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Veseljchak [2.6K]
3 years ago
5

A woman is randomly selected from the 18–24 age group. For women of this group, systolic blood pressures (in mm Hg) are normally

distributed with a mean of 114.8 and a standard deviation of 13.1. What is the probability this woman has a systolic blood pressure greater than 140?
Mathematics
1 answer:
Naddik [55]3 years ago
8 0

Answer:

X \sim N(114.8,13.1)  

Where \mu=114.8 and \sigma=13.1

We are interested on this probability

P(X>140)

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:

And we can find this probability using the complement rule:

P(z>1.924)=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".  

Solution to the problem

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

X \sim N(114.8,13.1)  

Where \mu=114.8 and \sigma=13.1

We are interested on this probability

P(X>140)

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>140)=P(\frac{X-\mu}{\sigma}>\frac{140-\mu}{\sigma})=P(Z>\frac{140- 1114.8}{2.6})=P(z>1.924)And we can find this probability using the complement rule:

P(z>1.924)=1-P(z

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