According to the Binomial probability distribution ,
Let x be the binomial variable .
Then the probability of getting success in x trials , is given by :
[tex]P(X=x)=^nC_xp^x(1-p)^{n-x}[/tex] , where n is the total number of trials or the sample size and p is the probability of getting success in each trial.
As per given , we have
n = 15
Let x be the number of defective components.
Probability of getting defective components = P = 0.03
The whole batch can be accepted if there are at most two defective components. .
The probability that the whole lot is accepted :
[tex]P(X\leq 2)=P(x=0)+P(x=1)+P(x=2)\\\\=^{15}C_0(0.03)^0(0.97)^{15}+^{15}C_1(0.03)^1(0.97)^{14}+^{15}C_2(0.03)^2(0.97)^{13}\\\\=(0.97)^{15}+(15)(0.03)^1(0.97)^{14}+\dfrac{15!}{2!13!}(0.03)^2(0.97)^{13}\\\\\approx0.63325+0.29378+0.06360=0.99063[/tex]
∴The probability that the whole lot is accepted = 0.99063
For sample size n= 2500
Expected value : [tex]\mu=np= (2500)(0.03)=75[/tex]
The expected value = 75
Standard deviation : [tex]\sigma=\sqrt{np(1-p)}=\sqrt{2500(0.03)(0.97)}\approx8.53[/tex]
The standard deviation = 8.53
