machine a produces components with holes whose diameter is normally distributed

Question

machine a produces components with holes whose diameter is normally distributed with a mean 56,000 and a standard deviation 10. machine B produces components with holes whose diameter is normally distributed with a mean 56,005 and a standard deviation 10. machine C produces components with holes whose diameter is normally distributed with a mean 55,980 and a standard deviation 10. machine D produces components with holes whose diameter is normally distributed with a mean 55,985 and a standard deviation 9. Machine A produces components with holes whose diameter is normally distributed with a mean 56,000 and a standard deviation 10. Machine B produces components with holes whose diameter is normally distributed with a mean 56,005 and a standard deviation 8. Machine C produces pins whose diameter is normally distributed with a mean 55,980 and a standard deviation 10. Machine D produces pins whose diameter is normally distributed with a mean 55,985 and a standard deviation 9 (a) What is the probability that a pin from machine C will have a larger diameter than a pin from machine D? (b) What is the probability that a pin from machine C will fit inside the hole of a component from machine A? (c) If a component is taken from machine A and a component is taken from machine B, what is the probability that both holes will be smaller than 55,995?

Explanation / Answer

let X be the diameter of hole produced from machine A, X~N(56000, 100)

let Y be the diameter of hole produced from machine B, Y~N(56005, 64)

let Z be the diameter of pin produced from machine C , Z~N(55980, 100)

let W be the diameter of pin produced from machine D , W~N(55985, 81)

(a)

P(Z>W)=P(Z-W>0)= 1-P(Z-W<=0) = 0.489

Note that Z-W~ N(-5, 181)

(b)

P(Z<X)=P(Z-X<0)

Z-X~ N(-20,200)

P(Z-X<0)= 0.54

(c)

P(X<55995,Y<55995)=P(X<55995)*P(Y<<55995)

P(X<55995)=0.48

P(Y<55995)=0.4379

So, P(X<55995,Y<55995)=0.21

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