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 120 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

Using normal approximation to the binomial:

n = 120, p = 0.40

Mean, µ = n*p = 120 * 0.4 = 48

Standard deviation, σ = √(n*p*(1-p)) = √(120 * 0.4 * 0.6) = 5.3666

a) Probability that half or more of the claims have been padded, P(X ≥ 60) =

Using continuity correction :

P(X ≥ 60-0.5)

= P((X - µ)/σ ≥ (59.5 - 48)/5.3666)

= P(z ≥ 2.1429)

= 1 - P(z < 2.1429)

Using excel function:

= 1 - NORM.S.DIST(2.1429, 1)

= 0.0161

b) Probability that fewer than 45 of the claims have been padded, P(X < 45) =

Using continuity correction :

P(X < 45-0.5)

= P((X - µ)/σ < (44.5 - 48)/5.3666)

= P(z < -0.6522)

Using excel function:

= NORM.S.DIST(-0.6522, 1)

= 0.2571

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

Using continuity correction :

P(40-0.5 ≤ X ≤ 64+0.5)

= P((39.5 - 48)/5.3666 ≤ (X - µ)/σ ≤ (64.5 - 48)/5.3666)

= P( -1.5839 ≤ z ≤ 3.0746)

= P(z < 3.0746) - P(z < -1.5839)

Using excel function:

= NORM.S.DIST(3.0746, 1) - NORM.S.DIST(-1.5839, 1)

= 0.9423

d) Probability that more than 80 of the claims have not been padded = probability that less than or equal to 40 claims have been padded, P(X ≤ 40) =

Using continuity correction :

P(X ≤ 40+0.5)

= P((X - µ)/σ ≤ (40.5 - 48)/5.3666)

= P(z ≤ -1.3975)

Using excel function:

= NORM.S.DIST(-1.3975, 1)

= 0.0811

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Using normal distribution :

n = 120, p = 0.40

Mean, µ = n*p = 120 * 0.4 = 48

Standard deviation, σ = √(n*p*(1-p)) = √(120 * 0.4 * 0.6) = 5.3666

a) Probability that half or more of the claims have been padded, P(X ≥ 60) =

= P((X - µ)/σ ≥ (60 - 48)/5.3666)

= P(z ≥ 2.2361)

= 1 - P(z < 2.2361)

Using excel function:

= 1 - NORM.S.DIST(2.2361, 1)

= 0.0127

b) Probability that fewer than 45 of the claims have been padded, P(X < 45) =

= P((X - µ)/σ < (45 - 48)/5.3666)

= P(z < -0.559)

Using excel function:

= NORM.S.DIST(-0.559, 1)

= 0.2881

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

= P((40 - 48)/5.3666 ≤ (X - µ)/σ ≤ (64 - 48)/5.3666)

= P( -1.4907 ≤ z ≤ 2.9814)

= P(z < 2.9814) - P(z < -1.4907)

Using excel function:

= NORM.S.DIST(2.9814, 1) - NORM.S.DIST(-1.4907, 1)

= 0.9305

d) Probability that more than 80 of the claims have not been padded = probability that less than or equal to 40 claims have been padded, P(X ≤ 40) =

= P((X - µ)/σ ≤ (40 - 48)/5.3666)

= P(z ≤ -1.4907)

Using excel function:

= NORM.S.DIST(-1.4907, 1)

= 0.0680