Question

Every day a factory produces 5000 light bulbs, of which 2500 are type 1 and 2500 are type 2. if a sample of 40 light bulbs is selected at random on a particular day, and examined for defects, what is the approximate probability that this sample contains at least 18 light bulbs of each type?

Answer 1

The solution: = 40, p = q = 0.5 P[x] = nCx *p^x *q^(n-x) when p = q = 0.5, the formula simplifies to P[x] = nCx/2^n = 40Cx/2^40 at least 18 of each type means 18 to 22 of (say) type I P(18 <= X <= 22) = 0.5704095 <------- qb mean = 40*0.5 = 20 SD = sqrt(npq) = sqrt(40*0.5*0.5) = 3.1623 z1= (18-20)/3.1623 = -0.63 , z2 = (22-20)/3.1623 = 0.63 P(-0.63 < z < 0.63) = 0.4713 <-------