Question

About 40% of all US adults will try to pad their insurance claims. Suppose that you are the director of an insurance adjustment office. Your office has just received 128 insurance claims to be processed int the next few days. What is the probability that -

half or more claims have been padded?

fewer than 45 of the claims have been padded:

From 40 to 64 of the claims have been padded?

More than 80 of the claims have not been padded?

Answer 1

Probability Mass function of Binomial distribution is

Mean = n * P = 128 * 0.4 = 51.2

Standard deviation =

P ( X > 63.5 ) = 1 - P ( X < 63.5 )

Using continuity correction

P ( X >= 64) = P ( X > 64 - 0.5 ) = P ( X >= 63.5)
Standardizing the value

Z = ( 63.5 - 51.2 ) / 5.5426
Z = 2.22

P ( Z > 2.22 )
P ( X > 63.5 ) = 1 - P ( Z < 2.22 )
P ( X > 63.5 ) = 1 - 0.9868
P ( X > 63.5 ) = 0.0132

P ( X < 45 )

Using continuity correction

P ( X < 45) = P ( X < 45 - 0.5 ) = P ( X < 44.5)

P ( X < 44.5 )
Standardizing the value

Z = ( 44.5 - 51.2 ) / 5.5426
Z = -1.21

P ( X < 44.5 ) = P ( Z < -1.21 )
P ( X < 44.5 ) = 0.1131

P ( 40 < X < 64 )

Using continuity correction

P ( a < X < b ) = P ( a - 0.5 < X < b + 0.5)
P ( 39.5 < X < 64.5 )
Standardizing the value

Z = ( 39.5 - 51.2 ) / 5.5426
Z = -2.11
Z = ( 64.5 - 51.2 ) / 5.5426
Z = 2.4
P ( -2.11 < Z < 2.4 )
P ( 39.5 < X < 64.5 ) = P ( Z < 2.4 ) - P ( Z < -2.11 )
P ( 39.5 < X < 64.5 ) = 0.9918 - 0.0174
P ( 39.5 < X < 64.5 ) = 0.9744

P ( X > 80)

Using continuity correction

P ( X > 80 ) = P ( X > 80 + 0.5 ) = P ( X > 80.5)

P ( X > 80.5 ) = 1 - P ( X < 80.5 )
Standardizing the value

Z = ( 80.5 - 51.2 ) / 5.5426
Z = 5.29

P ( Z > 5.29 )
P ( X > 80.5 ) = 1 - P ( Z < 5.29 )
P ( X > 80.5 ) = 1 - 1
P ( X > 80.5 ) = 0