The mean percent of childhood asthma prevalence in 43 cities is 2.22% . 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 2.5%? Interpret this probability. Assume that sigma equals 1.39%. nterpret 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. What % of samples of 43 cities will have a mean childhood asthma prevalence greater than 2.5%? (B. What % of samples of 30 cities will have a mean childhood asthma prevalence greater than 2.22%? (C. What% of samples of 30 cities will have a mean childhood asthma prevalence greater than 2.5?

C.

Step-by-step 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:

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!

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer

idk

step-by-step explanation: