Question 413514: 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 128 insurance claims to be processed in the next few days. What is the probability that: (n =128, p = .40, q = .60) Binomial problem
I DO NOT HAVE A CALCULATOR TO DO IT BY binomcdf AND NEED TO DO THIS BY PAPER AND PENCIL.
(a)Half or more of the claims have been padded?
(b) 44 or less of the claims have been padded?
(c) From 40 to 64 of the claims have been padded?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
If you mean that you don't have a TI-83+/84+ Graphing Calculator so that you can't do the cumulative binomial function directly, then I can show you how to do this on a regular calculator -- though it is still a bunch of work.
On the other hand if you really mean that you cannot use a calculator of any kind and that you must do this with pencil and paper, then go put on the coffee now because you have a L-O-N-G night of arithmetic ahead of you.
The probability of  successes in  trials where  is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
Where ) is the number of combinations of things taken at a time and is calculated by !}) . Also note that for your situation 
A cumulative probability is just the sum of probabilities. Here we have the calculation of the probability of  or fewer occurances]
\ =\ \sum_{i=0}^k\left(n\cr i\right\)p^i\left(1\,-\,p\right)^{n\,-\,i})
Note that the probability of or fewer occurances is the complement of  or more occurences, in other words:
\ =\ 1\ -\ P_n(\leq{k},p))
For your first situation, which can be restated as AT LEAST 64 out of 128 occurences with a probability of an individual occurence of 0.4 we need to calculate 1 minus the probability of AT MOST 63 out of 128.
\ =\ 1\ -\ P_{128}(\leq{63},0.4))
\ =\ 1\ -\ P_n(\leq{63},0.4)\ =\ 1\ -\ \sum_{i=0}^{63}\left(128\cr\ \ i\right\)\(0.4)^i\left(0.6\right)^{128\,-\,i})
Since you have communicated with this website using a computer, either Windows or a Mac, then you have scientific calculator built in to the computer. I suggest you use it.
For the first problem, you need to calculate the value of 64 terms and then add them together.
\ =\cr\ \ \ \ \ \ \ \ \ \ \ 1\ -\ \left(\left(128\cr\ \ 0\right\)(0.4)^0\left(0.6\right)^{128}\ +\ \left(128\cr\ \ 1\right\)(0.4)^1\left(0.6\right)^{127}\ +\ \cdots\ +\ \left(128\cr\ 62\right\)(0.4)^{62}\left(0.6\right)^{66}\ +\ \left(128\cr\ 63\right\)(0.4)^{63}\left(0.6\right)^{65}\right))
You understand why I say you have a very long night ahead if you are doing this by pencil and paper?
The answer to the second problem is found by:
\ =\ \sum_{i=0}^{44}\left(128\cr\ \ i\right\)(0.4)^i\left(0.6\right)^{128\,-\,i})
The third problem, if you include the probability of exactly 40 and exactly 64:
=\ \sum_{i=40}^{64}\left(128\cr\ \ i\right\)(0.4)^i\left(0.6\right)^{128\,-\,i})
Or if you exclude the end points:
=\ \sum_{i=41}^{63}\left(128\cr\ \ i\right\)(0.4)^i\left(0.6\right)^{128\,-\,i})
Good luck. If you have the ability to use MS Excel or the Numbers application on a Mac, write back and I'll show you how to do these three problems in about 2 minutes total.
John
My calculator said it, I believe it, that settles it
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