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 40% 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 140 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

Answer 1

Let X be the number of US adults who will pad their insurance claims.

Here, p=0.4 (close to 0.5), and n=140 (large sample), hence we can approximate the binomial distribution X ~ B(n,p) by

X ~ N(np, npq) => X ~ N(140*0.4, 140*0.4*0.6) => X ~ N(56, 33.6)

a) P(half or more claims padded) = P(X>=140/2) = P(X>=70).

Let's calculate by converting the X value into Z-score

z-score of 70, z* = (70-56)/33.6^1/2 = 2.415

Hence, P(X>=70) = P(Z>=z*) = P(Z>=2.415) = 0.0078 (looked up from z-score table)

b) Similarly, P(X<=45) = P(Z<=(45-56)/33.6^1/2 ) = P(Z <= -1.898) = 0.0287

c) P(from 40 to 64 claims padded) = P(40 <= X <= 64)

= P((40-56)/33.6^1/2 <= Z <= (64-56)/33.6^1/2 )

=P(-2.76 <= Z <= 1.38 ) = 0.9133

d) More than 80 claims have not been padded => (140-80) = 60 or lesser claims have been padded

Hence, P(More than 80 claims have not been padded)

= P(X<=60) = P(Z <= (60-56)/33.6^1/2 ) = P(Z <= 0.69) = 0.7549