Question

# 1. Use the given values of n and p to find the minimum usual value muμminus−2sigmaσ...

1. Use the given values of n and p to find the minimum usual value

muμminus−2sigmaσ

and the maximum usual value

muμplus+2sigmaσ.

Round to the nearest hundredth unless otherwise noted.

nequals=10141014​;

pequals=0.860.86

 x ​P(x) 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. 0 0.0350.035 1 0.1510.151 2 0.3140.314 3 0.3140.314 4 0.1510.151 5 0.0350.035

Does the table show a probability​ distribution? Select all that apply.

A.

​Yes, the table shows a probability distribution.

B.

​No, the random variable x is categorical instead of numerical.

C.

​No, the random variable​ x's number values are not associated with probabilities.

D.

​No, the sum of all the probabilities is not equal to 1.

E.

​No, not every probability is between 0 and 1 inclusive.

Find the mean of the random variable x. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.

muμequals=nothing

​child(ren) ​(Round to one decimal place as​ needed.)

B.

The table does not show a probability distribution.

Find the standard deviation of the random variable x. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.

sigmaσequals=nothing

​child(ren) ​(Round to one decimal place as​ needed.)

B.

The table does not show a probability distribution.

A. ​Yes, the table shows a probability distribution.

 x P(X=x) xP(x) x2P(x) 0 0.035 0.00000 0.00000 1 0.151 0.15100 0.15100 2 0.314 0.62800 1.25600 3 0.314 0.94200 2.82600 4 0.151 0.60400 2.41600 5 0.035 0.17500 0.87500 total 2.5000 7.5240 E(x) =μ= ΣxP(x) = 2.5000 E(x2) = Σx2P(x) = 7.5240 Var(x)=σ2 = E(x2)-(E(x))2= 1.274000 std deviation= σ= √σ2 = 1.12872

from above:

mean  μ=2.5

σ= 1.11

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