Question

According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with a standard deviation of 23 mmHg (Kuulasmaa, Hense & Tolonen, 1998). Assume that blood pressure is normally distributed. a.) State the random variable. b.) Find the probability that a person in China has blood pressure of 135 mmHg or more.

Answer 1

Answer:

a) Let X the random variable that represent the blood pressure for people of a population, and for this case we know the distribution for X is given by:

$X \sim N(128,23)$  

Where $\mu=128$ and $\sigma=23$

b) $P(X\geq 135)=P(\frac{X-\mu}{\sigma}\geq \frac{135-\mu}{\sigma})=P(Z\geq \frac{135-128}{23})=P(Z\geq 0.304)$

And we can find this probability using the complement rule:

$P(Z\geq 0.304)=1-P(Z$

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

$P(Z\geq 0.304)=1-P(Z$

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".  

Part a

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

$X \sim N(128,23)$  

Where $\mu=128$ and $\sigma=23$

Part b

We are interested on this probability

$P(X\geq 135)$

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:

$P(X\geq 135)=P(\frac{X-\mu}{\sigma}\geq \frac{135-\mu}{\sigma})=P(Z\geq \frac{135-128}{23})=P(Z\geq 0.304)$

And we can find this probability using the complement rule:

$P(Z\geq 0.304)=1-P(Z$

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

$P(Z\geq 0.304)=1-P(Z$