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 55% 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 = 0.55? Might it make you suspect that p is less than 0.55? Explain.

Yes. This is unlikely to happen if the true value of p is 0.55.

Yes. This is likely to happen if the true value of p is 0.55.

No. This is unlikely to happen if the true value of p is 0.55.

No. This is likely to happen if the true value of p is 0.55.

(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 0.55? Explain.

Yes. This is unlikely to happen if the true value of p is 0.55.

Yes. This is likely to happen if the true value of p is 0.55.

No. This is unlikely to happen if the true value of p is 0.55.

No. This is likely to happen if the true value of p is 0.55.

Answer 1

Solution:

(a)
P(at least 3)

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

= (5C3 * .55^3 * .45^2) + (5C4 * .55^4 * .45^1) + (5C5 * .55^5 * .45^0)

= 0.5931

b)

Mean = Expected Value = μ_x̅ = n•p = (10)(0.55) = 5.5

n•p•(1 - p) = (10)(0.55)(1 - 0.55) = 2.475

Standard Deviation = σ_x̅ = √ n•p•(1 - p) = √2.475 = 1.5732132722552274


Finding probability of getting at least 6 success(es):


Using Binomial Distribution: P(X ≥ 6) = 0.5044

c)

Mean = Expected Value = μ_x̅ = n•p = (10)(0.55) = 5.5

n•p•(1 - p) = (10)(0.55)(1 - 0.55) = 2.475

Standard Deviation = σ_x̅ = √ n•p•(1 - p) = √2.475 = 1.5732132722552274


Finding probability of getting less than 5 success(es):


Using Binomial Distribution: P(X < 5) = 0.2615

d)

Mean = Expected Value = μ_x̅ = n•p = (20)(0.55) = 11

n•p•(1 - p) = (20)(0.55)(1 - 0.55) = 4.95

Standard Deviation = σ_x̅ = √ n•p•(1 - p) = √4.95 = 2.224859546128699


Finding probability of getting less than 6 success(es):


Using Binomial Distribution: P(X < 6) = 0.0064

No. This is unlikely (but still possible) to happen if the true value of p is 0.55

e) Mean = Expected Value = μ_x̅ = n•p = (20)(0.55) = 11

n•p•(1 - p) = (20)(0.55)(1 - 0.55) = 4.95

Standard Deviation = σ_x̅ = √ n•p•(1 - p) = √4.95 = 2.224859546128699


Finding probability of getting greater than 17 success(es):


Using Binomial Distribution P(X > 17) = 0.0009

No. This is unlikely to happen if the true value of p is 0.50.