Five males with a particular genetic disorder have one child each. The random variable x is the number of children among the five who inherit the genetic disorder. Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation.
x 
0 
1 
2 
3 
4 
5 

P(x) 
0.00020.0002 
0.00530.0053 
0.04500.0450 
0.19190.1919 
0.40890.4089 
0.34870.3487 
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
(Round to one decimal place as needed.)
B.
The table is not 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
(Round to one decimal place as needed.)
B.
The table is not a probability distribution.
Click to select and enter your answer(s).
To be a valid probability distribution, the sum of all probability must be equal to one.
Sum = 0.0002 + 0.0053 + 0.045 + 0.1919 + 0.4089 + 0.3487 = 1
µ = E(X) = 0 * 0.0002 + 1 * 0.0053 + 2 * 0.045 + 3 * 0.1919 + 4 * 0.4089 + 5 * 0.3487 = 4.0501 = 4.1 (ans)
OptionA) µ = 4.1
E(X^{2}) = 0^{2} * 0.0002 + 1^{2} * 0.0053 + 2^{2} * 0.045 + 3^{2} * 0.1919 + 4^{2} * 0.4089 + 5^{2} * 0.3487 = 17.1723
Variance = E(X^{2})  (E(X))^{2} = 17.1723  4.1^{2} = 0.3623
Standard deviation = sqrt(0.3623) = 0.6
OptionA) σ = 0.6
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