Trevor is interested in purchasing the local hardware/sporting goods store in th

Question

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. (Round your answers to three decimal places.)

(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 out of the next 20 business days

e) more than 17 out of the next 20 business days

Explanation / Answer

BINOMIAL PROBABILITY
P = n!/[r! * (n - r)] * p^r * (1 - p)^(n - r)

n = TRIALS
r = SUCCESSES
p = PROBABILITY OF SUCCESS [0.60] (60%)

a)

P (r 3)
P (r 3) = P (r = 3or r = 4 or r = 5)
Now that we are armed with the proper technique we will use table 2 in the appendix and a calculator to solve for P (r 3).
P (r 3) = P (.346 + .259 + .078)
P (r 3) = .346 + .259 + .078 =
P (r 3) = 0.683 % chance that at least 3 out of 5 business days grossed over $850.00.

b) at least 6 out of 10 business days
P (r 6)
P (r 6) = P (r = 6 or r = 7 or r = 8 or r = 9 or r = 10)
Now that we are armed with the proper technique we will use table 2 in the appendix and a calculator to solve for P (r 6).
P (r 6) = P (.251 + .215 + .121 + .040 + .006)
P (r 6) = .251 + .215 + .121 + .040 + .006 =
P (r 6) = 0.633 % chance that at least 6 out of 10 business days grossed over $850.00.

c) fewer than 5 out of 10 business days

P (r < 5)
P (r < 5) = P (r = 0 or r = 1 or r = 2 or r = 3 or r = 4 or r = 5)
Now that we are armed with the proper technique we will use table 2 in the appendix and a calculator to solve for P (r < 5).
P (r < 5) = P (.000 + .002 + .011 + .042 + .111 + .201)
P (r < 5) = .000 + .002 + .011 + .042 + .111 + .201 =
P (r < 5) = 0.367 % chance that fewer than 5 out of 10 business days grossed over $850.00.

Similarly all other parts can be found out using the formula given in the start.

For d) part n = 20

p = 0.6 and q = 0.4

We have to find values by taking r = 0,1,2,3,4,5 and add them

P (r 3)
P (r 3) = P (r = 3or r = 4 or r = 5)
Now that we are armed with the proper technique we will use table 2 in the appendix and a calculator to solve for P (r 3).
P (r 3) = P (.346 + .259 + .078)
P (r 3) = .346 + .259 + .078 =
P (r 3) = 0.683 % chance that at least 3 out of 5 business days grossed over $850.00.

b) at least 6 out of 10 business days
P (r 6)
P (r 6) = P (r = 6 or r = 7 or r = 8 or r = 9 or r = 10)
Now that we are armed with the proper technique we will use table 2 in the appendix and a calculator to solve for P (r 6).
P (r 6) = P (.251 + .215 + .121 + .040 + .006)
P (r 6) = .251 + .215 + .121 + .040 + .006 =
P (r 6) = 0.633 % chance that at least 6 out of 10 business days grossed over $850.00.

c) fewer than 5 out of 10 business days

P (r < 5)
P (r < 5) = P (r = 0 or r = 1 or r = 2 or r = 3 or r = 4 or r = 5)
Now that we are armed with the proper technique we will use table 2 in the appendix and a calculator to solve for P (r < 5).
P (r < 5) = P (.000 + .002 + .011 + .042 + .111 + .201)
P (r < 5) = .000 + .002 + .011 + .042 + .111 + .201 =
P (r < 5) = 0.367 % chance that fewer than 5 out of 10 business days grossed over $850.00.

Similarly all other parts can be found out using the formula given in the start.

For d) part n = 20

p = 0.6 and q = 0.4

We have to find values by taking r = 0,1,2,3,4,5 and add them

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