Mathematics, 26.06.2020 15:01 Bryoh

"National survey released in 2003 showed that among U. S. adults ages 70 and older, 21.1% had been told by a doctor that they had some form of cancer. Use this result as valid for the population of U. S. adults, 70 yrs. old and older. What is the probability that among 40 adults (ages 70 and older) chosen at random more than 22 percent will have been told by their doctor that they have cancer

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44.43% probability that among 40 adults (ages 70 and older) chosen at random more than 22 percent will have been told by their doctor that they have cancer

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question, we have that:

p = 0.211, n = 40

$\mu = 0.211, s = \sqrt{\frac{0.211*0.789}{40}} = 0.0645$

What is the probability that among 40 adults (ages 70 and older) chosen at random more than 22 percent will have been told by their doctor that they have cancer

This is 1 subtracted by the pvalue of Z when X = 22. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.22 - 0.211}{0.0645}$

Z = 0.14

Z = 0.14 has a pvalue of 0.5557

1 - 0.5557 = 0.4443

44.43% probability that among 40 adults (ages 70 and older) chosen at random more than 22 percent will have been told by their doctor that they have cancer

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