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 If the outcome described in part (e) actually occurred, might you suspect that p is greater 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.

A) P(X ≥ 3 ) = 0.7648

B) P(X ≥ 6) = 0.7515

C) P(X < 5) = 0.0949

D) P(X < 6) = 0.00031113

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

E) P(X ≥ 17) = 0.012118

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

Step-by-step explanation:

Solving this question means we have to deal with binomial probability distribution which follows a formula;

P(X = x) = C(n, x) * p^(x) * (1 - p)^(n - x)

We are given p = 65% = 0.65

A) probability of getting At least 3 out of the 5 business days, we have;

P(X ≥ 3 ) = P(X = 3) + P(X = 4) + P(X = 5)

And n = 5

Thus;

P(X = 3) = C(5, 3) × 0.65³ × (1 - 0.65)^(5 - 3)

P(X = 3) = 0.3364

P(X = 4) = C(5, 4) × 0.65⁴ × (1 - 0.65)^(5 - 4)

P(X = 4) = 0.3124

P(X = 5) = C(5, 5) × 0.65^(5) × (1 - 0.65)^(5 - 5)

P(X = 5) = 0.116

Thus;

P(X ≥ 3 ) = 0.3364 + 0.3124 + 0.116

P(X ≥ 3 ) = 0.7648

B) Probability of getting at least 6 out of the 10 business days.

P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

From online binomial calculator, we have;

P(X = 6) = 0.2377

P(X = 7) = 0.2522

P(X = 8) = 0.1757

P(X = 9) = 0.0725

P(X = 10) = 0.0134

P(X ≥ 6) = 0.2377 + 0.2522 + 0.1757 + 0.0725 + 0.0134

P(X ≥ 6) = 0.7515

C) probability of getting fewer than 5 out of the 10 business days.

P(X < 5) = P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1)

From online binomial calculator, we have;

P(X = 4) = 0.0689

P(X = 3) = 0.0212

P(X = 2) = 0.0043

P(X = 1) = 0.0005

P(X < 5) = 0.0689 + 0.0212 + 0.0043 + 0.0005

P(X < 5) = 0.0949

D)probability of getting fewer than 6 out of the next 20 business days;

P(X < 6) = P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1)

From online binomial calculator, we have;

P(X = 5) = 0.000261

P(X = 4) = 0.000044

P(X = 3) = 0.0000056

P(X = 2) = 0.0000005

P(X = 1) = 0.00000003

P(X < 6) = 0.000261 + 0.000044 + 0.0000056 + 0.0000005 + 0.00000003

P(X < 6) = 0.00031113

Yes, in this scenario, It is likely to happen and I would think p is less than 0.65 because probability is less than 1 out of 1000.

E) probability of getting more than 17 out of the next 20 business days.

P(X ≥ 17) = P(X = 18) + P(X = 19) + P(X = 20)

From online binomial calculator, we have;

P(X = 18) = 0.0099846

P(X = 19) = 0.00195187

P(X = 20) = 0.00018125

P(X ≥ 17) = 0.0099846 + 0.00195187 + 0.00018125

P(X ≥ 17) = 0.012118

Probability is around 12 in 100 and not up to 50%. Thus, This is unlikely to happen if the true value of p is 0.65

do you have the measurements?

898.1666667

step-by-step explanation:

5389/6 = 898.1666667