(1) If random variable X follows the Binomial distribution X~Binomi.pdf

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(1) If random variable X follows the Binomial distribution: X~Binomial(8,0.7), which among the following statements is/are correct? P(X<6)=p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4)+p(x=5)+p(x=6) False P(X<6) = P(0) + P(1) + ... + P(5) ** You do NOT include 6. P(X<=20)=1 True Since n = 8, X can\'t be 9 or larger, so P(9 < = X < = 20) = 0. P(X< = 8) + P(9 < = X < = 20) = 1 + 0 = 1 X can take 8 different values. False X can take on 9 values: 0, 1, 2, 3, 4, 5, 6, 7 and 8 P(X>=3)+P(X<=3)=1 False P(X > = 3) + P(X < = 2) = 1 P(X>6)=1-P(X<7) True P(X > 6) = P(X > = 7) = 1 - P(X < = 6) = 1 - P(X < 7) P(2 Unknown - this statement is incomplete --------------- Note: The following notation varies from book-to-book. Given: n = 12 and p = .5 P(X) = nCx * p^x * (1-p)^(n-x) P(X = 5) = 12C5 * .5^5 * .5^7 P(X = 5) = 0.19336 (b) P(X > 4) = P(5) + P(6) + ... + P(12) <-- 8 probabilities to find = 1 - [P(4) + P(3) + ... + P(0)] <-- 5 probabilities to find = 8C4 * 0.5^4 * 0.5^8 = 0.12085 + 8C3 * 0.5^3 * 0.5^9 = 0.05371 + 8C2 * 0.5^2 * 0.5^10 = 0.01611 + 8C1 * 0.5^1 * 0.5^11 = 0.00293 + 8C0 * 0.5^0 * 0.5^12 = 0.00024 = 1 - [0.12085 + 0.05371 + ... + 0.00024] = 0.806 --------------- (1) Given: n = 20 and p = .01 (a) P(X = 0) = 20C0 * 0.01^0 * 0.99^20 = 0.818 (b) P(at most 1) = P(1) + P(0) = 20C1 * 0.01^1 * 0.99^19 = 0.16523 + 20C0 * 0.01^0 * 0.99^20 = 0.81791 = 0.983 (c) P(more than 1) = P(2) + P(3) + ... + P(20) = 1 - [P(1) + P(0)] = 1 - 0.983 = 0.017 --------------- (1) Given: n = 20 and p = .01 P(X = 0) = 20C0 * 0.01^0 * 0.99^20 = 0.818 Given: n = 20 and p = .02 P(X = 0) = 20C0 * 0.02^0 * 0.98^20 = 0.668 Given: n = 20 and p = .05 P(X = 0) = 20C0 * 0.05^0 * 0.95^20 = 0.358 --------------- (1) (a) mean = (6+2)/2 = 4 (b) P(X > 4) = P(4 < X < 6) = (6-4)/(6-2) = 1/2 = .5 Note: P(a < X < b) = (b - a)/(Max - Min) (c) = P(3 < X < 5) = (5-3)/(6-2) = 1/2 = .5 --------------- I hope this helped. If you have any questions, please ask them in the comment section. :) Solution (1) If random variable X follows the Binomial distribution: X~Binomial(8,0.7), which among the following statements is/are correct? P(X<6)=p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4)+p(x=5)+p(x=6) False P(X<6) = P(0) + P(1) + ... + P(5) ** You do NOT include 6. P(X<=20)=1 True Since n = 8, X can\'t be 9 or larger, so P(9 < = X < = 20) = 0. P(X< = 8) + P(9 < = X < = 20) = 1 + 0 = 1 X can take 8 different values. False X can take on 9 values: 0, 1, 2, 3, 4, 5, 6, 7 and 8 P(X>=3)+P(X<=3)=1 False P(X > = 3) + P(X < = 2) = 1 P(X>6)=1-P(X<7) True P(X > 6) = P(X > = 7) = 1 - P(X < = 6) = 1 - P(X < 7) P(2 Unknown - this statement is incomplete --------------- Note: The following notation varies from book-to-book. Given: n = 12 and p = .5 P(X) = nCx * p^x * (1-p)^(n-x) P(X = 5) = 12C5 * .5^5 * .5^7 P(X = 5) = 0.19336 (b) P(X > 4) = P(5) + P(6) + ... + P(12) <-- 8 probabilities to find = 1 - [P(4) + P(3) + ... + P(0)] <-- 5 probabilities to find = 8C4 * 0.5^4 * 0.5^8 = 0.12085 + 8C3 * 0.5^3 * 0.5^9 = 0.05371 + 8C2 * 0.5^2 * 0.5^10 .

