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 se
lected 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?
<span>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 <-------</span>
Step-by-step explanation: no because Lauren will receive 26 because 10 = (8 x 2) which equals 26 and Migule is 20 + ( 1 x 10) which = 30 so not the same this and simple and cute reason hope it right and helped.
