Question: 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 and 7/8 inches and a standard deviation of 1/16 inch. Assume independence. Round your answers to three decimal places (e.g. 98.765).

(a) Determine the mean and standard deviation of the difference between the width of the casing and the width of the door.  Edit

Answer: Given that the width of a casing (X) for a door is normally distributed with mean of 24 in and SD of 1/8 in. The width of a door is normally distributed with mean of 23.875 in and SD of 1/16 in.
i.e. X~N(µx = 24 σx = 1/8) and Y~N(µy = 23.875 σy = 1/16)
Also X and Y are independent.

a) The mean and SD of the difference between the width of casing and the width of door (X-Y)
E(X-Y) = E(X) – E(Y)
= µx - µy
=24 – 23.875
=0.125 (Ans.)

SD(X-Y) = √ ( σx^2 + σy2)
= √ [(1/8)^2 + (1/16)^2]
=√ 50/64
= 0.1397  Edit