A hole is drilled in a sheet-metal component, and then a shaft is inserted through the hole. The shaft clearance is equal to the difference between the radius of the hole and the radius of the shaft. Let the random variable X denote the clearance, in millimetres. The probability density function of X is: fx) = {1.25(1 – x*), 0

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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A hole is drilled in a sheet-metal component, and then a shaft is inserted through the hole. The
shaft clearance is equal to the difference between the radius of the hole and the radius of the
shaft. Let the random variable X denote the clearance, in millimetres.
The probability density function of X is:
1.25(1 x), 0<x <1
0, otherwise
f(x) =
Components with clearance larger than 0.8mm must be scrapped.
a) What proportion of components is scrapped?
b) Find the mean clearance
c) Variance of the clearance
Transcribed Image Text:A hole is drilled in a sheet-metal component, and then a shaft is inserted through the hole. The shaft clearance is equal to the difference between the radius of the hole and the radius of the shaft. Let the random variable X denote the clearance, in millimetres. The probability density function of X is: 1.25(1 x), 0<x <1 0, otherwise f(x) = Components with clearance larger than 0.8mm must be scrapped. a) What proportion of components is scrapped? b) Find the mean clearance c) Variance of the clearance
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