# Homework 2 ISyE 6420 solution

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Original Work ?

## 1. k-out-of-n and Weibull Lifetime.

Engineering system of type k-out-of-n is operational if at least k out of n components are operational. Otherwise, the system fails.

Suppose
that a k-out-of-n system consists of n identical and independent elements for which the lifetime has Weibull distribution with parameters r and λ. More precisely, if T is a lifetime of
a component,
P(T ≥ t) = e
−λtr
, t ≥ 0.

Time t is in units of months, and consequently, rate parameter λ is in units (month)−1
.
Parameter r is dimensionless.

Assume that n = 8, k = 4, r = 3/2 and λ = 1/10.
(a) Find the probability that a k-out-of-n system is still operational when checked at
time t = 3.

(b) At the check up at time t = 3 the system was found operational. What is the
probability that at that time exactly 5 components were operational?

Hint: For each component the probability of the system working at time t is p = e
−0.1 t
3/2
.

The probability that a k-out-of-n system is operational corresponds to the tail probability
of binomial distribution: IP(X ≥ k), where X is the number of components working.

You
can do exact binomial calculations or use binocdf in Octave/MATLAB (or dbinom in R,
or scipy.stats.binom.cdf in Python when scipy is imported). Be careful with ≤ and <,
because of the discrete nature of binomial distribution. Part (b) is straightforward Bayes
formula.

## 2. Precision of Lab Measurements.

The error X in measuring the weight of a chemical
sample is a random variable with PDF
f(x) = (
3x
2
16 , −2 < x < 2
0, otherwise

(a) A measurement is considered to be accurate if |X| < 0.5. Find the probability that a
randomly chosen measurement can be classified as accurate.
(b) Find the cumulative distribution function F(x) and sketch its graph.

(c) The loss in thousands of dollars, which is caused by measurement error, is Y = X2
.
Find the mean of Y (expected loss).

(d) Compute the probability that the loss is less than \$3.
f(x, y) = (
x + y, 0 ≤ x ≤ 1; 0 ≤ y ≤ 1
0, else

Find:
(a) marginal distribution fX(x) .
(b) conditional distribution f(y|x).

4. From the first page of Rand’s book A Million Random Digits with 100,000 Normal
Deviates.

10097 32533 76520 13586 34673 54876 80959 09117 39292 74945
37542 04805 64894 74296 24805 24037 20636 10402 00822 91665
08422 68953 19645 09303 23209 02560 15953 34764 35080 33606
99019 02529 09376 70715 38311 31165 88676 74397 04436 27659
12807 99970 80157 36147 64032 36653 98951 16877 12171 76833

The first 50 five-digit numbers form the Rand’s “A Million Random Digits with 100,000
Normal Deviates” book (shown above) are rescaled to [0, 1] (by dividing by 100,000) and
then all numbers < 0.6 are retained. We can consider the n = 34 retained numbers as a
random sample from uniform U(0, 0.6) distribution.

0.10097 0.32533 0.13586 0.34673 0.54876 0.09117
0.39292 0.37542 0.04805 0.24805 0.24037 0.20636
0.10402 0.00822 0.08422 0.19645 0.09303 0.23209
0.02560 0.15953 0.34764 0.35080 0.33606 0.02529
0.09376 0.38311 0.31165 0.04436 0.27659 0.12807
0.36147 0.36653 0.16877 0.12171

Pretend now that the threshold 0.6 is not known to us, that is, we are told that the
sample is from uniform U(0, θ) distribution, with θ to be estimated.

Let M be the maximum of the retained sample u1, . . . , u34, in our case M = 0.54876.
The likelihood is
f(u1, . . . , u34|θ) = Y
34
i=1
1
θ
1(θ > ui) = θ
−34 1(θ > M),
2

Figure 1: First page of RAND’s book.
where 1(A) is 1 if A is true, and 0 if A is false.
Assume noninformative (Jeffreys’) prior on θ,
π(θ) = 1
θ
1(θ > 0).

Posterior depends on data via the maximum M and belongs to the Pareto family,
Pa(c, α), with a density
αcα
θ
α+1 1(θ > c).

(a) What are α and c?
(b) Estimate θ and calculate 95% equitailed credible set. Is the true value of parameter
(0.6) in the credible set?
Hint: Expectation of the Pareto Pa(c, α) is αc
α−1
and CDF is F(θ) = [1 − (c/θ)
α
] 1(θ > c).
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