Respuesta :
Answer:
The approximate distribution of the number who carry this gene is approximately normal with mean [tex]\mu = 5[/tex] and standard deviation [tex]\sigma = 2.23[/tex]
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:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer.
This means that [tex]p = \frac{1}{500} = 0.002[/tex]
In a sample of 2500 individuals, what is the approximate distribution of the number who carry this gene?
[tex]n = 2500[/tex]
So
[tex]\mu = E(X) = np = 2500*0.002 = 5[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2500*0.002*0.998} = 2.23[/tex]
So the approximate distribution of the number who carry this gene is approximately normal with mean [tex]\mu = 5[/tex] and standard deviation [tex]\sigma = 2.23[/tex]
