II. Probability and Statistics for Reliability
A. Basic concepts
3. Discrete and continuous probability distributions (Analyze)
Compare and contrast various distributions (binomial, Poisson, exponential, Weibull, normal, log-normal, etc.) and their functions (e.g., cumulative distribution functions (CDFs), probability density functions (PDFs), hazard functions), and relate them to the bathtub curve.
This lesson takes a close look at the four common functions used with statistical distributions.
Additional References
The Four Functions (article)
Reliability Function (article)
Reliability from Hazard step function (article)
PDF to CDF with Brief Calculus Refresher (article)
Quick Quiz
1-12. Identify which of the following are normally accepted reliability engineering tools.
I. probability density function
II. hazard rate function
III. conditional reliability function
IV. mean life function
V. cumulative distribution function
(A) I and II only
(B) I, II, and III only
(C) II, III, IV, and V only
(D) I, II, III, IV, and V
(D) I, II, III, IV, and V
All five terms represent either ways to describe distributions (PDF, hazard rate, CDF, and conditional reliability) or the mean of a set of life data.
1-20. The hazard rate function h(t) as a function of time t in hours for a device is
h(t) = 0, for t ≤ 0 and 0.66t, for t > 0
Determine the reliability of this device at t = 3 hours?
(A) 0.0.00263
(B) 0.0.0513
(C) 0.138
(D) 0.372
(B) 0.0.0513
Given a distributions PDF you can derive the reliability function with
$$ R(t)={{e}^{-\int_{-\infty }^{x}{h\left( \tau \right)d\tau }}}$$
Then substitute the given hazard rate function and simplify
$$ R(t)={{e}^{-\int_{0}^{t}{0.66\left( \tau \right)d\tau }}}={{e}^{-0.66\int_{0}^{t}{\left( \tau \right)d\tau }}}$$
We can move the constant out of the integral due to the Constant Multiple Property, the solve the integral
$$ R(t)={{e}^{-0.66\int_{0}^{t}{\left( \tau \right)d\tau }}}={{e}^{-0.66\left( \frac{{{t}^{2}}}{2} \right)}}={{e}^{-0.33{{t}^{2}}}}$$
Finally, solve when t = 3
$$ R(3)={{e}^{-0.33{{\left( 3 \right)}^{2}}}}=0.0513$$
The distractors are answers when the integral is done incorrectly a few different ways.
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