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Central Limit Theorem The theorem states that the sum of a large number of independent observations from the same distribution has, under certain general conditions, an approximate normal distribution.
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Central Limit Theorem The theorem states that the sum of a large number of independent observations from the same distribution has, under certain general conditions, an approximate normal distribution. Moreover, the approximation steadily improves as the number of observations increases. The theorem is considered to be the heart of probability theory . The central limit theorem is one of the most remarkable results of the theory of probability.
(iii) Rayleigh distribution The envelop (instantaneous amplitude) of a narrowband noise follows a Rayleigh distribution, given by The distribution is similar to Gaussian but is not symmetrical. The envelop cannot be less than zero but has no upper limit. In amplitude modulation, the envelop carries information, but noise perturbs the envelop.
Revision: Binomial distribution: There are two outcomes each trial, with probability for two outcomes given by p, (1-p) The probability of r successes in n trials is given by Mean variance
Example: A multiple choice examination have 100 questions, each having one correct answer, three incorrect answers. (i) Find the mean and the standard deviation for the distribution of the correct answers for one who answers the questions by random guess.
(ii) What is the probability of getting 50% correct answers by guessing answers for all of the questions? n=100, p=1/4, r=50. …
If for each question, the exam taker knows that two answers are wrong, but have to guess one from the two others. What is the probability of getting 70% correct answers? • n=100, p=1/2, r=70. ….
(iii) If one knows the answers to 90 questions, but have to purely guess answers for the remaining 10 questions. What is the probability of getting more than 95 correct answers? • Solution: He (she) needs to guess correctly more than 5 answers out of 10 questions. P(r>5) • So use n =100-90=10, p=1/4. • and then sum up the probabilities for p(r) for r= 6, r= 7, r= 8, r= 9, r= 10.
Example: A telecommunication system sends a binary sequence using two levels of ±1V. A sequence of “A=101010” was sent through a channel which is disturbed by random noise of mean squares voltage 0.1V. • Find the error rate, assuming that an instantaneous decision is made at the centre of each received pulse with decision level zero. • What is the probability of correct transmission of the sequence A?
1. p(1->0) s(t) = +1V but s (t)+ n(t) < 0 So want to find p( n(t) < -1) n(t) is zero mean and mean square of 0.1
2. p(0->1) s(t) = -1V but s (t)+ n(t) > 0 So want to find p( n(t) > 1) n(t) is zero mean and mean square of 0.1
(ii) A has 6 bits, each bit has error probability of p= 0.008. For all 6 bits to be correctly transmitted, using binomial distribution to calculate p(0),