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Five males with an​ x-linked genetic disorder have one child each. the random variable x is the number of children among the five who inherit the​ x-linked genetic disorder. determine whether a probability distribution is given. if a probability distribution is​ given, find its mean and standard deviation. if a probability distribution is not​ given, identify the requirements that are not satisfied.

Respuesta :

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With a known frequency f of incidence of the x-linked genetric disorder alleles, the probability p of a child from a father with the disorder can be calculated, and is constant throughout the experiment.

We further assume that the five cases are independent of each other (e.g. no common female parents, etc.)
The number of trials (cases) is known (n=5) and remains constant.
Each trial is bernoulli (i.e. either success=disorder, or failure=no disorder)

The above conditions together satisfy the requirements for a binomial distribution, where the mean is np, and the variance is npq=np(1-p).
Hence 
mean = np
standard deviation = sqrt(np(1-p))

It will be a probability distribution if:

  • The probability of a value of x is never negative.
  • The sum of the probabilities is always 1.

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A sequence (x, P(X = x)) is a probability distribution if:  

  • The sum of all values of P(X = x) equals 1, that is, considering n values [tex]\sum_{i = 1}{n} P(X_i) = 1[/tex]
  • There are no negative values of P(X = x).

A similar problem is given at https://brainly.com/question/24858659

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