(1) If random variable X follows the Binomial distribution X~Binomi.pdf

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(1) If random variable X follows the Binomial distribution: X~Binomial(8,0.7), which among the following statements is/are correct? P(X<6)=p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4)+p(x=5)+p(x=6) False P(X<6) = P(0) + P(1) + ... + P(5) ** You do NOT include 6. P(X<=20)=1 True Since n = 8, X can\'t be 9 or larger, so P(9 < = X < = 20) = 0. P(X< = 8) + P(9 < = X < = 20) = 1 + 0 = 1 X can take 8 different values. False X can take on 9 values: 0, 1, 2, 3, 4, 5, 6, 7 and 8 P(X>=3)+P(X<=3)=1 False P(X > = 3) + P(X < = 2) = 1 P(X>6)=1-P(X<7) True P(X > 6) = P(X > = 7) = 1 - P(X < = 6) = 1 - P(X < 7) P(2 Unknown - this statement is incomplete --------------- Note: The following notation varies from book-to-book. Given: n = 12 and p = .5 P(X) = nCx * p^x * (1-p)^(n-x) P(X = 5) = 12C5 * .5^5 * .5^7 P(X = 5) = 0.19336 (b) P(X > 4) = P(5) + P(6) + ... + P(12) <-- 8 probabilities to find = 1 - [P(4) + P(3) + ... + P(0)] <-- 5 probabilities to find = 8C4 * 0.5^4 * 0.5^8 = 0.12085 + 8C3 * 0.5^3 * 0.5^9 = 0.05371 + 8C2 * 0.5^2 * 0.5^10 = 0.01611 + 8C1 * 0.5^1 * 0.5^11 = 0.00293 + 8C0 * 0.5^0 * 0.5^12 = 0.00024 = 1 - [0.12085 + 0.05371 + ... + 0.00024] = 0.806 --------------- (1) Given: n = 20 and p = .01 (a) P(X = 0) = 20C0 * 0.01^0 * 0.99^20 = 0.818 (b) P(at most 1) = P(1) + P(0) = 20C1 * 0.01^1 * 0.99^19 = 0.16523 + 20C0 * 0.01^0 * 0.99^20 = 0.81791 = 0.983 (c) P(more than 1) = P(2) + P(3) + ... + P(20) = 1 - [P(1) + P(0)] = 1 - 0.983 = 0.017 --------------- (1) Given: n = 20 and p = .01 P(X = 0) = 20C0 * 0.01^0 * 0.99^20 = 0.818 Given: n = 20 and p = .02 P(X = 0) = 20C0 * 0.02^0 * 0.98^20 = 0.668 Given: n = 20 and p = .05 P(X = 0) = 20C0 * 0.05^0 * 0.95^20 = 0.358 --------------- (1) (a) mean = (6+2)/2 = 4 (b) P(X > 4) = P(4 < X < 6) = (6-4)/(6-2) = 1/2 = .5 Note: P(a < X < b) = (b - a)/(Max - Min) (c) = P(3 < X < 5) = (5-3)/(6-2) = 1/2 = .5 --------------- I hope this helped. If you have any questions, please ask them in the comment section. :) Solution (1) If random variable X follows the Binomial distribution: X~Binomial(8,0.7), which among the following statements is/are correct? P(X<6)=p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4)+p(x=5)+p(x=6) False P(X<6) = P(0) + P(1) + ... + P(5) ** You do NOT include 6. P(X<=20)=1 True Since n = 8, X can\'t be 9 or larger, so P(9 < = X < = 20) = 0. P(X< = 8) + P(9 < = X < = 20) = 1 + 0 = 1 X can take 8 different values. False X can take on 9 values: 0, 1, 2, 3, 4, 5, 6, 7 and 8 P(X>=3)+P(X<=3)=1 False P(X > = 3) + P(X < = 2) = 1 P(X>6)=1-P(X<7) True P(X > 6) = P(X > = 7) = 1 - P(X < = 6) = 1 - P(X < 7) P(2 Unknown - this statement is incomplete --------------- Note: The following notation varies from book-to-book. Given: n = 12 and p = .5 P(X) = nCx * p^x * (1-p)^(n-x) P(X = 5) = 12C5 * .5^5 * .5^7 P(X = 5) = 0.19336 (b) P(X > 4) = P(5) + P(6) + ... + P(12) <-- 8 probabilities to find = 1 - [P(4) + P(3) + ... + P(0)] <-- 5 probabilities to find = 8C4 * 0.5^4 * 0.5^8 = 0.12085 + 8C3 * 0.5^3 * 0.5^9 = 0.05371 + 8C2 * 0.5^2 * 0.5^10 .

