Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 128, \sigma = 23[/tex]
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:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{135 - 128}{23}[/tex]
[tex]Z = 0.3[/tex]
[tex]Z = 0.3[/tex] 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:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{141 - 128}{23}[/tex]
[tex]Z = 0.565[/tex]
[tex]Z = 0.565[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{125 - 128}{23}[/tex]
[tex]Z = -0.13[/tex]
[tex]Z = -0.13[/tex] has a p-value of 0.4483.
X = 120
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{120 - 128}{23}[/tex]
[tex]Z = -0.35[/tex]
[tex]Z = -0.35[/tex] 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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 128}{23}[/tex]
[tex]X - 128 = 1.28*23[/tex]
[tex]X = 157.44[/tex]
90% of all people in China have a blood pressure of less than 157.44 mmHg.
