Question

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 1

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 :

$P(X=x)=^nC_xp^x(1-p)^{n-x}$ , 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 :

$P(X\leq 2)=P(x=0)+P(x=1)+P(x=2)\\\\=^{15}C_0(0.03)^0(0.97)^{15}+^{15}C_1(0.03)^1(0.97)^{14}+^{15}C_2(0.03)^2(0.97)^{13}\\\\=(0.97)^{15}+(15)(0.03)^1(0.97)^{14}+\dfrac{15!}{2!13!}(0.03)^2(0.97)^{13}\\\\\approx0.63325+0.29378+0.06360=0.99063$

∴The probability that the whole lot is accepted = 0.99063

For sample size n= 2500

Expected value : $\mu=np= (2500)(0.03)=75$

The expected value = 75

Standard deviation :  $\sigma=\sqrt{np(1-p)}=\sqrt{2500(0.03)(0.97)}\approx8.53$

The standard deviation = 8.53