Answer
$μ_x=4.9$
If the manager watches the line a large number of times, we would expect the mean number of people waiting in line during lunch to be about $4.9$
Work Step by Step
$μ_x=[x.P(x)]=0[P(0)]+1[P(1)]+2[P(2)]+3[P(3)]+4[P(4)]+5[P(5)]+6[P(6)]+7[P(7)]+8[P(8)]+9[P(9)]+10[P(10)]+11[P(11)]+12[P(12)]=0(0.011)+1(0.035)+2(0.089)+3(0.150)+4(0.186)+5(0.172)+6(0.132)+7(0.098)+8(0.063)+9(0.035)+10(0.019)+11(0.004)+12(0.006)=4.87\approx4.9$
