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2015-05-05T23:05:19+05:30

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      We assume a known probability distribution for this kind of problem.  The population is big enough (100 pages).   We can say that the number of defective pages is a random variable X.  The probability of any randomly selected page being defective is equal and  = 100/1000 = 0.1

      In the first 50 pages, the expected number of defective pages is 50 * 0.1 = 5.  Let X denote the number of defective pages in the first 50 pages.  

    We can have 0 defective pages, 1 or 2 or 5 or 10 or 19 or 20 or up to 50 defective pages in the first 50 pages of the book.  That means,

    0 <= X <= 50         and  we have that     E(X) = λ = mean value or expected value = 5

      We can assume the Poission probability distribution or the Normal probability distribution to solve this.  We don’t know the standard deviation for the normal distribution.   We assume that  X follows the Poisson's probability distribution  function.  It is a discrete probability distribution.

P(X=k)=\frac{e^{-\lambda}\ \lambda^k}{k!}\\\\and,\ \ P(X \ge k)=\frac{e^{-\lambda}\ \lambda^k}{k^k}

Let us calculate the probabilities, with λ = 5.

    P(X=10) = 0.0181 = the probability that the number of defective pages is 10.

     P(X >= 10)  <= 0.145.    This is the upper limit on the probability that the number of defective pages is equal to or more than 10.
   P(X <= 10)  >= 0.855,    so probability that  the number of defective pages is less than  or equal to10,  is  0.855

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for info:
we can also calculate  P(X=0) = 0.006738
 P(X=1)=0.03369         P(X=2) = 0.0842          P(X=3) = 0.1404
 P(X=4) = 0.1755         P(X=5) = 0.1755          P(X=6) =  0.1462
 P(X=7) = 0.1044         P(X=8) = 0.0653          P(X=9) = 0.0362


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