Question

the mean percent of childhood asthma prevalence in 43 cities is 2.14%. a random sample of 32 of these cities is selected. what is the probability that the mean childhood asthma prevalence for the sample is greater than 2.5%? interpret this probability. assume that $sigma = 1.30%$. the probability is . (round to four decimal places as needed.)

Explanation:

Step1: Identify the formula for z - score of the sample mean

The formula for the z - score of the sample mean $\bar{x}$ is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\mu$ is the population mean, $\bar{x}$ is the sample - mean, $\sigma$ is the population standard deviation, and $n$ is the sample size.

Step2: Identify the given values

We are given that $\mu = 2.14\%=0.0214$, $\bar{x}=2.5\% = 0.025$, $\sigma=1.30\%=0.013$, and $n = 32$.

Step3: Calculate the z - score

Substitute the values into the z - score formula:
\[

$$\begin{align*} z&=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\\ &=\frac{0.025 - 0.0214}{\frac{0.013}{\sqrt{32}}}\\ &=\frac{0.0036}{\frac{0.013}{\sqrt{32}}}\\ &=\frac{0.0036}{\frac{0.013}{5.656854}}\\ &=\frac{0.0036}{0.0023}\\ &\approx1.57 \end{align*}$$

\]

Step4: Find the probability

We want to find $P(\bar{X}>0.025)$, which is equivalent to $P(Z > 1.57)$ using the standard normal distribution. Since the total area under the standard - normal curve is 1, $P(Z>1.57)=1 - P(Z\leq1.57)$.
From the standard normal table, $P(Z\leq1.57) = 0.9418$. So, $P(Z>1.57)=1 - 0.9418=0.0582$.

Answer:

$0.0582$