# n the following problem, check

n the following problem, check that it is appropriate to use thenormal approximation to the binomial. Then use the normaldistribution 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 insuranceclaims! Suppose that you are the director of an insuranceadjustment office. Your office has just received 120 insuranceclaims to be processed in the next few days. Find the followingprobabilities. (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:

Mean = n * P = ( 120 * 0.4 ) = 48Variance = n * P * Q = ( 120 * 0.4 * 0.6 ) = 28.8Standard deviation = = 5.3666

Part a)

P ( X >= 60 )Using continuity correctionP ( X > n – 0.5 ) = P ( X > 60 – 0.5 ) =P ( X > 59.5 )

P ( X > 59.5 ) = 1 – P ( X < 59.5 )Standardizing the valueZ = ( 59.5 – 48 ) / 5.3666Z = 2.14P ( Z > 2.14 )P ( X > 59.5 ) = 1 – P ( Z < 2.14 )P ( X > 59.5 ) = 1 – 0.9838P ( X > 59.5 ) = 0.0162

Part b)

P ( X < 45 )Using continuity correctionP ( X < n – 0.5 ) = P ( X < 45 – 0.5 ) = P ( X < 44.5)

P ( X < 44.5 )Standardizing the valueZ = ( 44.5 – 48 ) / 5.3666Z = -0.65P ( X < 44.5 ) = P ( Z < -0.65 )P ( X < 44.5 ) = 0.2578

Part c)

P ( 40 < X < 64 )Using continuity correctionP ( n + 0.5 < X < n – 0.5 ) = P ( 40 + 0.5 < X < 64 -0.5 ) = P ( 40.5 < X < 63.5 )

P ( 40.5 < X < 63.5 )Standardizing the valueZ = ( 40.5 – 48 ) / 5.3666Z = -1.4Z = ( 63.5 – 48 ) / 5.3666Z = 2.89P ( -1.4 < Z < 2.89 )P ( 40.5 < X < 63.5 ) = P ( Z < 2.89 ) – P ( Z < -1.4)P ( 40.5 < X < 63.5 ) = 0.9981 – 0.0811P ( 40.5 < X < 63.5 ) = 0.9169

Part d)

Mean = n * P = ( 120 * 0.6 ) = 72Variance = n * P * Q = ( 120 * 0.6 * 0.4 ) = 28.8Standard deviation = = 5.3666

P ( X > 80 )Using continuity correctionP ( X > n + 0.5 ) = P ( X > 80 + 0.5 ) = P ( X > 80.5)

P ( X > 80.5 ) = 1 – P ( X < 80.5 )Standardizing the valueZ = ( 80.5 – 72 ) / 5.3666Z = 1.58P ( Z > 1.58 )P ( X > 80.5 ) = 1 – P ( Z < 1.58 )P ( X > 80.5 ) = 1 – 0.9429P ( X > 80.5 ) = 0.0571