I need help with my probabitly and statistics

Question

Need help The mean percent of childhood asthma in 43 cities is 2.25 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 2.6 interpret this probability assume that mean is 1.20

Answer 1

This sounds like a simple hypothesis test question, using a z-score. Reading your question, I am going to assume the last sentence is supposed to say "σ=1.20", not 'mean', as that wouldn't make sense. In this case, let's define a few variables:

Sample of Cities: n =32

Standard Deviation Population: σ_x=1.20

Mean of Population: μ_x=2.25

Assumed Sample Mean μ_n = 2.6

First, we would want to calculate the standard deviation of the sample: σ_n = (σ_x / Sqrt(n)) = 1.20/sqrt(32) = 0.2121.

Knowing that Z_n = (μ_n-μ_x)/σ_n , we want to calculate P[μ_n > 2.6] = P [ Z_n > Z_2.6 ] = 1 - P [ Z_n < Z_2.6 ]

This comes out to be P [ Z_n > Z_2.6 ] = 1 - P [ z_n < (μ_2.6-μ_x)/σ_2.6 ] = 1 - P [ Z_n < 1.64 ] = 1 - 0.9495

P [ Z_n > Z_2.6 ] = .0505 = 5.05%