posted by Lauren on .
According to the book Are you Normal?,40% of all US adults try to pad their auto insurance claims to cover their deductible. Your office has just received 128 in insurance claims to be processed in the next few days, what is the probability that fewer than 45 of the claims have been padded? Round your answer to the nearest thousandths place.
Let's assume the numbers are large enough so we can use a Normal approximation to the binomial distribution here. The mean and standard deviation of the binomial distribution will be Np and sqrt(Np(1-p)) respectively, where in this example N = 128 and p = 0.4, so the mean and standard deviation will be 51.2 and 5.543 respectively. You want the area to the left of 45 claims, so work out how many standard deviations 45 claims is below the mean of 51.2, and look that up in a set of Normal tables. The Z value you'll be looking up should be (44 - 51.2) / 5.543 = -1.299, but you'll want the area to the LEFT of that - which is the same as the area to the RIGHT of +1.299. That's 1 - 0.903, which is 0.097. However that's not one of your options, and I can't get one of those figures by trying the same calculation using either 44 or 44.5 instead of 45 by way of a continuity correction. However, that's essentially the way you tackle it, assuming a Normal approximation is considered good enough. I'm afraid you're just going to have to find out where I've made a mistake.