In the following problem, check that it is appropriate to use the normal approxi

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 134 insurance claims to be processed in the next few days. Find the following probabilities. (Round your answers to four decimal places.) fewer than 45 of the claims have been padded continuity correction P (x

Explanation / Answer

p = 0.4

q = 1 - p = 0.6

n = 134

mean = np = 134 X 0.4 = 53.6

SD = sqrt(npq) = sqrt(134X0.4X0.6) = 5.6710

(a) P(X<45):

Transforming to Stadard Normal Variate:

Z = (X - Mean)/SD = (45-53.6)/5.6710 = - 1.5165

Since Z is negative, it lies on LHS of mid value.

Table of area under stadard normal curve gives area from mid value to Z = 1.5165 as area = 0.4357.

So, P(X<45) = 0.5 - 0.4357 = 0.0643

(b) P(40 < X < 64):

For X = 40:

Z = (40-53.6)/5.6710 = - 2.3982

Since Z is negative, it lies on LHS of mid value. Table of Area Under Standard Normal Curve gives area from mid value to Z = - 2.3982 as area = 0.4918

For X = 64:
Z = (64-53.6)/5.6710 = 1.8834

Since Z is positive, it lies on RHS of mid value. Table of Area Under Standard Normal Curve area from mid value to Z = 1.8834 as area = 0.4699.

So, P(40 < X < 64) = 0.4918 + 0.4699 = 0.9617

(c) P(X>80):

Z = (80-53.6)/5.6710 = 4.6553

Since Z is positive, it lies on RHS of mid value.

Table of Area Under Standard Normal Curve gives area from mid value to Z = 4.6553 as area = 0.5 nearly.

So, P(X>80) = 0 nearly

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