Question

A hole is drilled in a sheet-metal component, and then a shaft is inserted through the hole. The shaft clearance is equal to 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 - x4) if 0 < x < 1
F(x) = 0 otherwise

A. Components with clearances larger than 0.8 mm must be scrapped. What proportion of components are scraped?
B. Find the cumulative distribution function F(x) and plot it.
C. Use the cumulative distribution to find the probability that the shaft clearance is less than 0.5 mm.
D. Find the mean clearance and the variance of the clearance.

Answer 1

Answer:

(A)

$P(X \geq 0.8) = \int\limits_{0.8}^{\infty} f(x) \, dx = \int\limits_{0.8}^{1} 1.25(1-x^4) \, dx = 0.08192$

(B)

Then the cumulative function would be

$CF(x) = 1.25x - 0.25x^5$       if   0<x<1

0 otherwise.

Step-by-step explanation:

(A)

We are looking for the probability that the random variable X is greater than 0.8.

$P(X \geq 0.8) = \int\limits_{0.8}^{\infty} f(x) \, dx = \int\limits_{0.8}^{1} 1.25(1-x^4) \, dx = 0.08192$

(B)

For any  $x$ you are looking for the probability $P(X \geq x)$  which is

$P(X \geq x) = \int\limits_{-\infty}^{x} 1.25(1-t^4) dt = \int\limits_{0}^{x} 1.25(1-t^4) dt = 1.25x - 0.25x^2$

Then the cumulative function would be

$CF(x) = 1.25x - 0.25x^5$       if   0<x<1

0 otherwise.