Question

the mean percent of childhood asthma prevalence in 43 cities is 2.45%. 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 2.8%? interpret this probability. assume that \\( \sigma=1.23 \\% \\).
the probability is \\( \square \\).
(round to four decimal places as needed.)
interpret this probability. select the correct choice below and fill in the answer box to complete your choice.
(round to two decimal places as needed.)
a. about \\( \square \\% \\) of samples of 43 cities will have a mean childhood asthma prevalence greater than 2.8%.
b. about \\( \square \\% \\) of samples of 32 cities will have a mean childhood asthma prevalence greater than 2.45%.
c. about \\( \square \\% \\) of samples of 32 cities will have a mean childhood asthma prevalence greater than 2.8%.

Explanation:

Step1: Calculate the standard error

The standard error $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 1.23\%$ and $n = 32$.
$\sigma_{\bar{x}}=\frac{1.23}{\sqrt{32}}\approx\frac{1.23}{5.65685}\approx0.2174$

Step2: Calculate the z - score

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, where $\bar{x}=2.8\%$, $\mu = 2.45\%$ and $\sigma_{\bar{x}}\approx0.2174$.
$z=\frac{2.8 - 2.45}{0.2174}=\frac{0.35}{0.2174}\approx1.61$

Step3: Find the probability

We want to find $P(\bar{X}>2.8)=P(Z > 1.61)$.
Since $P(Z>z)=1 - P(Z\leq z)$, and from the standard normal table $P(Z\leq1.61)=0.9463$.
$P(Z > 1.61)=1 - 0.9463=0.0537$

Answer:

The probability is $0.0537$.
For the interpretation:
We have a sample of size $n = 32$. So we look at the option related to sample size $32$.
The probability that the sample mean is greater than $2.8\%$ is $0.0537$. To convert this to a percentage, we multiply by $100$.
$0.0537\times100 = 5.37\%$
So, C. About $5.37\%$ of samples of 32 cities will have a mean childhood asthma prevalence greater than $2.8\%$.