Introduction to Statistics 5.4

Question

The mean percent of childhood asthma prevalence in 43 cities is 2.15​%. A random sample of 34 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 ​standard deviation is 1.30%.

The probability is?

Answer 1

The mean percent of childhood asthma prevalence in 43 cities is 2.15​%. A random sample of 34 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 ​standard deviation is 1.30%.

The probability is?

Population mean (mu) = 2,15%

Population Standard Deviation (sigma) = 1.30%

Calculate Z-score for 2.5%:

Z(2.5) = (2.5 - 2.15) / 1.30 = +0.2692 (to 4 decimal places)

Approach:

To get asthma prevalence for the sample > 2.5%, we have to find the area under the Standard Normal Curve to the right of Z = +0.2692.

Cumulative left area under the Standard Normal Curve for Z = +0.2692 is approximately 0.6061. Hence, the desired probability = 1.0000 - 0.6061 = 0.3939.

Answer: Probability that the mean childhood asthma prevalence for the sample is greater than ​2.5% is: 0.3939 or 39.39%