Question

A​ person's blood pressure is monitored by taking 3 readings daily. The probability distribution of his reading had a mean of 135 and a standard deviation of 3. 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 three 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

Answer:

a) The sampling distirbution has a mean of 135 and a standard deviation of 1.732.

b) The sampling distribution is also bell shaped, centered in the population mean. It has less spread that the population distribution because the standard deviation is smaller.

c) P(X>140)=0.002

Step-by-step explanation:

a) We are taking samples of size n=3, as we are taking 3 readings daily.

Each reading has a mean of 135 and a standard deviation of 3.

The sampling distribution depends on the population parameters and the sample size.

The mean of the sampling distribution is equal to the population mean. In this case, the mean of the sampling distribution is 135.

The standard deviation depends on the standard deviation of the population and the sample size, and is calculated as:

$\sigma_M=\dfrac{\sigma}{\sqrt{N}}=\dfrac{3}{\sqrt{3}}=1.732$

b) The sampling distribution is also bell shaped, centered in the population mean. It has less spread that the population distribution because the standard deviation is smaller.

c) The probability that the sample mean exceeds 140 can be calculated using the standard normal distribution.

First, we have to calculate the z-score for X=140. We use for this the mean and standard deviation of the sampling distribution.

$z=\dfrac{X-\mu_M}{\sigma_M}=\dfrac{140-135}{1.732}=\dfrac{5}{1.732}=2.887$

With this z-score, we can look up the probabilty in the standard normal distribution

$P(X>140)=P(z>2.887)=0.002$