Answer to Question #161031 in Statistics and Probability for Manoshithira

Question

Our subjects are 35-44-year-old males whose blood pressures are normally distributed with mean 80 and standard deviation 12 A borderline hypertensive is defined as a person whose diastolic blood pressure is between 90 and 95 mm Hg inclusive; what proportion of subjects are borderline hypertensive? A hypertensive is a person whose diastolic blood pressure is above 95 mm Hg; what proportion of subjects are hypertensive?

Answer 1

Solution:

1) To find the number of males from 35-44 years, that are bordeline hypertensive, we need to muptiply the number of tested males to the probability of them being in the borderline hypertensive $(90\le x\le95)$:

$n*p(x)$

The proportion of such people is the number of such people, divided by total number of people:

$\frac{n*p(x)}{n} = p(x)$

So, the only thing we need to find is the probability:

$P(90\le\xi\le95)$ , $\xi$ - normal distribution with

$E(\xi) = 80, \sigma(\xi)=12$

To calculate the probabilities, we will use the Standart Normal table.

To do this, we need to perform the following transformation:

$P(x_1\le\xi\le x_2) = P(\frac{x_1-a}{\sigma}\le\frac{\xi-a}{\sigma}\le\frac{x_2-a}{\sigma}) =\Phi(\frac{x_2-a}{\sigma})-\Phi(\frac{x_1-a}{\sigma})$

$a=E(\xi)=60,$

In our case:

$P\{90\le\xi\le95\} = P\{\frac{90-80}{12}\le\frac{\xi-80}{12}\le\frac{95-80}{12}\} =$

$=\Phi(\frac{15}{12}) - \Phi(\frac{10}{12}) = \Phi(1.25)-\Phi(0.83) = 0.89435 - 0.79673 = 0.09762$

2) Following the same logic as in 1) we need to calculate the following probability:

$P(95\le\xi\le+\infin)$

We will use the Standart Normal table again, but we will use one trick. Subtracting from or dividing from the infinity will result in infinity. And the value of $\Phi(+\infin) = 1$ :

$P\{95\le\xi\le+\infin\} = P\{\frac{95-80}{12}\le\frac{\xi-80}{12}\le+\infin\}=\Phi(+\infin) - \Phi(1.25)=$

$=1-0.89435=0.10565$

Answer:

1) $0.09762$

2) $0.10565$