11. 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 millimeters. The probability density function ofX is: f(x) = 1.25(1 – x*) ;0

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11. 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 millimeters. The
probability density function of X is:
f(x) = 1.25(1 – x*) ;0<x< 1, f(x) = 0 otherwise
Components with clearances larger than 0.8 mm must be scrapped.
a) What proportion of components are scrapped?
b) Find the mean clearance.
c) The standard deviation of the clearance.
Transcribed Image Text:11. 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 millimeters. The probability density function of X is: f(x) = 1.25(1 – x*) ;0<x< 1, f(x) = 0 otherwise Components with clearances larger than 0.8 mm must be scrapped. a) What proportion of components are scrapped? b) Find the mean clearance. c) The standard deviation of the clearance.
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