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.   x ​P(x)       0 0.031       1 0.146       2 0.323       3 0.323       4 0.146       5 0.031

Q: Five males with an X-linked genetic disorder have one child each. The random variable x is the…

A: Given PMF of X. Now will using this pmf to find required information

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Q: X P(x) 0 0.034 1 0.152 Five males with an X-linked genetic disorder have one child each. The random…

A: μ=Mean=E(X) E(x)=∑xP(X=x)

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Q: Five males with an​ X-linked genetic disorder have one child each. The random variable x is the…

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Q: Five males with an​ X-linked genetic disorder have one child each. The random variable x is the…

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Q: Five males with an X-linked genetic disorder have one child each. The random variable x is the…

A: Given PMF of X. Now will using this pmf to find required information

Q: Five males with an​ X-linked genetic disorder have one child each. The random variable x is the…

A: From the provided information, The probability distribution table can be constructed as: X 0…

Q: Value x of X P(x = x) -4 0.19 0.14 4 0.28 6.

A:

Q: Five males with an X-linked genetic disorder have one child each. The random variable x is the…

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Q: Five males with an X-linked genetic disorder have one child each. The random variable x is the…

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Q: Five males with an X-linked genetic disorder have one child each. The andom variable x is the number…

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Q: Five males with an X-linked genetic disorder have one child each. The random variable x is the…

A: According to the given information, we have The probability distribution table is given.

Q: Five males with an X-linked genetic disorder have one child each. The random variable x is the…

A: Mean(μ)=E(X)=∑X.P(X)E(X2)=∑X2.P(X)Variance(σ2)=E(X2)-(E(X))2Standard deviation(σ)=E(X2)-(E(X))2

Q: Groups of adults are randomly selected and arranged in groups of three. The random variable x is the…

A: X 0 1 2 3 P(X) 0.353 0.446 0.171 0.027 Yes , the  table shows a probability distribution.

Q: A sociologist randomly selects single adults for different groups of​ three, and the random variable…

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Q: hildren among the five who inherit the​ X-linked genetic disorder. Determine whether a probability…

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Q: P(x) Five males with an X-linked genetic disorder have one child each. The random variable x is the…

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Q: P(x) Five males with an X-linked genetic disorder have one child each. The random variable x is the…

A:

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