A TV company when purchasing thousands electronic components apply this sampling plan: randomly select 15 of them and then accept the whole batch if there are at most two defective components. If aparticular lot has a 3% of defective components, what is the probability that the whole lot is accepted? In addition,if they purchased2500electroniccomponents,whataretheexpectednumberofdefectivesinthislotof 2500?Find the standard deviation.

Answer with explanation: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 haven = 15 Let x be the number of defective components.Probability of getting defective components = P = 0.03The 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 = 0.99063For sample size n= 2500Expected value : [tex]\mu=np= (2500)(0.03)=75[/tex]The expected value = 75Standard deviation :  [tex]\sigma=\sqrt{np(1-p)}=\sqrt{2500(0.03)(0.97)}\approx8.53[/tex]The standard deviation = 8.53