Given:
The probability distribution for X = the number of people in line to use a vending machine has the following probability distribution:
X P(X = k )
0 0.10
1 0.10
2 0.40
3 0.30
4 0.10
The expected value of X is given by:
[tex]\begin{gathered} E(X)=\sum ^4_{k\mathop=1}x_kp(x_k) \\ =0\times_{}p(k=0)+1\times p(k=1)+2\times p(k=2)+3\times p(k=3)+4\times p(k=4) \\ =0+1\times0.10+2\times0.40+3\times0.30+4\times0.10 \\ =0+0.10+0.80+0.90+0.40 \\ =2.20 \end{gathered}[/tex]
Therefore,
expected value of X is 2.20.
