Question

The mean systolic blood pressure for people in the United States is reported to be 122 millimeters of mercury (mmHg) with a standard deviation of 22.8 millimeters of mercury. The wellness department of a large corporation is investigating if the mean systolic blood pressure is different from the national mean. A random sample of 200 employees in a company were selected and found to have an average systolic blood pressure of 124.7 mmHg.

a) What is the probability a random employees blood pressure is higher than 135 mmHg.

b) What is the probability that 200 randomly selected employees mean blood pressure is greater than 127 mmHg.

c) The wellness department is providing a new health program to their employees. Past studies have shown 9.9 % of their employees have high blood pressure. Find the probability that if the wellness department examines 200 randomly selected employees, less than 12 employees will have high systolic blood pressure. Do you think the new program significantly lowers the number of employees with a high blood pressure.

Answer 1

Given that

a)

The z-score for X = 135 is

The probability a random employees blood pressure is higher than 135 mmHg is

P(X > 135) = P(z > 0.57) = 1 - P(z <= 0.57) = 1 - 0.7157 = 0.2843

b)

The z-score for is

The probability that 200 randomly selected employees mean blood pressure is greater than 127 mmHg is

c)

let X is a random variable shows the number of employees have high blood pressure. Here X has binomial distribution with parameters as follows:

n=200 and p=0.099

The probability that if the wellness department examines 200 randomly selected employees, less than 12 employees will have high systolic blood pressure is

Answer: 0.0187

Excel function used to find the probability is "=BINOMDIST(11,200,0.099,TRUE)".

Since above probability is less than 0.05 so it seems that new program significantly lowers the number of employees with a high blood pressure.