You are performing a cohort study to look at risk factors for breast cancer in postmenopausal women. If the probability of developing breast cancer among smokers is .05 for the study duration, then if you sample 100 smokers, how many do you expect to develop the disease?

Respuesta :

Answer:

[tex] E(X) = np = 100 *0.05 = 5[/tex]

So we expect about 5 people with the disease in the random sample.

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]  

Where (nCx) means combinatory and it's given by this formula:  

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

Solution to the problem

For this case we can assume that our random variable who represent the number of students that develop the disease is distributed on this way [tex] X \sim Bin(n=100, p =0.05)[/tex]

And from definition the expected value is given by:

[tex] E(X) = np = 100 *0.05 = 5[/tex]

And the variance is given by:

[tex] Var (X) = np(1-p) = 100*0.05 *(1-0.05) =4.75[/tex]

And the deviation is given by:

[tex] Sd(X) = \sqrt{4.75}= 2.179[/tex]

So we expect about 5 people with the disease in the random sample.

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