Question

A​ person's blood pressure is monitored by taking 6 readings daily. The probability distribution of his reading had a mean of 130 and a standard deviation of 8. a. Each observation behaves as a random sample. Find the mean and the standard deviation of the sampling distribution of the sample mean for the six observations each day. b. Suppose that the probability distribution of his blood pressure reading is normal. What is the shape of the sampling​ distribution? c. Refer to​ (b). Find the probability that the sample mean exceeds 140.

Answer 1

a) For this case we have the following info:

`\mu = 130 , sigma= 8`

And we select a sample size of 6. Assuming that the data can be approximated to a normal distribution the sample mean have the following parameters:

`\mu_(\bar X) = 130`

`\sigma_(\bar X) = (8)/(√(6))=3.266`

b) For this case assuming a nomal distribution then the sample mean have the following distribution:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

c)
`P(\bar X>140)`

And we can use the z score formula given by:

`z = (140-130)/((8)/(√(6)))= 3.062`

And using the complement rule we have:

`P(Z>3.062) =1-P(Z<3.062) = 1-0.9989= 0.0011`

Explanation:

Part a

For this case we have the following info:

`\mu = 130 , sigma= 8`

And we select a sample size of 6. Assuming that the data can be approximated to a normal distribution the sample mean have the following parameters:

`\mu_(\bar X) = 130`

`\sigma_(\bar X) = (8)/(√(6))=3.266`

Part b

For this case assuming a nomal distribution then the sample mean have the following distribution:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

Part c

We want this probability:

`P(\bar X>140)`

And we can use the z score formula given by:

`z = (140-130)/((8)/(√(6)))= 3.062`

And using the complement rule we have:

`P(Z>3.062) =1-P(Z<3.062) = 1-0.9989= 0.0011`