That is P (µ - 2σ ≤ X ≤ µ + 2σ) = 0.95.
- 99.7% data falls within 3 standard-deviations of the mean.
That is P (µ - 3σ ≤ X ≤ µ + 3σ) = 0.997.
Compute the range of values that has a probability of 68% as follows:
[tex]P (\mu_{\hat p} - \sigma_{\hat p} \leq \hat p \leq \mu_{\hat p} + \sigma_{\hat p}) = 0.68\\P(0.236-0.067\leq \hat p \leq 0.236+0.067)=0.68\\P(0.169\leq \hat p \leq0.303)=0.68\\P(0.17\leq \hat p \leq0.30)=0.68[/tex]
Thus, there is a 68% chance that between 17% and 30% are smokers.
Compute the range of values that has a probability of 95% as follows:
[tex]P (\mu_{\hat p} - 2\sigma_{\hat p} \leq \hat p \leq \mu_{\hat p} + 2\sigma_{\hat p}) = 0.95\\P(0.236-2\times 0.067\leq \hat p \leq 0.236+2\times0.067)=0.95\\P(0.102\leq \hat p \leq 0.370)=0.95\\P(0.10\leq \hat p \leq0.37)=0.95[/tex]
Thus, there is a 95% chance that between 10% and 37% are smokers.
Compute the range of values that has a probability of 99.7% as follows:
[tex]P (\mu_{\hat p} - 3\sigma_{\hat p} \leq \hat p \leq \mu_{\hat p} + 3\sigma_{\hat p}) = 0.997\\P(0.236-3\times 0.067\leq \hat p \leq 0.236+3\times0.067)=0.997\\P(0.035\leq \hat p \leq 0.437)=0.997\\P(0.04\leq \hat p \leq0.44)=0.997[/tex]
Thus, there is a 99.7% chance that between 4% and 44% are smokers.
