In the following problem, check that it is appropriate to use the normal approximation to the binomial.
Then use the normal distribution to estimate the requested probabilities. Do you try to pad an insurance claim to cover your deductible? About 37% of all U.S. adults will try to pad their insurance claims! Suppose that you are the director of an insurance adjustment office. Your office has just received 128 insurance claims to be processed in the next few days. Find the following probabilities. (Round your answers to four decimal places.)
(a) half or more of the claims have been padded
(b) fewer than 45 of the claims have been padded
(c) from 40 to 64 of the claims have been padded
(d) more than 80 of the claims have not been padded
Here, n = 128 and p = 0.37.
We know, a binomial distribution can be approximated to a normal distribution if
(i) n*p > 5 and
(ii) n*(1-p) > 5
Here we see n*p = 128*0.37 = 47.36 >5 and n*(`1-p) = 128*0.63 = 80.64 >5.
So, we can approximate this binomial distribution to a normal distribution.
Parameters necessary for a normal distribution can be calculated using standard results of binomial distribution.
Mean = n*p = 128*0.37 = 47.36
Variance = n*p*(1-p) = 128*0.37*0.63 = 29.8368
Standard deviation = 5.46
Suppose, X denotes number of claims padded.
then,
Z is denoting the standard normal distribution.
(a)
= 0.5-0.4963 = 0.0037
(b)
= 0.5 - 0.1664 = 0.3336
(c)
= P(-1.35<Z<3.05)
= 0.4989 + 0.4115 = 0.9104
(d)
= 0.0478
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