PLEASE ANSWER WITH EXPLANATION. Trevor is interested in purchasing the local har

Question

PLEASE ANSWER WITH EXPLANATION.

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 65% 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

If the outcome described in part (d) actually occurred, might it shake your confidence in the statement p = 0.65? Might it make you suspect that p is less than 0.65? Explain. a.

a. Yes. This is unlikely to happen if the true value of p is 0.65.

b. Yes. This is likely to happen if the true value of p is 0.65.    

c. No. This is unlikely to happen if the true value of p is 0.65.

d. No. This is likely to happen if the true value of p is 0.65.

Explanation / Answer

Solution:

a) 5C3*(0.65)^3*(0.35)^2 + 5C4*(0.65)^4*(0.35)^1 + 5C5*(0.65)^5*(0.35)^0
=10*0.0337+5*0.0625+1*0.1160
~ 0.7655

b) 10C6*(0.65)^6*(0.35)^4 + 10C7*(0.65)^7*(0.35)^3 + 10C8*(0.65)^8*(0.35)^2 + 10C9* (0.65)^9*(0.35)^1 + 10C10*(0.65)^10*(0.35)^0
= 210*0.0011+120*0.0021+45*0.0039+10*0.0072+1*0.0135
~ 0.744

c) 10C4*(0.65)^4*(0.35)^6 + 10C3*(0.65)^3*(0.35)^7 + 10C2*(0.65)^2*(0.35)^8 + 10C1* (0.65)^1*(0.35)^9 + 10C0*(0.65)^0*(0.35)^10
= 210*0.0003+120*0.0002+45*9.51417353e-5+10*5.12301651e-5+1*2.75854735e-5
~ 0.0918

d) 20C5*(0.65)^5*(0.35)^15 + 20C4*(0.65)^4*(0.35)^16 + 20C3*(0.65)^3*(0.35)^17 + 20C2*0.65^2*0.35^18 + 20C1*0.65^1*0.35^19 + 20C0*0.65^0*0.35^20
= 15504*1.68107639e-8+4845*9.05194979e-9+1140*4.87412681e-9+190*2.62452982e- 9+20*1.41320836e-9+1*7.6095835e-10
~ 0.00031

e) 20C20*0.65^20*0.35^0 + 20C19*0.65^19*0.35^1 + 20C18*0.65^18*0.35^2
= 1*0.0002+ 20*9.75937083e-5+ 190*5.25504583e-5
~ 0.012136

a. Yes. This is unlikely to happen if the true value of p is 0.65.

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