(1) If random variable X follows the Binomial distribution X~Binomi.pdf

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(1) If random variable X follows the Binomial distribution: X~Binomial(8,0.7), which among the following statements is/are correct? P(X<6)=p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4)+p(x=5)+p(x=6) False P(X<6) = P(0) + P(1) + ... + P(5) ** You do NOT include 6. P(X<=20)=1 True Since n = 8, X can't be 9 or larger, so P(9 < = X < = 20) = 0. P(X< = 8) + P(9 < = X < = 20) = 1 + 0 = 1 X can take 8 different values. False X can take on 9 values: 0, 1, 2, 3, 4, 5, 6, 7 and 8 P(X>=3)+P(X<=3)=1 False P(X > = 3) + P(X < = 2) = 1 P(X>6)=1-P(X<7) True P(X > 6) = P(X > = 7) = 1 - P(X < = 6) = 1 - P(X < 7) P(2 Unknown - this statement is incomplete ---------------
Note: The following notation varies from book-to-book. Given: n = 12 and p = .5 P(X) = nCx * p^x * (1-p)^(n-x) P(X = 5) = 12C5 * .5^5 * .5^7 P(X = 5) = 0.19336 (b) P(X > 4) = P(5) + P(6) + ... + P(12) <-- 8 probabilities to find = 1 - [P(4) + P(3) + ... + P(0)] <-- 5 probabilities to find = 8C4 * 0.5^4 * 0.5^8 = 0.12085 + 8C3 * 0.5^3 * 0.5^9 = 0.05371 + 8C2 * 0.5^2 * 0.5^10 = 0.01611 + 8C1 * 0.5^1 * 0.5^11 = 0.00293 + 8C0 * 0.5^0 * 0.5^12 = 0.00024 = 1 - [0.12085 + 0.05371 + ... + 0.00024] = 0.806 --------------- (1) Given: n = 20 and p = .01 (a) P(X = 0) = 20C0 * 0.01^0 * 0.99^20 = 0.818 (b) P(at most 1) = P(1) + P(0) = 20C1 * 0.01^1 * 0.99^19 = 0.16523 + 20C0 * 0.01^0 * 0.99^20 = 0.81791 = 0.983
(c) P(more than 1) = P(2) + P(3) + ... + P(20) = 1 - [P(1) + P(0)] = 1 - 0.983 = 0.017 --------------- (1) Given: n = 20 and p = .01 P(X = 0) = 20C0 * 0.01^0 * 0.99^20 = 0.818 Given: n = 20 and p = .02 P(X = 0) = 20C0 * 0.02^0 * 0.98^20 = 0.668 Given: n = 20 and p = .05 P(X = 0) = 20C0 * 0.05^0 * 0.95^20 = 0.358 --------------- (1) (a) mean = (6+2)/2 = 4 (b) P(X > 4) = P(4 < X < 6) = (6-4)/(6-2) = 1/2 = .5
Note: P(a < X < b) = (b - a)/(Max - Min) (c) = P(3 < X < 5) = (5-3)/(6-2) = 1/2 = .5 --------------- I hope this helped. If you have any questions, please ask them in the comment section. :) Solution (1) If random variable X follows the Binomial distribution: X~Binomial(8,0.7), which among the following statements is/are correct? P(X<6)=p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4)+p(x=5)+p(x=6) False P(X<6) = P(0) + P(1) + ... + P(5) ** You do NOT include 6. P(X<=20)=1 True Since n = 8, X can't be 9 or larger, so P(9 < = X < = 20) = 0. P(X< = 8) + P(9 < = X < = 20) = 1 + 0 = 1 X can take 8 different values. False X can take on 9 values: 0, 1, 2, 3, 4, 5, 6, 7 and 8 P(X>=3)+P(X<=3)=1 False
