Question

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.

Do you try to pad an insurance claim to cover your deductible? About 37% 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

Using Normal Approximation to Binomial
Mean = n * P = ( 120 * 0.37 ) = 44.4
Variance = n * P * Q = ( 120 * 0.37 * 0.63 ) = 27.972
Standard deviation = √(variance) = √(27.972) = 5.2889

Part a)

P ( X >= 60 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 60 - 0.5 ) =P ( X > 59.5 )

X ~ N ( µ = 44.4 , σ = 5.2889 )
P ( X > 59.5 ) = 1 - P ( X < 59.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 59.5 - 44.4 ) / 5.2889
Z = 2.86
P ( ( X - µ ) / σ ) > ( 59.5 - 44.4 ) / 5.2889 )
P ( Z > 2.86 )
P ( X > 59.5 ) = 1 - P ( Z < 2.86 )
P ( X > 59.5 ) = 1 - 0.9979
P ( X > 59.5 ) = 0.0021

Part b)

P ( X < 45 )
Using continuity correction
P ( X < n - 0.5 ) = P ( X < 45 - 0.5 ) = P ( X < 44.5 )

X ~ N ( µ = 44.4 , σ = 5.2889 )
P ( X < 44.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 44.5 - 44.4 ) / 5.2889
Z = 0.02
P ( ( X - µ ) / σ ) < ( 44.5 - 44.4 ) / 5.2889 )
P ( X < 44.5 ) = P ( Z < 0.02 )
P ( X < 44.5 ) = 0.5080

Part c)

P ( 40 <= X <= 64 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 40 - 0.5 < X < 64 + 0.5 ) = P ( 39.5 < X < 64.5 )

X ~ N ( µ = 44.4 , σ = 5.2889 )
P ( 39.5 < X < 64.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 39.5 - 44.4 ) / 5.2889
Z = -0.93
Z = ( 64.5 - 44.4 ) / 5.2889
Z = 3.8
P ( -0.93 < Z < 3.8 )
P ( 39.5 < X < 64.5 ) = P ( Z < 3.8 ) - P ( Z < -0.93 )
P ( 39.5 < X < 64.5 ) = 0.9999 - 0.1762
P ( 39.5 < X < 64.5 ) = 0.8237

Part d)

Using Normal Approximation to Binomial
Mean = n * P = ( 120 * 0.63 ) = 75.6
Variance = n * P * Q = ( 120 * 0.63 * 0.37 ) = 27.972
Standard deviation = √(variance) = √(27.972) = 5.2889

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

X ~ N ( µ = 75.6 , σ = 5.2889 )
P ( X > 80.5 ) = 1 - P ( X < 80.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 80.5 - 75.6 ) / 5.2889
Z = 0.93
P ( ( X - µ ) / σ ) > ( 80.5 - 75.6 ) / 5.2889 )
P ( Z > 0.93 )
P ( X > 80.5 ) = 1 - P ( Z < 0.93 )
P ( X > 80.5 ) = 1 - 0.8238
P ( X > 80.5 ) = 0.1762