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 36% 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

P(an insurance claim is padded), p = 0.36

q = 1 - p = 0.64

Sample size, n = 140

np = 140x0.36 = 50.4

nq = 140x0.64 = 89.6

np and nq > 5. So, normal approximation to the binomial is appropriate.

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = np = 50.4

Standard deviation =

=

= 5.68

a) P(half or more of the claims have been padded) = P(X 70)

= 1 - P(X < 69.5)

= 1 - P(Z < (69.5 - 50.4)/5.68) {with continuity correction}

= 1 - P(Z < 3.36)

= 1 - 0.9996

= 0.0004

b) P(fewer than 45 claims are padded) = P(X < 45)

= P(Z < (44.5 - 50.4)/5.68)

= P(Z < -1.04)

= 0.1492

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

= P(X < 64.5) - P(X < 39.5)

= P(Z < (64.5 - 50.4)/5.68) - P(Z < (39.5 - 50.4)/5.68)

= P(Z < 2.48) - P(Z < -1.92)

= 0.9934 - 0.0274

= 0.9960

d) P(more than 80 of the claims have not been padded) = P(less than 40 claims have been padded)

= P(X < 39.5)

= 0.0274