Answer:
a) Mean blood pressure for people in China.
b) 38.21% probability that a person in China has blood pressure of 135 mmHg or more.
c) 71.30% probability that a person in China has blood pressure of 141 mmHg or less.
d) 8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.
e) Since Z when X = 135 is less than two standard deviations from the mean, it is not unusual for a person in China to have a blood pressure of 135 mmHg
f) 157.44mmHg
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 pvalue, we get the probability that the value of the measure is greater than X.
If X is two standard deviations from the mean or more, it is considered unusual.
In this question:
![\mu = 128, \sigma = 23](https://tex.z-dn.net/?f=%5Cmu%20%3D%20128%2C%20%5Csigma%20%3D%2023)
a.) State the random variable.
Mean blood pressure for people in China.
b.) Find the probability that a person in China has blood pressure of 135 mmHg or more.
This is 1 subtracted by the pvalue of Z when X = 135.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{135 - 128}{23}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B135%20-%20128%7D%7B23%7D)
![Z = 0.3](https://tex.z-dn.net/?f=Z%20%3D%200.3)
has a pvalue of 0.6179
1 - 0.6179 = 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 pvalue of Z when X = 141.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{141 - 128}{23}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B141%20-%20128%7D%7B23%7D)
![Z = 0.565](https://tex.z-dn.net/?f=Z%20%3D%200.565)
has a pvalue of 0.7140
71.30% 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 pvalue of Z when X = 125 subtracted by the pvalue of Z when X = 120. So
X = 125
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{125 - 128}{23}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B125%20-%20128%7D%7B23%7D)
![Z = -0.13](https://tex.z-dn.net/?f=Z%20%3D%20-0.13)
has a pvalue of 0.4483
X = 120
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{120 - 128}{23}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B120%20-%20128%7D%7B23%7D)
![Z = -0.35](https://tex.z-dn.net/?f=Z%20%3D%20-0.35)
has a pvalue of 0.3632
0.4483 - 0.3632 = 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 b), when X = 135, Z = 0.3
Since Z when X = 135 is less than two standard deviations from the mean, 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?
This is the 90th percentile, which is X when Z has a pvalue of 0.28. So X when Z = 1.28. Then
X = 120
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![1.28 = \frac{X - 128}{23}](https://tex.z-dn.net/?f=1.28%20%3D%20%5Cfrac%7BX%20-%20128%7D%7B23%7D)
![X - 128 = 1.28*23](https://tex.z-dn.net/?f=X%20-%20128%20%3D%201.28%2A23)
![X = 157.44](https://tex.z-dn.net/?f=X%20%3D%20157.44)
So
157.44mmHg