In the following problem check that it is appropriate to use

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 40% 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 122 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

Solution

Normal Approximation to Binomial Distribution
Mean ( np ) =122 * 0.4 = 48.8
Standard Deviation ( npq )= 122*0.4*0.6 = 5.4111
Normal Distribution = Z= X- u / sd                   
a)
P(X < 61) = (61-48.8)/5.4111
= 12.2/5.4111= 2.2546
= P ( Z <2.2546) From Standard NOrmal Table
= 0.9879                  
P(half or more of the claims have been padded) = 1 - P(X < 61) = 1 - 0.9879 = 0.0121

b)
P(X < 45) = (45-48.8)/5.4111
= -3.8/5.4111= -0.7023
= P ( Z <-0.7023) From Standard NOrmal Table
= 0.2413                  

c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 40) = (40-48.8)/5.4111
= -8.8/5.4111 = -1.6263
= P ( Z <-1.6263) From Standard Normal Table
= 0.05194
P(X < 64) = (64-48.8)/5.4111
= 15.2/5.4111 = 2.809
= P ( Z <2.809) From Standard Normal Table
= 0.99752
P(40 < X < 64) = 0.99752-0.05194 = 0.9456                  

d)
claims have not been padded are 60%

Mean ( np ) =122 * 0.6 = 73.2
Standard Deviation ( npq )= 122*0.6*0.4 = 5.4111

Normal Distribution = Z= X- u / sd                   
P(X > 80) = (80-73.2)/5.4111
= 6.8/5.4111 = 1.2567
= P ( Z >1.257) From Standard Normal Table
= 0.1044