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Please help with the rest of these? Use Problem I1 to answer questions 7-13 belo

ID: 3061814 • Letter: P

Question

Please help with the rest of these? Use Problem I1 to answer questions 7-13 below Problem sample of 20 dish owners. /l:20 Sores al-5s 7. Are these data discrete or continuous?2 8. How do you know? 8. II. An industry representative claims that 50 percent of all satellite dish owners subscribe to at least mium movie channel. In an attempt to justify this claim, the representative will poll a randomly selected Continous 7. Assuming independence, 1) show the binomial formula for this situation and 2) compute the following binomial probabilities using technology (table, calculator, Excel): 10, P(x = 10)- 10. 11. P(x 211)- 12. P(2 s x s6)- 12. 13, P(x 10) = 13.

Explanation / Answer

n = 20

p = 0.5

9) P(X = 8) = 20C8 * (0.5)^8 * (0.5)^12 = 0.1201

10) P(x = 10) = 20C10 * (0.5)^10 * (0.5)^10 = 0.1762

11) P(X > 11) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

   = 20C11 * (0.5)^11 * (0.5)^9 + 20C12 * (0.5)^12 * (0.5)^8 + 20C13 * (0.5)^13 * (0.5)^7 + 20C14 * (0.5)^14 * (0.5)^6 + 20C15 * (0.5)^15 * (0.5)^5 + 20C16 * (0.5)^16 * (0.5)^4 + 20C17 * (0.5)^17 * (0.5)^3 + 20C18 * (0.5)^18 * (0.5)^2 + 20C19 * (0.5)^19 * (0.5)^1 + 20C20 * (0.5)^20 * (0.5)^0 = 0.4119

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

                        = 20C2 * (0.5)^2 * (0.5)^18 + 20C3 * (0.5)^3 * (0.5)^17 + 20C4 * (0.5)^4 * (0.5)^16 + 20C5 * (0.5)^5 * (0.5)^15 + 20C6 * (0.5)^6 * (0.5)^14 = 0.0576

13) P(X < 10) = 1 - P(X > 10)

                      = 1 - ( P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20))

                     = 1 - ( 20C11 * (0.5)^11 * (0.5)^9 + 20C12 * (0.5)^12 * (0.5)^8 + 20C13 * (0.5)^13 * (0.5)^7 + 20C14 * (0.5)^14 * (0.5)^6 + 20C15 * (0.5)^15 * (0.5)^5 + 20C16 * (0.5)^16 * (0.5)^4 + 20C17 * (0.5)^17 * (0.5)^3 + 20C18 * (0.5)^18 * (0.5)^2 + 20C19 * (0.5)^19 * (0.5)^1 + 20C20 * (0.5)^20 * (0.5)^0)

                   = 1 - 0.4119

                   = 0.5881