Question

Diastolic blood pressure is a measure of the pressure when arteries rest between heartbeats. Suppose diastolic blood pressure levels in women are normally distributed with a mean of 69.3mm Hg and a standard deviation of 8.8mm Hg. Complete parts​ (a) and​ (b) below.
A. A diastolic blood pressure level above 90 mm Hg is considered to be hypertension. What percentage of women have​ hypertension?

B. If we randomly collect samples of women with 16 in each​ sample, what percentage of those samples have a mean above 90 mm​ Hg?

Answer 1

Answer:

a) $P(X>90)=P(Z>\frac{90-69.3}{8.8})=P(Z>2.35)=1-P(Z$

$P(X>90)=1-\phi(2.35)=1-0.9906=0.0094$

b) $P(\bar X >90)=P(Z>\frac{90-69.3}{\frac{8.8}{\sqrt{16}}}=9.409)$

And using a calculator, excel ir the normal standard table we have that:

$P(Z>9.409)=1-P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  The letter $\phi(b)$ is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: $\phi(b)=P(z$

Let X the random variable that represent the blood pressure for women, and for this case we know the distribution for X is given by:

$X \sim N(69.3,8.8)$  

Where $\mu=69.3$ and $\sigma=8.8$

Part a

We are interested on this probability

$P(X>90)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>90)=P(\frac{X-\mu}{\sigma}>\frac{90-\mu}{\sigma})$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$P(X>90)=P(Z>\frac{90-69.3}{8.8})=P(Z>2.35)=1-P(Z$

$P(X>90)=1-\phi(2.35)=1-0.9906=0.0094$

Part b

The distribution for the sample mean $\bar X$ on this case is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we want to calculate the following probability:

$P(\bar X >90)=P(Z>\frac{90-69.3}{\frac{8.8}{\sqrt{16}}}=9.409)$

And using a calculator, excel ir the normal standard table we have that:

$P(Z>9.409)=1-P(Z$