Question

5. The probability distribution for X = the number of people in line to use a vending machine has thefollowing probability distribution:kP(X=k)00.1010.102.0.4030.3040.10What is the expected value of X?

Answer 1

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:

`\begin{gathered} E(X)=\sum ^4_(k\mathop=1)x_kp(x_k) \\ =0*_{}p(k=0)+1* p(k=1)+2* p(k=2)+3* p(k=3)+4* p(k=4) \\ =0+1*0.10+2*0.40+3*0.30+4*0.10 \\ =0+0.10+0.80+0.90+0.40 \\ =2.20 \end{gathered}`

Therefore,

expected value of X is 2.20.