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natulia
1 month ago
5

A hole is drilled in a sheet-metal component, and then a shaft is inserted through the hole. The shaft clearance is equal to dif

ference 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.
Mathematics
1 answer:
Leona [12.6K]1 month ago
5 0

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)

Thus, the cumulative function can be expressed as

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

0 otherwise.

Step-by-step explanation:

(A)

We want to determine the likelihood that the random variable X exceeds 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're seeking the probability P(X \geq x)  which can be expressed as

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

The cumulative function will therefore be

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

0 otherwise.

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