Answer:
C.
Explanation:
Hello!
Given the variable:
X: childhood asthma prevalence
With mean μ= 2.22%
and standard deviation σ= 1.39%
You have to calculate the probability of the sample average of childhood asthma prevalence in a sample of n= 30 cities is greater than 2.5%
We don't know the distribution of the variable, but remember that thanks to the central limit theorem, since the n ≥ 30, we can approximate the sampling distribution to normal:
X[bar]≈N(μ;σ²/n)
And use the standard normal distribution to calculate the asked probability:
P(X[bar]>2.5)= 1 - P(X≤2.5)
Calculate the Z value for the given X[bar] value:
![Z= (X[bar]-Mu)/((Sigma)/(√(n) ) ) = (2.5-2.22)/((1.39)/(√(30) ) )= 1.10](https://img.qammunity.org/2021/formulas/mathematics/college/gh6gjfo8yyngcna95trlzfs7mdr56afrxj.png)
Using the Z-tables you have to look for the value of
P(Z≤1.10)= 0.86433
1 - 0.86433= 0.13567
Then P(X[bar]>2.5)= 1 - P(X≤2.5)= 1 - P(Z≤1.10)= 1 - 0.86433= 0.13567
13.567% of the 30 cities will have a mean childhood asthma prevalence greater than 2.5%
I hope this helps!