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
SD(X-Y) = √ ( σx^2 + σy2)
= √ [(1/8)^2 + (1/16)^2]
= 0.1397 Edit
TutorTeddy.com & Boston Predictive Analytics
[ Email your Statistics or Math problems to firstname.lastname@example.org (camera phone photos are OK) ]
Boston Office (Near MIT/Kendall 'T'):
Cambridge Innovation Center,
One Broadway, 14th Floor,
Cambridge, MA 02142,
Dallas Office (Near Galleria):
15950 Dallas Parkway,
Dallas, TX 75248,