P(X > = 3) + P(X < = 2) = 1 P(X>6)=1-P(X<7) True P(X > 6) = P(X > = 7) = 1 - P(X < = 6) = 1 - P(X < 7) P(2 Unknown - this statement is incomplete --------------- Note: The following notation varies from book-to-book. Given: n = 12 and p = .5 P(X) = nCx * p^x * (1-p)^(n-x) P(X = 5) = 12C5 * .5^5 * .5^7 P(X = 5) = 0.19336 (b) P(X > 4) = P(5) + P(6) + ... + P(12) <-- 8 probabilities to find = 1 - [P(4) + P(3) + ... + P(0)] <-- 5 probabilities to find = 8C4 * 0.5^4 * 0.5^8 = 0.12085 + 8C3 * 0.5^3 * 0.5^9 = 0.05371 + 8C2 * 0.5^2 * 0.5^10 = 0.01611 + 8C1 * 0.5^1 * 0.5^11 = 0.00293 + 8C0 * 0.5^0 * 0.5^12 = 0.00024 = 1 - [0.12085 + 0.05371 + ... + 0.00024] = 0.806 ---------------
(1) Given: n = 20 and p = .01 (a) P(X = 0) = 20C0 * 0.01^0 * 0.99^20 = 0.818 (b) P(at most 1) = P(1) + P(0) = 20C1 * 0.01^1 * 0.99^19 = 0.16523 + 20C0 * 0.01^0 * 0.99^20 = 0.81791 = 0.983 (c) P(more than 1) = P(2) + P(3) + ... + P(20) = 1 - [P(1) + P(0)] = 1 - 0.983 = 0.017 --------------- (1) Given: n = 20 and p = .01 P(X = 0) = 20C0 * 0.01^0 * 0.99^20 = 0.818 Given: n = 20 and p = .02 P(X = 0) = 20C0 * 0.02^0 * 0.98^20 = 0.668 Given: n = 20 and p = .05 P(X = 0) = 20C0 * 0.05^0 * 0.95^20 = 0.358
--------------- (1) (a) mean = (6+2)/2 = 4 (b) P(X > 4) = P(4 < X < 6) = (6-4)/(6-2) = 1/2 = .5 Note: P(a < X < b) = (b - a)/(Max - Min) (c) = P(3 < X < 5) = (5-3)/(6-2) = 1/2 = .5 --------------- I hope this helped. If you have any questions, please ask them in the comment section. :)

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(1) If random variable X follows the Binomial distribution X~Binomi.pdf

  • 1. (1) If random variable X follows the Binomial distribution: X~Binomial(8,0.7), which among the following statements is/are correct? P(X<6)=p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4)+p(x=5)+p(x=6) False P(X<6) = P(0) + P(1) + ... + P(5) ** You do NOT include 6. P(X<=20)=1 True Since n = 8, X can't be 9 or larger, so P(9 < = X < = 20) = 0. P(X< = 8) + P(9 < = X < = 20) = 1 + 0 = 1 X can take 8 different values. False X can take on 9 values: 0, 1, 2, 3, 4, 5, 6, 7 and 8 P(X>=3)+P(X<=3)=1 False P(X > = 3) + P(X < = 2) = 1 P(X>6)=1-P(X<7) True P(X > 6) = P(X > = 7) = 1 - P(X < = 6) = 1 - P(X < 7) P(2 Unknown - this statement is incomplete ---------------
  • 2. Note: The following notation varies from book-to-book. Given: n = 12 and p = .5 P(X) = nCx * p^x * (1-p)^(n-x) P(X = 5) = 12C5 * .5^5 * .5^7 P(X = 5) = 0.19336 (b) P(X > 4) = P(5) + P(6) + ... + P(12) <-- 8 probabilities to find = 1 - [P(4) + P(3) + ... + P(0)] <-- 5 probabilities to find = 8C4 * 0.5^4 * 0.5^8 = 0.12085 + 8C3 * 0.5^3 * 0.5^9 = 0.05371 + 8C2 * 0.5^2 * 0.5^10 = 0.01611 + 8C1 * 0.5^1 * 0.5^11 = 0.00293 + 8C0 * 0.5^0 * 0.5^12 = 0.00024 = 1 - [0.12085 + 0.05371 + ... + 0.00024] = 0.806 --------------- (1) Given: n = 20 and p = .01 (a) P(X = 0) = 20C0 * 0.01^0 * 0.99^20 = 0.818 (b) P(at most 1) = P(1) + P(0) = 20C1 * 0.01^1 * 0.99^19 = 0.16523 + 20C0 * 0.01^0 * 0.99^20 = 0.81791 = 0.983
