1 (a) A manufacturer has found from experience that 3% of his product is rejecte

Question

1 (a) A manufacturer has found from experience that 3% of his product is rejected because of flaws. A new lot of 900 units comes up for inspection. What is the approximate probability that less than 30 units will be rejected? I DON'T HAVE AN IDEA ABOUT (a). (b) Assuming that the number of white blood cells per unit of volume of diluted blood counted under a microscope follows a Poisson distribution with u=150, what is the probability, using a normal approximation, that a count of 140 or less will be observed? For (b) my problem is the final stage. Using a normal approximation to Poisson = = 150 = = sqr. 150 for x=140 z= (x-) / = (140-150) / sqr. 150 = -0.816496581 Finding the probability when z <= -0.816496581 P(z<=-0.816496581)= 1- P(z<= -0.816496581) = 1- ???? = MY MAIN PROBLEM IS THIS LAST PART (WHERE THE ???? QUESTION MARK IS AND ONWARD). How do I solve for Z? This has been my problem. Please also check if I did everything right. Please help solve the (a) part also. I have no idea. Thanks

Explanation / Answer

Ans:

a)Normal approximation:

mean=np=900*0.03=27

standard deviation=sqrt(900*0.03*0.97)=5.12

use continuity corretion factor +/-0.5

z=(29.5-27)/5.12=0.49

P(x<29.5)=P(z<0.49)=0.6879

Uisng Binomial distribution,P(x<30)=P(x<=29)=binomdist(29,900,0.03,true)=0.6956

b)mean=150

standard deviation=sqrt(150)=12.25

z=(140-150)/12.25=-0.82

P(x<=140)=P(z<=-0.82)=0.2061

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.