Question

4. (15 points) 4. (15 points) A hole is drilled in a sheet-metal component, and then a rod is inserted through the hole. The rod clearance is equal to the difference between the radius of the hole and the radius of the rod. Let the random variable X denote the clearance, in millimeters. The probability density function of X is given by: f(x) = {c(1 – x4) 0<x< 1 otherwise Components with clearances larger than 0.8mm must be scrapped. What proportion of components are scrapped? What is the expected clearance? What is the variance of rod clearance?

Answer 1

function Since f(x)= e(1-x) jocx c) 1 0 Orhesuise we first determine the constant e Smee fix) is the probability density | Hx) dez 1 le(1-x) dx = 1 es e (b-20) dr - 1 -e (x-x57 = 1 - e { 1 – 1/5 ] = 1 - 4 er 1 UD N OLX 2 i tx) = {/4(1-x); OLALI Therwise The proportion of components that are scrapped is given by PLA> 0.8] = f(x) de Jo.8

" da "plx>0.8] = 1 4 (1-x) dx -% (24) - %%s -(0.0–160,0)) - | 0-09) - 10.0819 . . is given The Expected clearance ELS) 124m) dx = *6-2") de - %1-21 - % (-) =% (s)-15/2 = 0.417 The variance of rod clearance is given by VLX) - ELX²) - Elx) 2 Now NEL% -) dx = 5( 2), - Ellos - ] = 5.91) 5 = 0.336 (x) = 0.20 – (0.417) = 10.064) 0.238 2 ) thank you, please Thumbs Up!