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.
c.) Find the probability that a person in China has blood pressure of 141 mmHg or less.
d.) Find the probability that a person in China has blood pressure between 120 and 125 mmHg.
e.) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not?
f.) What blood pressure do 90% of all people in China have less than?

Answer 1

Answer:

a) Mean blood pressure for people in China, which has mean 128 and standard deviation 23.

b) 0.3821 = 38.21% probability that a person in China has blood pressure of 135 mmHg or more.

c) 0.714 = 71.4% probability that a person in China has blood pressure of 141 mmHg or less.

d) 0.0851 = 8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.

e) Since |Z| = 0.3 < 2, it is not unusual for a person in China to have a blood pressure of 135 mmHg.

f) 90% of all people in China have a blood pressure of less than 157.44 mmHg.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The mean blood pressure for people in China is 128 mmHg with a standard deviation of 23 mmHg

This means that $\mu = 128, \sigma = 23$

a.) State the random variable.

Mean blood pressure for people in China, which has mean 128 and standard deviation 23.

b.) Find the probability that a person in China has blood pressure of 135 mmHg or more.

This is 1 subtracted by the p-value of Z when X = 135, so:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{135 - 128}{23}$

$Z = 0.3$

$Z = 0.3$ has a p-value of 0.6179.

1 - 0.6179 = 0.3821

0.3821 = 38.21% probability that a person in China has blood pressure of 135 mmHg or more.

c.) Find the probability that a person in China has blood pressure of 141 mmHg or less.

This is the p-value of Z when X = 141, so:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{141 - 128}{23}$

$Z = 0.565$

$Z = 0.565$ has a p-value of 0.7140.

0.714 = 71.4% probability that a person in China has blood pressure of 141 mmHg or less.

d.) Find the probability that a person in China has blood pressure between 120 and 125 mmHg.

This is the p-value of Z when X = 125 subtracted by the p-value of Z when X = 120, so:

X = 125

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{125 - 128}{23}$

$Z = -0.13$

$Z = -0.13$ has a p-value of 0.4483.

X = 120

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{120 - 128}{23}$

$Z = -0.35$

$Z = -0.35$ has a p-value of 0.3632.

0.4483 - 0.3632 = 0.0851

0.0851 = 8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.

e.) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not?

From item b, when X = 135, Z = 0.3.

Since |Z| = 0.3 < 2, it is not unusual for a person in China to have a blood pressure of 135 mmHg.

f.) What blood pressure do 90% of all people in China have less than?

The 90th percentile, which is X when Z has a p-value of 0.9, so X when Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 128}{23}$

$X - 128 = 1.28*23$

$X = 157.44$

90% of all people in China have a blood pressure of less than 157.44 mmHg.