Mathematics, 12.03.2020 21:23 noeltan12031

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2500 individuals, what is the approximate distribution of the number who carry this gene?

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The approximate distribution of the number who carry this gene is approximately normal with mean $\mu = 5$ and standard deviation $\sigma = 2.23$

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they carry the defective gene that causes inherited colon cancer, or they do not. The probability of a person carrying this gene is independent from other people. So the binomial probability distribution is used to solve this question.

A sample of 2500 individuals is quite large, so we use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer.

This means that $p = \frac{1}{500} = 0.002$

In a sample of 2500 individuals, what is the approximate distribution of the number who carry this gene?

n = 2500

$\mu = E(X) = np = 2500*0.002 = 5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2500*0.002*0.998} = 2.23$

So the approximate distribution of the number who carry this gene is approximately normal with mean $\mu = 5$ and standard deviation $\sigma = 2.23$

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the answer to this question is 3/19

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Answer from: Quest

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