Question

A corporate Web site contains errors on 50 of 1000 pages. If 100 pages are sampled randomly, without replacement, approximate the probability that at least 2 of the pages in error are in the sample. (Use normal approximation to the binomial distribution.)

Answer 1

Answer:

The approximate probability that at least 2 pages with errors are in the sample is 0.6406

Step-by-step explanation:

The probability that 1 random page contains errors is 50/1000 = 0.02. Thus, if we take 100 pages, the mean amount of errors is 0.02*100 = 2. The standard deviation will be √2*(1-0.02) = 1.4. Lets call X the number of errors in 100 pages. We can assume that X has normal distribution (otherwise we approximate it to a normal random variable). Lets work with the standarization of X, W, given by

$W = \frac{X-\mu}{\sigma} = \frac{X-2}{1.4}$

We want to know the probability of X being greater than or equal to 2, for that we will use $\phi$ , the cummulative distribution function of W, whose values can be found in the attached file. We will compute instead the probability of X being greater than 1.5, so that we correct the variable from continuity$P(X \geq 1.5) = P(\frac{X-2}{1.4} \geq \frac{1.5-2}{1.4}) = P(W \geq -0.3571) = 1-\phi(-0.3571) = 1-(1-\phi(0.3571) ) = \phi(0.3571) = 0.6406$

The approximate probability that at least 2 of the pages have errors in the sample is 0.6406.