The mean percent of childhood asthma prevalence in 43 cities is 2.25
​%. 
A random sample of 30
 
of these cities is selected. What is the probability that the mean childhood asthma prevalence for the sample is greater than 2.7
​%? 
Interpret this probability. Assume that sigma
equals1.39
​%. 
Question content area bottom
Part 1
The probability is enter your response here
.
​(Round to four decimal places as​ needed.)

To find the probability that the mean childhood asthma prevalence for a random sample of 30 cities is greater than 2.7%, we can use the **Central Limit Theorem** (CLT). Here's how we can solve this step-by-step:

### Given information:
1. **Population mean (\(\mu\))** = 2.25%
2. **Population standard deviation (\(\sigma\))** = 1.39%
3. **Sample size (n)** = 30
4. **Sample mean threshold** = 2.7%

We want to find \( P(\bar{X} > 2.7) \), where \( \bar{X} \) is the mean prevalence of asthma in the sample.

### Step 1: Calculate the Standard Error of the Mean (SEM)
The standard error of the mean (SEM) is given by:
\[
\text{SEM} = \frac{\sigma}{\sqrt{n}}
\]

Substitute the values:
\[
\text{SEM} = \frac{1.39}{\sqrt{30}} \approx 0.2538
\]

### Step 2: Convert the Problem to a Standard Normal Distribution
Now, we convert the threshold (2.7%) to a z-score using the formula:
\[
z = \frac{\bar{X} - \mu}{\text{SEM}}
\]

Substitute the values:
\[
z = \frac{2.7 - 2.25}{0.2538} \approx 1.7734
\]

### Step 3: Find the Probability
Now we look up the probability corresponding to \( z = 1.7734 \) in the standard normal distribution table.

The area to the left of \( z = 1.7734 \) is approximately 0.9616, which represents the probability that the sample mean is **less than 2.7%**. To find the probability that the sample mean is greater than 2.7%, we subtract this value from 1:
\[
P(\bar{X} > 2.7) = 1 - 0.9616 = 0.0384
\]

### Conclusion
The probability that the mean childhood asthma prevalence for the sample is greater than 2.7% is approximately **0.0384**.

### Interpretation
This probability of 0.0384 (or 3.84%) indicates that there is a low likelihood of observing a sample mean of 2.7% or higher if the true population mean is 2.25%.

---

Would you like more details on this calculation, or have any questions?

#### Related Questions:
1. How does the sample size affect the standard error of the mean?
2. What is the probability that the sample mean is between 2.0% and 2.5%?
3. How would the probability change if the population standard deviation was different?
4. Can the Central Limit Theorem apply to non-normal distributions with small sample sizes?
5. Why do we use the z-score in probability calculations?

#### Tip:
Remember, the Central Limit Theorem allows you to approximate the distribution of the sample mean as normal if the sample size is large enough, typically \( n > 30 \) is a good rule of thumb.

The mean percent of childhood asthma prevalence in 43 cities is 2.25%. A random sample of 30 of these cities is selected. What is the probability that the mean childhood asthma prevalence for the sample is greater than 2.7%? Interpret this probability. Assume that sigma equals 1.39%.

Explore the probability calculation for a sample mean of childhood asthma prevalence exceeding 2.7% in a group of 30 cities, given a population mean of 2.25% and standard deviation of 1.39%. This problem utilizes the Central Limit Theorem and involves calculating the standard error and Z-score.

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