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 70% 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 = 0.70? Might it make you suspect that p is less than 0.70? Explain. Yes. This is unlikely to happen if the true value of p is 0.70. Yes. This is likely to happen if the true value of p is 0.70. No. This is unlikely to happen if the true value of p is 0.70. No. This is likely to happen if the true value of p is 0.70. (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 0.70? Explain. Yes. This is unlikely to happen if the true value of p is 0.70. Yes. This is likely to happen if the true value of p is 0.70. No. This is unlikely to happen if the true value of p is 0.70. No. This is likely to happen if the true value of p is 0.70.

Answer 1

(a)

Let X be the number of days that the store will gross over $850 for 5 business days. Then X ~ Binomial(n = 5, p = 0.7)

Using Normal approximation of Binomial distribution, X follows Normal distribution with mean = 5 * 0.7 = 3.5 and standard deviation = = 1.024695

Probability that at least 3 out of 5 business days, the store will gross over $850 = P(X 3)

= P(X > 2.5) (Using continuity correction)

= P[Z > (2.5 - 3.5) /  1.024695]

= P[Z > -0.9759]

= 0.835

(b)

Let X be the number of days that the store will gross over $850 for 10 business days. Then X ~ Binomial(n = 10, p = 0.7)

Using Normal approximation of Binomial distribution, X follows Normal distribution with mean = 10 * 0.7 = 7 and standard deviation = = 1.449138

Probability that at least 6 out of 10 business days, the store will gross over $850 = P(X 6)

= P(X > 5.5) (Using continuity correction)

= P[Z > (5.5 - 7) /  1.449138]

= P[Z > -1.0351]

= 0.850

(c)

Probability that at fewer than 5 business days, the store will gross over $850 = P(X < 5)

= P(X < 4.5) (Using continuity correction)

= P[Z < (4.5 - 7) /  1.449138]

= P[Z < -1.7251]

= 0.042

(d)

Let X be the number of days that the store will gross over $850 for 20 business days. Then X ~ Binomial(n = 20, p = 0.7)

Using Normal approximation of Binomial distribution, X follows Normal distribution with mean = 20 * 0.7 = 14 and standard deviation = = 2.04939

Probability that fewer than 6 business days, the store will gross over $850 = P(X < 6)

= P(X < 5.5) (Using continuity correction)

= P[Z < (5.5 - 14) /  2.04939]

= P[Z < -4.1476]

= 0.00002

Since the probability for part (d) is very close to zero and less than 0.05, this is unlikely to happen if the true value of p is 0.70.

Yes. This is unlikely to happen if the true value of p is 0.70.

(e)

Probability that more than 17 business days, the store will gross over $850 = P(X > 17)

= P(X > 17.5) (Using continuity correction)

= P[Z > (17.5 - 14) /  2.04939]

= P[Z > 1.7078]

= 0.04383

Since the probability for part (d) is less than 0.05, this is unlikely to happen if the true value of p is 0.70.

Yes. This is unlikely to happen if the true value of p is 0.70.