The width of a casing for a door is normally distributed with a mean of 24 in an
ID: 3150373 • Letter: T
Question
The width of a casing for a door is normally distributed with a mean of 24 in and a standard deviation of 0.125 in. The width of a door is normally distributed with a mean of 23.875 in and a standard deviation of 0.0625 in. Assume independence of the two widths. Determine the mean and standard deviation of the difference between the width of the casing and the width of the door. What is the probability that the width of the casing minus the width of the door exceeds 0.25 in? What is the probability that the door does not fit in the casing?Explanation / Answer
a)
Let X = width of casing
Y = width of door
The mean is
u(X-Y) = u(X) - u(Y) = 24 - 23.875 = 0.125 in [ANSWER]
while the standard deviation is
sigma(X-Y) = sqrt[sigma^2(X) + sigma^2(Y)] = sqrt(0.125^2 + 0.0625^2) = 0.139754249 [ANSWER]
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b)
We first get the z score for the critical value. As z = (a - u) / s, then as
a = critical value = 0.25
u = mean = 0.125
s = standard deviation = 0.139754249
Thus,
z = (a - u) / s = 0.894427188
Thus, using a table/technology, the right tailed area of this is
P(z > 0.894427188 ) = 0.185546685 [ANSWER]
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c)
That means X - Y < 0.
We first get the z score for the critical value. As z = (a - u) / s, then as
a = critical value = 0
u = mean = 0.125
s = standard deviation = 0.139754249
Thus,
z = (a - u) / s = -0.894427188
Thus, using a table/technology, the left tailed area of this is
P(z < -0.894427188 ) = 0.185546685 [ANSWER]
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