The width of a casing for a door is normally distributed wit

淡泊如水Leo2022-10-04 11:39:541条回答

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?

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3390784 共回答了15个问题 | 采纳率100%
(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
1年前

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