Harsh P. answered 08/27/20
A hard working and dedicated person trying to motivate young kids
(a) Using the binomial distribution:
The question asks specifically for the probability of at least 3 in 50 defects. So we must take the whole probability and remove the cases that do not fit that. Then, we have that our answer must be 1-p0-p1-p2, or the total probability less the probability or 0, 1, or 2 defects.
Then,
p0= (0 choose 50)*(0.12)^0*(0.88)^50=0.001675
p1= (1 choose 50)*(0.12)^1*(0.88)^49=0.011424
p2= (2 choose 50)*(0.12)^2*(0.88)^48=0.038165
After we subtract these values we are left with 94.87%.
(b) To apply the normal distribution. We must find the mean and variance of the sample set first.
mean=np=(50)(0.12)=6
variance=np(1-p)=50*0.12*0.88=5.28
Then, we have Pr(X<=3)=Pr(Z<= (3-6)/√5.28)=Pr(Z<=-1.31)= 0.0951.
Then, the probability that X>=3 is 1-0.0951=0.9049.