Question

A person’s blood pressure is monitored by taking 3 readings daily. The probability distribution of his reading had a mean of 132 and a standard deviation of 5. Each observation behaves as a random sample. Find the mean of the sampling distribution of the sample mean for the three observations each day.

Answer 1

The mean of the sampling distribution of the sample mean for the three observations each day is 132.

Explanation:

We are given the following information in the question:

Mean, μ = 132

Standard Deviation, σ = 5

Each observation behaves as a random sample.

a) Mean of the sampling distribution

`\bar{x} = \mu = 132`

b) Standard deviation of the sampling distribution

Sample size,n = 3

`s = (\sigma)/(√(n)) = (5)/(√(3)) = 2.887`

Thus, the mean of the sampling distribution of the sample mean for the three observations each day is 132.