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Hypergeometric Distribution
Consider the previous example of the batch of light bulbs. Suppose the Bernoulli experiment is repeated without replacement. That is, once a bulb is tested, it will be set aside.
If X = number of successes, then
It can be shown that for the Hypergeometric Distribution
It should be intuitively clear that if nM/N is small enough, the hypergeometric distribution may be replaced by the binomial distribution. In our development of the subject, we will do so if n/N £ 0.05.
Case 1: Suppose we have two terms 7ab and 3ab. When we multiply these two terms, we get 7ab x 3ab = (7 x 3) a 1 + 1 . b 1 + 1 ( Therefore, x m . x n = x m +
Testing the hypothesis equality of two variances The test for equality of two population variances is based upon the variances in two independently chosen random samples drawn
without a calculator give the exact value of each of the following logarithms. (a) (b) log1000 (c) log 16 16 (d) log 23 1 (e) Solution (b) log10
Evaluate the log function: Calculate 3log 10 2. Solution: Rule 3. log (A n ) = nlog b A 3log 10 2 = log 10 (2 3 ) = log 10 8 = 0.903
Binormal Vector - Three Dimensional Space Next, is the binormal vector. The binormal vector is illustrated to be, B → (t) = T → (t) * N → (t) Since the binormal vecto
what is the product of the solutions to the equation: x2+4x=-4
bananas are on sale for 3 pounds for $2. At that price how many pounds can you buy for $22
r=asin3x
If y 1 (t) and y 2 (t) are two solutions to y′′ + p (t ) y′ + q (t ) y = 0 So the Wronskian of the two solutions is, W(y 1 ,y 2 )(t) = =
how can we prove that an absolute convergent series is convergent but the converse is not true.
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