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 :

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 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:

[tex]\mu = 128, \sigma = 23[/tex]

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.

[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 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.

[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 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

[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 pvalue 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 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

[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]

So

157.44mmHg

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