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

Question

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.)

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4 months 2022-01-31T08:03:41+00:00 1 Answer 0 views 0

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:

$$\mu_{\hat p}=p\\$$

The standard deviation of the sampling distribution of sample proportion is:

$$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$$

Given:

n = 40

p = 0.236

Compute the mean and standard deviation of this sampling distribution of sample proportion as follows:

$$\mu_{\hat p}=p=0.236$$

$$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.236(1-0.236)}{40}}=0.067$$

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:

$$P (\mu_{\hat p} – \sigma_{\hat p} \leq \hat p \leq \mu_{\hat p} + \sigma_{\hat p}) = 0.68\\P(0.236-0.067\leq \hat p \leq 0.236+0.067)=0.68\\P(0.169\leq \hat p \leq0.303)=0.68\\P(0.17\leq \hat p \leq0.30)=0.68$$

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:

$$P (\mu_{\hat p} – 2\sigma_{\hat p} \leq \hat p \leq \mu_{\hat p} + 2\sigma_{\hat p}) = 0.95\\P(0.236-2\times 0.067\leq \hat p \leq 0.236+2\times0.067)=0.95\\P(0.102\leq \hat p \leq 0.370)=0.95\\P(0.10\leq \hat p \leq0.37)=0.95$$

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:

$$P (\mu_{\hat p} – 3\sigma_{\hat p} \leq \hat p \leq \mu_{\hat p} + 3\sigma_{\hat p}) = 0.997\\P(0.236-3\times 0.067\leq \hat p \leq 0.236+3\times0.067)=0.997\\P(0.035\leq \hat p \leq 0.437)=0.997\\P(0.04\leq \hat p \leq0.44)=0.997$$

Thus, there is a​ 99.7% chance that between 4​% and 44​% are​ smokers.