## The most recent public health statistics available indicate that 23.6% of American adults smoke cigarettes. Using the 68-95-99.7 Rule, des

The most recent public health statistics available indicate that 23.6% of American adults smoke cigarettes. Using the 68-95-99.7 Rule, describe the sampling distribution model for the proportion of smokers among a randomly selected group of 40 adults. Be sure to discuss your assumptions and conditions. Describe the sampling distribution. There is a 68% chance that between nothing% and nothing% are smokers, a 95% chance that between nothing% and nothing% are smokers, and a 99.7% chance that between nothing% and nothing% are smokers. (Round to one decimal place as needed.)

## Answers ( )

Answer:There is a 68% chance that between

17%and30%are smokers.There is a 95% chance that between

10%and37%are smokers.There is a 99.7% chance that between

4%and44%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= 40p= 0.236Compute 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

µand standard deviationσ, nearly all the data will fall within 3 standard deviations of the mean. The empirical rule can be divided into three parts:That is P (µ – σ ≤ X ≤ µ + σ) = 0.68.

That is P (µ – 2σ ≤ X ≤ µ + 2σ) = 0.95.

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%and30%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%and37%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%and44%are smokers.