Question

Problem 2:

Problem 2: (20 points) Exam #3A ong all the computer chips produced by a certain factory, 6 percent are defective. A sample of 400 chips a. (10 points) What is the probability that this sample contains between 20 and 38 defective b. (10 points) Suppose that each of 40 inspectors collects a sample of 400 chips. What is the is selected for inspection 4 xt.S -np chips (including 20 and 38)? probability that at least 30 inspectors will find between 20 and 38 defective chips in their samples? n仁400106)=24 AS 24 nh-22.56 0 24 35-24 22.54 ) PC :30) np= 40C-8278)-33.112- Z a 245-3.12 g 5.701 215 33 112 5-7o

Answer 1

a)

$np=400*0.06=24$

$npq=400*0.06*0.94=22.56$

we have to calculate probablity between 20 and 38

$p( z\leq \frac{19.5-24}{\sqrt{22.56}})=p(z\leq -0.95)=0.1711$

$p( z\leq \frac{38.5-24}{\sqrt{22.56}})=p(z\leq 3.05)=0.9989$

$p( -0.95\leq z\leq 3.05)=p( z\leq 3.5)-p( z\geq -0.95)$

$=0.9989-0.1711$

probablity between 20 and 38 is 0.8278

b)

$n=40$

$np=40*0.8278=31.112$

$npq=40*0.8278*0.1722=5.701$

we have to calculate $p(x\geq 30)$

$p(x\geq 30) = p(z\geq \frac{29.5-33.112}{\sqrt{5.701}} )$

$= p(z\geq -1.51)$

   $= 1-p(z\leq -1.51)$