Let "X=" the number of the claims have been padded: "X\\sim B(n, p)"
The mean and variance for the binomial variate are
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
"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
(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
"=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
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