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According to the Centers for Disease Control and Prevention (CDC) , 47% of adults in the United States have hypertension. In a random sample of 500 U.S adults, find the probability that sample the proportion of adults with hypertension is greater than 0.5.

Sagot :

MrRoyalidn

The probability that sample the proportion of adults with hypertension is greater than 0.5 is 0.090

How to determine the probability?

The given parameters are:

p = 47%

Sample size, n =500

The standard deviation is calculated as:

[tex]\sigma = \sqrt{np(1 - p)[/tex]

So, we have:

[tex]\sigma = \sqrt{500 * 47\%(1 - 47\%)[/tex]

Evaluate

σ = 11.16

The number of adults with hypertension is calculated as:

x = 500 * 0.5 = 250

And the mean is:

μ = 500 * 0.47 = 235

Start by calculating the z-score

[tex]z = \frac{x - \mu}{\sigma}[/tex]

This gives

[tex]z = \frac{250 - 235}{11.16}[/tex]

Evaluate

z = 1.34

The probability is then calculated as:

P(z >1.34) = ??

From z table of probabilities, we have:

P(z >1.34) =  0.090

Hence, the probability that sample the proportion of adults with hypertension is greater than 0.5 is 0.090

Read more about probability at:

https://brainly.com/question/251701

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