Question

can you check if my answers are correct and can you please type the correct answers for each question

1. 5 points) Trevor is interested in purchasing the local hardware/electronic goods store in a small town in South Ohio. After examining accounting records for the past several years, he found that the store has been grossing over $850 per day about 60% of the business days it is open.

Estimate the probability that the store will gross over $850

1. at least 6 out of 10 business days.   0.633

2. at most 3 of the 10 business days. 0.0547

3. fewer than 7 out of 20 business days. 0.0065

4. more than 16 out of 20 business days. 0.0159

5. What is the expected (mean) number of business days for which the store would gross over $850 in a month (30 days)? 18 days

Answer 1

Answer:

p = .6 and q = 1 - p= .4

a.

=> P(at least 6 out of 10 business days)

=> P(x=6) + P(x=7) + P(x=8) + P(x=9) + P(x=10)

=> (10 C 6) * (0.6)^6 * (0.4)^4 + (10 C 7) * (0.6)^7 * (0.4)^3 + (10 C 8) * (0.6)^8 * (0.4)^2 + (10 C 9) * (0.6)^9 * (0.4)^1 + (10 C 10) * (0.6)^10 * (0.4)^0

=> 210 * (0.6)^6 * (0.4)^4 + 120* (0.6)^7 * (0.4)^3 + 45 * (0.6)^8 * (0.4)^2 + 10 * (0.6)^9 * (0.4)^1 + 1 * (0.6)^10 * (0.4)^0

=> 0.25 + 0.214+ 0.12+0.0403+0.006

=0.63

b.

P(at most 3 of the 10)

=> P(x=0) + P(x=1) + P(x=2) + P(x=3)

=> (10 C 0) * .6^1 * 0.4^10 + (10 C 1) * 0.6^1 * .4^9 + (10 C 2) * 0.6^2 * 0.4 ^ 8 + (10 C 3) * 0.6 ^ 3 * 0.4^7

=0.058

c.

=> P(fewer than 7 out of 20)

=> P(x=0) + P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x=5) + P(x=6)

=> (20 C 0) * 0.6^0 * 0.4^20 + (20 C 1) * 0.6^1 * 0.4^19 + (20 C 2) * 0.6^2 * 0.4^18 + (20 C 3) * 0.6^3 * 0.4^17 +(20 C 4) * 0.6^4 * 0.4^16 + (20 C 5) * 0.6^5 * 0.4^15 + (20 C 6) * 0.6^6 * 0.4^14

=> 0.00653

d.

P(more than 16 out of 20)

=> P(x=17) +  P(x=18) + P(x=19) + P(x=20)

=> (20 C 17) * 0.6^17 * .4^3 + (20 C 18) * 0.6^18* .4^2 + (20 C 19) * 0.6^19 * .4^1 + (20 C 20) * 0.6^20 * 0.4^0

=> 0.016