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 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.

Explanation / Answer

let A,B,C,D denotes the diameter of the holes of the corresponding machines

by question A~N(56000,102) B~N(56005,82)   C~N(55980,102)      D~N(55985,92)      

and A,B,C,D are independent since the manufacture of one machine is not affected by the other.

a) we need to find the probability that the hole made from machine C will have a larger diameter then that of made from machine D.

hence we need to find P[C>D] or P[C-D>0]

now since C and D are normally distributed C-D will also be normally distributed with mean E[C]-E[D]=55980-55985=-5 and since C and D are independent hence the variance of C-D will be V[C]+V[D]=102+92=181

so C-D~N(-5,181)

so P[C-D>0]=1-P[C-D<0]=1-P[(C-D+5)/sqrt(181)<(0+5)/sqrt(181)]=1-P[Z<0.371647]   where Z~N(0,1)

=1-0.644922=0.355078 [answer]   [using minitab]

b) the pin made by C will fit inside the hole of pin made by A means the diameter of pin made by C is less than that of made by A.

so P[C<A] or P[C-A<0]

by similar logic as in part a)

C-A~N(55980-56000,102+102) or C-A~N(-20,200)

so P[C-A<0]=P[(C-A+20)/sqrt(200)<(0+20)/sqrt(200)]=P[Z<1.1412]   where Z~N(0,1)

=0.921350 [answer]   [suing MINITAB]

c) a component is selected from A and another from B. we need to find the probability that both holes will be less than 55995

hence required probability

P[A<55995,B<55995]=P[A<55995]*P[B<55995]   since A and B are independent

=P[(A-56000)/10<(55995-56000)/10]*P[(B-56005)/8<(55995-56005)/8]

=P[Z<-0.5]*P[Z<-1.25] where Z~N(0,1)

=0.308538*0.105650=0.032597 [answer] [using MINITAB]

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