Question

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 44% 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 134 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

Any help would be appreciated. Thanks!

Answer 1

Let X be the random variable that denotes the number of the claims that have been padded.

About 44% of all U.S adults will try to pad their insurance claims. p = 0.44

The office has just received 134 insurance claims to be processed in the next few days. n = 134

X Binomial (n = 134, p = 0.44)

The binomial distribution is approximated by normal distribution if the following condition is met :

If X Binomial(n, p) and if n is large ( 30) and p is close to 0.5, then X Normal(np , npq) approximately where (q = 1 - p)

The above condtions are met.

E(X) = n * p = 134 * 0.44 = 58.96

V(X) = n * p * q = 134 * 0.44 * 0.56 = 33.0176

X Normal (58.96, 33.0176)

a) Answer :

P(half or more claims have been padded) = P(X 134 / 2)

P(X 67) = P(X > 66.5) ........(using continuity correction)

                 = P(Z > (66.5 - 58.96) / )

                 = P(Z > 1.31) = 0.09510

The probability that half or more of the claims have been padded is 0.0951.

b) Answer :

P(fewer than 45 of the claims have been padded) = P(X < 45)

P(X < 45) = P(X < 44.5) ........(using continuity correction)

                = P(Z < (44.5 - 58.96) / )

                = P(Z < -2.52) = P(Z > 2.52)

                = 0.00587

The probability that fewer than 45 claims have been padded is 0.0059.

c) Answer :

P(40 to 64 of the claims have been padded) = P(40 X 64)

P(40 X 64) = P(39.5 < X < 64.5) ............(using continuity correction)

                         = P((39.5 - 58.96) / < Z < (64.5 - 58.96) / )

                         = P(-3.39 < Z < 0.96)

                         = P(Z < 0.96) - P(Z < -3.39)

                         = P(Z < 0.96) - P(Z > 3.39)

                         = 0.83147 - 0.00035

                         = 0.83112

The probability that from 40 to 64 of the claims have been padded is 0.8311.

d) Answer :

P(more than 80 of the claims have not been padded) = P(less than (134 - 80) of the claims have been padded)

P(X < 54) = P(X < 53.5) ..........(using continuity correction)

                = P(Z < (53.5 - 58.96) / )

                = P(Z < -0.95) = P(Z > 0.95)

                = 0.17106

The probability that more than 80 of the claims have not been padded is 0.1711.