The width of a casing for a door is normally distributed with a mean of 24 inches andThe width of a casing for a door is normally distributed with a mean of 24 inches and a standard deviation of 1/8 inch.The width of a door is normally distributed with a mean of 23-7/8 inches and a standard deviation of 1/16 inch.Assume independence.(a) Determine the mean and standard deviation of the difference between the width of the casing and the width of the doo
The width of a casing for a door is normally distributed with a mean of 24 inches and
The width of a casing for a door is normally distributed with a mean of 24 inches and a standard deviation of 1/8 inch.The width of a door is normally distributed with a mean of 23-7/8 inches and a standard deviation of 1/16 inch.Assume independence.(a) Determine the mean and standard deviation of the difference between the width of the casing and the width of the door.(b) What is the probability that the width of the casing minus the width of the door exceeds 1/4 inch?(c) What is the probability that the door does not fit in the casing?
(a) Let C be the width of a casing for a door,and D be the width of a door
Then,N(24,1/8) and (23-7/8,1/16)
E(C-D)=E(C)-E(D)=24-(23-7/8)=1+7/8=15/8
Assume independence,the variation D(C-D)=D(C)+D(D)=(1/8)^2+(1/16)^2=5/256
Then,the standard deviation of C-D is (√5)/16
Therefore,C-D~N(15/8,(√5)/16)
(b) E(C-D)-1/4=13/8>10*[(√5)/16]=10σ
Therefore,you don't really need to caculate the probability for sure.
P(C-D>1/4)≈1
(c) P(C-D