Question

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

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	0.0350.035
		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.

Click to select and enter your answer(s).

Answer 1

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