Math Problem Statement

A diastolic blood pressure level above 90 mm Hg is considered to be hypertension. What percentage of females have hypertension? Suppose diastolic blood pressure levels in females are normally distributed with a mean of 69.6 mm Hg and a standard deviation of 11.8 mm Hg. If we randomly collect samples of females with 25 in each sample, what percentage of those samples have a mean above 90 mm Hg?

Solution

To solve this problem, we need to apply the concepts of the normal distribution, as the diastolic blood pressure levels of females are normally distributed with a given mean and standard deviation.

Part (a)

The problem states:

  • The mean diastolic blood pressure is 69.6 mm Hg.
  • The standard deviation is 11.8 mm Hg.
  • Hypertension is defined as a diastolic blood pressure above 90 mm Hg.

We are asked to find the percentage of females whose diastolic blood pressure is above 90 mm Hg. This can be done by finding the corresponding z-score for 90 mm Hg and then using the standard normal distribution table.

The z-score formula is: z=xμσz = \frac{x - \mu}{\sigma} Where:

  • x=90x = 90 mm Hg (the threshold for hypertension),
  • μ=69.6\mu = 69.6 mm Hg (the mean),
  • σ=11.8\sigma = 11.8 mm Hg (the standard deviation).

Part (b)

In this case, we are asked to randomly collect samples of 25 females and determine what percentage of these samples will have a mean diastolic blood pressure above 90 mm Hg. Here, we will use the sampling distribution of the sample mean.

For the sampling distribution, the standard deviation of the sample mean (also called the standard error) is: σxˉ=σn\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} Where n=25n = 25 is the sample size.

After calculating the standard error, we can find the z-score for a sample mean of 90 mm Hg and proceed in the same way as part (a).

Let me calculate both (a) and (b).### Results:

  • Part (a): The percentage of females with diastolic blood pressure above 90 mm Hg is approximately 4.19%.
  • Part (b): The percentage of samples (with 25 females each) having a mean diastolic blood pressure above 90 mm Hg is 0%. This means it's extremely unlikely for a sample mean to exceed 90 mm Hg given the distribution.

Would you like more details on the calculations or any clarifications?

Here are five related questions to expand on this topic:

  1. What is the interpretation of a z-score in a normal distribution?
  2. How would the percentage change if the standard deviation were higher?
  3. How does sample size affect the standard error in a sampling distribution?
  4. What are the implications of a very small probability in statistical inference?
  5. How can the Central Limit Theorem help in understanding sample means?

Tip: The larger the sample size, the smaller the standard error, which makes extreme sample means (far from the population mean) less likely.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Z-Scores
Sampling Distribution
Standard Error

Formulas

z = (x - μ) / σ
σ_x̄ = σ / sqrt(n)

Theorems

Central Limit Theorem

Suitable Grade Level

College Level Statistics