Trevor is interested in purchasing the local hardware/sporting goods store in the small town of Dove Creek, Montana. After examining accounting records for the past several years, he found that the store has been grossing over $850 per day about 60% of the business days it is open. Estimate the probability that the store will gross over $850 for the following.

a. at least 3 out of 5 business days

b. at least 6 out of 10 business days

c. fewer than 5 out of 10 business days

d. fewer than 6 outof the next 20 business days. If thisactually happened might it shake your confidence inthe statementp=0.60? Might it make you suspect that p is less than 0.60?Explain.

e. more than 17 out of the next 20 business days. Ifthis accually happened might you suspect that p is greater than0.60? Explain.

Please refer to Explanation

Explanation:

Firstly we have to remember that this is a Binomial Distribution.

We have p = 0.6 and you either have over $850 gross or you do not.

A binomial is described by (n choose k) p^k (1-p)^n-k

a) At least 3 out of 5 business days

So what this means is that we should find the probability when there are 3, 4, and 5business days greater than $850 and add them up.

Therefore we will have,

= (5 choose 3) .6^3 (.4)^2 + (5 choose 4) .6^4 (.4)^1 + (5 choose 5).6^5 (.4)^0

= 0.68256

b) At least 6 out of 10 business days.

Repeating the method above,

We will find p(6) + p(7) + p(8) +p(9) + p(10)

Which is,

= (10 choose 6) 0.6^6(0.4)^4 + (10 choose 7) 0.6^7(0.4)^3 + (10choose 8) 0.6^8(0.4)^2 + (10 choose 9) 0.6^9(0.4)^1 + (10 choose 10)0.6^10 (.4)^0

= 0.63310

c) fewer than 5 out of 10

From the previous question, we found at least 6 which meant we found probability when there are 6, 7, 8, 9, or 10 business days that will gross over $850.

Now, fewer than 5 means 4,3, 2, 1 , 0 business days grossing over 850) and seeing as we have already found at least 6 business days, we just need to find p(5) and then add it to p(at least 6) and subtract it from 1.

= (10 choose 5)(.6^5)(.4^5)

= .2006

p(at least 5) = p(at least 6) + p(5)

=0.63310 + 0.2006

= 0.83376

Subtracting from 1,

= 1-0.83376

= 0.1662

d) fewer than 6 out of the next 20 business days.

This is the same as finding p(0) + p(1) + p(2) + p(3) + p(4) +p(5) with n = 20

= (20 choose 0)(0.6^0)(0.4^20) + (20 choose 1)(0.6^1)(0.4^19) + ... +(20 choose 5)(0.6^5)(0.4^15)

= 0.0016

e) More than 17 out of the next 20 business days.

More than 17 means the same as p(18) + p(19) + p(20)

= (20 choose 18)(0.6^18)(0.4^2) + (20 choose 19)(0.6^19)(0.4) + (20choose 20)(0.6^20)(0.4^0)

= .0036

If you need any clarification do comment. Cheers.

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