(8) 1. A manufacturer of electronics products expects that 1% of units fail during the warranty period. A sample of 600 independent units is tracked for warranty performance.

a) What is the expected number of failures during the warranty period?

b) What is the probabilitythat none fails during the warranty period?

c) What is the probability that more than three units fail during the warranty period?

d) What is the probability that the number of failures will be between 1, 2, 3 standard deviations?

(4) 2. A binary message is transmitted as a signal S, which is either ? 1 or + 1. The communication

channel corrupts the transmission with additive normal noise with = .1 and variancecr2 2.

The receiver concludes that the signal ? 1 was transmitted if the value received is < 0.

What is the probability of error?

(7) 5. A system consists of two components connected a) in series ; b) in parallel.

Suppose that i-th component has a lifetime that is exponentially distributed with … = .01 and

.. = .02, and that components fail independently. Let X = the time at which the system fails.

Calculate the cdf F(x) = P(X ≤ x), the pdf of X and the expected value of X for cases a) and b).

Hint: use notation X = lifetime of i-th component, A = {X2 <x}, A = {X <x}, i = 1, 2.

(5) 6.* A stick of a length one is broken at random. Find the expected value of the difference between the larger part and the smaller part.

A manufacturer of electronics products expects that p = 1% of units fail during the warranty period. A sample of n = 600 independent units is tracked.

a) This is a binomial distribution, and the expected number of failures during the warranty period is

E(X) = p*n = 1%*600 = 6

b) The probability of X units fail during the warranty period is calculated by:

Pr(X)=N!/X!(N-X)! * p^x (1-p)^ (N-X)

where Pr(X) is the probability of exactly X fail, N=600 is the number of samples, and p= 1% is the probability of failure. This formula assumes that the events are dichotomous (fall into only two categories), mutually exclusive, independent and randomly selected

Then Pr(0)=600!/ 0!(600-0)! * 1%^0* (1-1%)^ (600-0) = 0.24%

c) The probability of more than 3 units fail during the warranty period = 1 – probability (0, 1, 2, 3 units fails)

By the above formula, we can compute:

X Pr(X)

0 0.0024

1 0.0146

2 0.0441

3 0.0888

Total 0.1499

Pr(X<=3) = …

Standard Deviation, Exponential Distribution and Expected Values are investigated. The solution is detailed and well presented.