Here,
P(a random insurance claim is padded)= p =0.39
n= 132,q=1-p= 0.61
Since, np =132*0.39 = 51.48
And n(1-p) = 132*0.61 = 80.52
Both of which are greater than 5 as needed.
Hence, normal approximation can be used here.
Mean = np = 51.48
Std dev = sqrt(npq) = 5.60382
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