1A textile manufacturing process finds that on average 3 fla

1.A textile manufacturing process finds that on average, 3 flaws occur per every 220 yards of material produced.

a.

What is the probability of exactly 4 flaws in a 220-yard piece of material? (Round your intermediate calculations and final answer to 4 decimal places.)

What is the probability of no more than 2 flaws in a 220-yard piece of material? (Round your intermediate calculations and final answer to 4 decimal places.)

c.

What is the probability of no flaws in a 110-yard piece of material? (Round your intermediate calculations and final answer to 4 decimal places.

2.Assume that X is a hypergeometric random variable with N = 10, S = 3, and n = 3. Calculate the following probabilities. (Round your answers to 4 decimal places.)

Solution

(1)(a) Given X~Poisson(mean=3)

P(X=x)=3^x*exp(-3)/x!

So the probability of exactly 4 flaws in a 220-yard piece of material is

P(X=4) = 3^4*exp(-3)/4! = 0.1680

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(b) So the probability of no more than 2 flaws in a 220-yard piece of material is

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

=3^0*exp(-3)/0! + 3^1*exp(-3)/1! +3^2*exp(-3)/2!

=0.4232

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(c) Given X~Poisson(mean=3/2=1.5)

P(X=x)=(1.5^x)*exp(-1.5)/x!

So the probability of no flaws in a 110-yard piece of material is

P(X=0)= (1.5^0)*exp(-1.5)/1 = 0.2231

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(2)(a) P(X = 1) = 3C1* 7C2/ 10C3

=3*(7*6/2) / (10*9*8/6)

=0.525

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(b)P(X = 2) =3C2* 7C1/ 10C3

=3*7/(10*9*8/6)

=0.175

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(c)P(X > 2) = 1-P(X=0)-P(X=1)-P(X=2)

=1-3C0* 7C3/ 10C3 - 0.525 - 0.175

=1- 1*(7*6*5/6) / (10*9*8/6) - 0.525 - 0.175

=0.0083