Question

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 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

Let $X=$ the number of the claims have been padded: $X\sim B(n, p)$

$$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$

The mean and variance for the binomial variate are

$$mean=E(X)=np,Var(X)=\sigma^2=np(1-p)$$

The probability of success (US adults pad their insurance claims) $p=0.44$

The total number of insurance claims received is $n=120$ (Which is large in size).

The mean and variance of the variable $X$  (the number of padded claims) are 

$$mean=E(X)=np=120(0.44)=52.8>10,$$

$$Var(X)=\sigma^2=np(1-p)=120(0.44)(1-0.44)=29.568>10$$

Normal approximation to binomial distribution:

If $X$  follows binomial distribution with parameters $np$  and $np(1-p)$, the binomial variate tends to normal as follows:$X\sim N(np, np(1-p)) .$ Then

$$Z={X-np \over \sqrt{np(1-p)}}\sim N(0,1)$$

(a) half or more of the claims have been padded 

$$P(X\geq60)=P(Z\geq{60-52.8 \over \sqrt{29.568}})=$$ $$=1-P(Z<{7.2\over \sqrt{29.568}})\approx1-0.907265\approx0.0927$$

(b) fewer than 45 of the claims have been padded 

$$P(X<45)=P(Z<{45-52.8 \over \sqrt{29.568}})=$$ $$=P(Z<{-7.8\over \sqrt{29.568}})\approx0.0757$$

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

$$P(40<X<64)=P({40-52.8 \over \sqrt{29.568}}<Z<{64-52.8 \over \sqrt{29.568}})=$$

$$=P(Z<{64-52.8 \over \sqrt{29.568}})-P({40-52.8 \over \sqrt{29.568}})\approx$$

$$\approx0.980287-0.009287\approx0.9710$$

(d) more than 80 of the claims have not been padded

Continuous CorrectionFactor

$$P(X<20-0.5)=P(Z<{19.5-52.8 \over \sqrt{29.568}})=$$ $$=P(Z<{-33.3\over \sqrt{29.568}})\approx4.56\times10^{-10}\approx0.0000$$