Answer: Option C
[tex]E =1.06\ siblings[/tex]
Step-by-step explanation:
Let X be a discrete random variable that counts the number of siblings a randomly selected student has. Then the expected value of X is defined as:
[tex]E =\sum_{i=0}^{i=n} X_iP(X_i)[/tex]
Where [tex]P(X_i)[/tex] is the probability that a randomly selected student has [tex]X_i[/tex] siblings
With [tex]i = \{0,1, 2, 3, 4, 5\}[/tex]
So in this case we know that
#of siblings 0 1 2 3 4 5
probability 0.20 0.65 0.08 0.04 0.02 0.01
Therefore:
[tex]E =\sum_{i=0}^{i=5} X_iP(X_i)[/tex]
[tex]E =0*(0.20)+1*(0.65)+2*(0.08)+3*(0.04)+4*(0.02)+5*(0.01)[/tex]
[tex]E =1.06[/tex]
The answer is the option C.
