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 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

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.

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

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

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

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

(e) more than 17 out of the next 20 business days Incorrect: .

Explanation / Answer

Solution:

Part a

We are given n = 5, p = 0.65

We have to find P(X3)

P(X3) = 1 – P(X<3)

P(X3)= 1 – P(X2)

P(X3) = 1 – 0.235169 (by using binomial table or excel)

P(X3) = 0.764831

Required probability = 0.765

Part b

We are given n = 10, p = 0.65

We have to find P(X6)

P(X6) = 1 – P(X<6)

P(X6) = 1 – P(X5)

P(X6) = 1 - 0.248504 (by using binomial table or excel)

P(X6) = 0.751496

Required probability = 0.751

Part c

We are given n = 10, p = 0.65

We have to find P(X<5)

P(X<5) = P(X4)

P(X<5) = 0.094934 (by using binomial table or excel)

Required probability = 0.095

Part d

We are given n = 20, p = 0.65

We have to find P(X<6)

Here, we have to use normal approximation to binomial distribution because n*p = 20*0.65 = 13 > 5 and n*q = 20*0.35 = 7 > 5

Mean = n*p = 20*0.65 = 13

SD = sqrt(n*p*q) = sqrt(20*0.65*0.35) = 2.133073

Z = (X – mean) / SD

Z = (6 – 13) / 2.133073

Z = -3.28165

P(Z<-3.28165) = 0.000516 (by using z-table or excel)

Required probability = 0.001

Yes, this is unlikely to happen if the true value of p is 0.65.

Part e

Here, we have to find P(X>17)

P(X>17) = 1 – P(X17)

We are given n = 20, p = 0.65

Here, we have to use normal approximation to binomial distribution because n*p = 20*0.65 = 13 > 5 and n*q = 20*0.35 = 7 > 5

Mean = n*p = 20*0.65 = 13

SD = sqrt(n*p*q) = sqrt(20*0.65*0.35) = 2.133073

Z = (X – mean) / SD

Z = (17 – 13) / 2.133073

Z = 1.875229

P(Z< 1.875229) = 0.969619 (by using z-table or excel)

P(X17) = 0.969619

P(X>17) = 1 – P(X17)

P(X>17) = 1 – 0.969619

P(X>17) = 0.030381

Required probability = 0.030

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