  • 3. (c) P(more than 1) = P(2) + P(3) + ... + P(20) = 1 - [P(1) + P(0)] = 1 - 0.983 = 0.017 --------------- (1) Given: n = 20 and p = .01 P(X = 0) = 20C0 * 0.01^0 * 0.99^20 = 0.818 Given: n = 20 and p = .02 P(X = 0) = 20C0 * 0.02^0 * 0.98^20 = 0.668 Given: n = 20 and p = .05 P(X = 0) = 20C0 * 0.05^0 * 0.95^20 = 0.358 --------------- (1) (a) mean = (6+2)/2 = 4 (b) P(X > 4) = P(4 < X < 6) = (6-4)/(6-2) = 1/2 = .5
  • 4. Note: P(a < X < b) = (b - a)/(Max - Min) (c) = P(3 < X < 5) = (5-3)/(6-2) = 1/2 = .5 --------------- I hope this helped. If you have any questions, please ask them in the comment section. :) Solution (1) If random variable X follows the Binomial distribution: X~Binomial(8,0.7), which among the following statements is/are correct? P(X<6)=p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4)+p(x=5)+p(x=6) False P(X<6) = P(0) + P(1) + ... + P(5) ** You do NOT include 6. P(X<=20)=1 True Since n = 8, X can't be 9 or larger, so P(9 < = X < = 20) = 0. P(X< = 8) + P(9 < = X < = 20) = 1 + 0 = 1 X can take 8 different values. False X can take on 9 values: 0, 1, 2, 3, 4, 5, 6, 7 and 8 P(X>=3)+P(X<=3)=1 False
  • 5. P(X > = 3) + P(X < = 2) = 1 P(X>6)=1-P(X<7) True P(X > 6) = P(X > = 7) = 1 - P(X < = 6) = 1 - P(X < 7) P(2 Unknown - this statement is incomplete --------------- Note: The following notation varies from book-to-book. Given: n = 12 and p = .5 P(X) = nCx * p^x * (1-p)^(n-x) P(X = 5) = 12C5 * .5^5 * .5^7 P(X = 5) = 0.19336 (b) P(X > 4) = P(5) + P(6) + ... + P(12) <-- 8 probabilities to find = 1 - [P(4) + P(3) + ... + P(0)] <-- 5 probabilities to find = 8C4 * 0.5^4 * 0.5^8 = 0.12085 + 8C3 * 0.5^3 * 0.5^9 = 0.05371 + 8C2 * 0.5^2 * 0.5^10 = 0.01611 + 8C1 * 0.5^1 * 0.5^11 = 0.00293 + 8C0 * 0.5^0 * 0.5^12 = 0.00024 = 1 - [0.12085 + 0.05371 + ... + 0.00024] = 0.806 ---------------
  • 6. (1) Given: n = 20 and p = .01 (a) P(X = 0) = 20C0 * 0.01^0 * 0.99^20 = 0.818 (b) P(at most 1) = P(1) + P(0) = 20C1 * 0.01^1 * 0.99^19 = 0.16523 + 20C0 * 0.01^0 * 0.99^20 = 0.81791 = 0.983 (c) P(more than 1) = P(2) + P(3) + ... + P(20) = 1 - [P(1) + P(0)] = 1 - 0.983 = 0.017 --------------- (1) Given: n = 20 and p = .01 P(X = 0) = 20C0 * 0.01^0 * 0.99^20 = 0.818 Given: n = 20 and p = .02 P(X = 0) = 20C0 * 0.02^0 * 0.98^20 = 0.668 Given: n = 20 and p = .05 P(X = 0) = 20C0 * 0.05^0 * 0.95^20 = 0.358
  • 7. --------------- (1) (a) mean = (6+2)/2 = 4 (b) P(X > 4) = P(4 < X < 6) = (6-4)/(6-2) = 1/2 = .5 Note: P(a < X < b) = (b - a)/(Max - Min) (c) = P(3 < X < 5) = (5-3)/(6-2) = 1/2 = .5 --------------- I hope this helped. If you have any questions, please ask them in the comment section. :)