Question

3. Identifying Binomial Distribution: Determine whether the given procedure resalts in a binommial distribution (or a distribution that can be treated as binomial). For those that are not bisomial, identify at least one requirement that is not satisfied. a) In a U.S. Cellular survey of 500 smartphone users, subjects are asked if they find abbreviations (sach as LOL, or BFF) annoying, and each response was recorded as Yes" or "Other In a U.S. Cellular survey of 500 amartphone users, subjects are asked if they find abbreviations (such as LOL, or BFF) annoying, and each response was recorded as "Yes "No" or "not sure" b) In an AARP Bulletin Sarvey, 1019 different adults were randomly selected without replacement. Respondents c) were asked if they have cne or more credit cards, and responses were recorded as "Yes" and "No" Answer: a) b c) 4. SAT TEST Assume that random guesses are made for eight maltiple choice questions on an SAT test, so that n-8 trials, each with probability of success ( correct) given by p-0.20. Find the indicated probability for the nunber of correct answers. a) Find the probability that the number x of correct answers is exactly 6. b) Find the probability that x is at least 4. c) Find the probability that x is fewer than 3. d) Find the probability that at least one answer is correct. e) Find the probability of no correct answer. Answer: a) bi 5.SAT TEST Assume n-10. a) Use the range rule of thumb to identify the limits separating values that are that are significantly high. Based on the result, is x-8 significantly high? b) Find the probability that x is exactly 6. c) Find the probability that x is at most 6. d) Find the probability that at least one answer is correct e) Find the probability of no correct answer. significantly low and those Answer: a)

Answer 1

3) a) It is a binomial distribution

b) lt is a binomial distribution

C) It is not a binomial distribution. Because the selection of the adults should be done by with replacement process.

4) n = 8

P = 0.2

P(X = x) = nCx * p^x * (1 - p)^{n - x}

a) P(X = 6) = 8C6 * (0.2)^6 * (0.8)^2 = 0.0011

b) P(X > 4) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 8C4 * (0.2)^4 * (0.8)^4 + 8C5 * (0.2)^5 * (0.8)^3 + 8C6 * (0.2)^6 * (0.8)^2 + 8C7 * (0.2)^7 * (0.8)^1 + 8C8 * (0.2)^8 * (0.8)^0 = 0.0563

c) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= 8C0 * (0.2)^0 * (0.8)^8 + 8C1 * (0.2)^1 * (0.8)^7 + 8C2 * (0.2)^2 * (0.8)^6 = 0.7969

d) P(X > 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - (8C0 * (0.2)^0 * (0.8)^8)

= 1 - 0.1678 = 0.8322

e) P(X = 0) = 8C0 * (0.2)^0 * (0.8)^8 = 0.1678

5) n = 10

P = 0.2

a) $\mu$ = n * P = 10 * 0.2 = 2

$\sigma$ = sqrt(np(1 - p))

= sqrt(10 * 0.2 * 0.8) = 1.2649

= 2 - 2 * 1.2649 = -0.5298

-20

= 2 + 2 * 1.2649 = 4.5298

+20

Yes, X = 8 is significantly high, because it is greater than .

+20

b) P(X = 6) = 10C6 * (0.2)^6 * (0.8)^4 = 0.0055

c) P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

= 10C0 * (0.2)^0 * (0.8)^10 + 10C1 * (0.2)^1 * (0.8)^9 + 10C2 * (0.2)^2 * (0.8)^8 + 10C3 * (0.2)^3 * (0.8)^7 + 10C4 * (0.2)^4 * (0.8)^6 + 10C5 * (0.2)^5 * (0.8)^5 + 10C6 * (0.2)^6 * (0.8)^4

= 0.9991

d) P(X > 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - (10C0 * (0.2)^0 * (0.8)^10)

= 1 - 0.1074 = 0.8926

e) P(X = 0) = 10C0 * (0.2)^0 * (0.8)^10 = 0.1074