The mean percent of childhood asthma prevalence in 43 cities is 2.14​%. A random sample of 31 of these cities is selected.

Question

The mean percent of childhood asthma prevalence in 43 cities is 2.14​%. A random sample of 31 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 ​standard deviation is 1.20%. The probability is?

Answer 1

The sampling distribution of the sample mean will be a Normal distribution with a mean of 2.14 and a standard deviation of 1.20 divided by the square root of 31 (the sample size).

Define a random variable X as the mean percent of childhood asthma in a sample of 31 cities. This random variable has a Normal distribution with a mean of 2.14 and a standard deviation of 1.20/sqrt(31) = 0.2155

We need to find the probability that X > 2.6. We can find the z-score by subtracting the mean and dividing by the standard deviation. The z-score is (2.6-2.14) / 0.2155 = 2.135. We now want to find the probability that Z (which is now a standard Normal random variable) is greater than 2.135. This can be interpreted as asking what is the probability that this z-score is greater than 2.135 standard deviations above the mean.

A table or calculator can be used to look up the probability. P(Z > 2.135) = 0.0164