In: Statistics and Probability
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 36% 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 140 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:
We are given
n = 140
p = 0.36
q = 1 – p = 1 - 0.36 = 0.64
np = 140*0.36 = 50.4 > 5
nq = 140*0.64 = 89.6 > 5
np > 5, nq > 5, so we can use normal approximation.
Mean = np = 140*0.36 = 50.4
SD = sqrt(npq) = sqrt(140*0.36*0.64) = 5.679437
Z = (X – mean) / SD
Part a
Here, we have to find P(X≥70)
P(X≥70) = P(X>69.5) (by using continuity correction)
P(X>69.5) = 1 – P(X<69.5)
Z = (69.5 - 50.4)/ 5.679437
Z = 3.363009
P(Z<3.363009) = 0.999615
(by using z-table)
P(X>69.5) = 1 – P(X<69.5)
P(X>69.5) = 1 – 0.999615
P(X>69.5) = 0.000385
Required probability = 0.0004
Part b
Here, we have to find P(X<45)
P(X<45) = P(X<44.5) (by using continuity correction)
Z = (44.5 - 50.4)/ 5.679437
Z = -1.03884
P(Z<-1.03884) = 0.149441
(by using z-table)
Required probability = 0.1494
Part c
Here, we have to find P(40<X<64)
P(40<X<64) = P(X<64) – P(X<40)
Find P(X<64)
Z = (64 - 50.4)/ 5.679437
Z = 2.394604
P(Z<2.394604) = 0.991681
(by using z-table)
Now find P(X<40)
Z = (40 - 50.4)/ 5.679437
Z =-1.83117
P(Z<-1.83117) = 0.033538
(by using z-table)
P(40<X<64) = P(X<64) – P(X<40)
P(40<X<64) = 0.991681 – 0.033538
P(40<X<64) = 0.958143
Required probability = 0.9581
Part d
Here, we have to find P(X>80 of the claims not been padded) = P(X>(140 – 80) claims have padded)
That is, we have to find P(X>60)
P(X>60) = P(X>60.5) (by using continuity correction)
P(X>60.5) = 1 – P(X<60.5)
Z = (60.5 - 50.4)/ 5.679437
Z =1.778345
P(Z<1.778345) = 0.962326
(by using z-table)
P(X>60.5) = 1 – P(X<60.5)
P(X>60.5) = 1 – 0.962326
P(X>60.5) = 0.037674
Required probability = 0.0377