Question

A company that manufactures large LCD screens knows that not all pixels on their screen light, even if they spend great care when making them. In a sheet 66 ft by 1212 ft that will be cut into smaller screens, they find an average of 5.35.3 blank pixels. They believe that the occurrences of blank pixels are independent. Their warranty policy states that they will replace any screen sold that shows more than 2 blank pixels. Complete parts a through d below.

c) What is the probability that a 2 ft by 3 ft screen will have at least one defect?

(Round to three decimal places as needed.)

d) What is the probability that a 2 ft by 3 ft screen will be replaced because it has too many defects?

(Round to three decimal places as needed.)

Answer #1

c)

expected defects in 2 ft by 3 ft screen =5.3*(2*3)/(6*12)=0.4417

hence probability that a 2 ft by 3 ft screen will have at least one defect =P(X>=1)=1-P(X=0)

=1-e^{-0.4417}*0.4417^{0}/0! =1-0.6429
=**0.357** ( please try 0.356 if this comes wrong due
to rounding error)

d)

probability that a 2 ft by 3 ft screen will be replaced because it has too many defects=P(X>2)

=1-P(X<=2)=1-(P(X=0)+P(X=1)+P(X=2))

=1-(e^{-0.4417}*0.4417^{0}/0!+e^{-0.4417}*0.4417^{1}/1!+e^{-0.4417}*0.4417^{2}/2!)

=**0.010**

A company that manufactures large LCD screens knows that not all
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