Five males with as X-linked genetic disorder have one child each. The random var

sjeikdom0

sjeikdom0

Answered question

2021-10-23

Five males with as 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.
XP(X)00.03110.15620.31330.31340.15650.031

Answer & Explanation

jlo2niT

jlo2niT

Skilled2021-10-24Added 96 answers

Step 1 
Prerequisites for a probability distribution:
p(x)>0 
p(x)=1 
Step 2 
Given that X-linked genetic disorder is inherited by x% of offspring.
Since the sum of the probabilities for a random variable x is equal to one and all the probabilities are greater than zero. It is a probability distribution as a result.
p(x)>0 
xp(x)=0.031+0.156+0.313+0.313+0.156+0.031=1 
Step 3 
The mean is computed as follows:
A probability distribution's mean is 2.5. 
E(x)=xx×p(x) 
=[0(0.031)+1(0.156)+2(0.313)+3(0.313)+4(0.156)+5(0.031)] 
=0+0.156+0.626+0.939+0.624+0.155 
=2.5 
Step 4 
The following is how the variance can be found:
V(x)=E(x2)(E(x))2 
Step 5 
Thus,
E(x2)=xx2p(x) 
=0+12(0.156)+22(0.313)+32(0.313)+42(0.156)+52(0.031) 
=0.156+1.252+2.817+2.496+0.775 
=7.496 
Step 6 
V(x)=E(x2)(E(x))2 
=7.496(2.5)2 
=7.4966.25 
=1.246

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