An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2500 individuals, what i

Question

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?

in progress 0
Athena 3 hours 2021-09-15T09:35:09+00:00 1 Answer 0

Answers ( )

    0
    2021-09-15T09:36:39+00:00

    Answer:

    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

    So

    \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

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )