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

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

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:

$P(Z>9.409)=1-P(Z<9.409) \approx 0$

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: $\phi(b)=P(z<b)$

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.  

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

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

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:

$P(Z>9.409)=1-P(Z<9.409) \approx 0$

Answer 2

Very few women have​ hypertension.

The z score is given by:

$z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation$

Given that μ = 69.3 mmHg, σ = 8.8 mmHg

a) For x > 90 mmHg:

$z=\frac{90-69.3}{8.8}=2.35$

P(x > 90) = P(z > 2.35) = 1 - P(z < 2.35) = 1 - 0.9906 = 0.04%

a) For x > 90 mmHg, n = 16:

$z=\frac{90-69.3}{8.8/\sqrt{16} }=9.4$

P(x > 90) = P(z > 9.9) = 1 - P(z < 9.4) = 0%

Therefore, very few women have​ hypertension.

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