Answer to Question #112972 in Statistics and Probability for Cody Q. Hart

Question

suppose that 1 out of 10 plasma televisions shipped with a defective speaker. out of a shipment of n=400 plasma televisions. find the probability (as a %) that there are more than 28 with defective speakers

Answer 1

Let $X=$ the number of plasma televisions with defective speakers: $X\sim Bin(n, p)$

Given $n=400, p=0.1$

Use the dishonest-coin principle with p=0.1 to find

$$\mu=np=400(0.1)=40$$

$$\sigma=\sqrt{np(1-p)}=\sqrt{400(0.1)(1-0.1)}=6$$

$$28=40-2\times6=\mu-2\sigma$$

By the 68–95–99.7 rule

$$Pr(\mu-2\sigma\leq X\leq \mu+2\sigma)\approx 0.9545$$

The probability that there are more than 28 with defective speakers

$$0.9545+{1-0.9545\over 2}\approx0.97725$$

Or

$n=400\geq30, np=40\geq5, n(1-p)=360\geq5$

Then, $X$  has an approximately normal distribution with mean $\mu=np=40$ and standard deviation 

$\sigma=\sqrt{np(1-p)}=6$

$$P(X>28)=P(X>28+0.5)=1-P(X\leq28.5)=$$

$$1-P(Z\leq{28.5-40\over6}) \approx1-0.02764\approx0.97236$$

By Binomial distribution

$$P(X>28)\approx0.976515$$