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# Binomial lecture

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Binomial Lecture

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### Binomial lecture

1. 1. 1
2. 2. The Binomial Experiment • Repeated n times(trials) under identical conditions • Each trial can result in only one out of two outcomes – Success – probability success p – Failure – probability failure q = 1 – p • Trials are independent • Measure number of successes, x, in n trails 2
3. 3. The Binomial Experiment Typical cases where the binomial experiment applies: – A coin flipped results in heads or tails – A party wins or loses election – An employee is male or female – A car uses leaded, or unleaded fuel 3
4. 4. The Binomial Experiment • Binomial distribution is the probability distribution that applies to the binomial experiment • Displayed in the form of a table where the first row (or column) displays all possible number of successes, second row (or column) displays the probability associated with number of successes 4
5. 5. The Binomial ExperimentCalculating the Binomial ProbabilityDetermining x successes in n trials: P ( X  x)  P ( x)  n x Cx p q n-xwhere, n = number of trails p = probability of a success q = probability of a failure x = number of successes n! n Cx  x !(n - x )! 5
6. 6. The Binomial Experiment - Example• 10% of students are late for the early morning class• In a sample of 5 students, find the probability distribution of the number students that are lateAre the conditions required for the binomial experiment met?• Repeated n = 5 times• Each trial can result in only one out of two outcomes – Success – late for class → p = 0.10 – Failure – not late for class → q = 1 - 0.10 = 0.90• Students are independent 6
7. 7. The Binomial Experiment - Example • Let X be the binomial random variable indicating the number of late studentsCalculate the probability that threestudents are lateCalculate the probability that zerostudent is late late one students are P ( X  x)  p( x)  n C x p x (1 - p ) n - x P(X = 0) = P(0) = 5 C 0 (0.10) 0 (0.90) 5-0 = 0.5905 P(X = 1) = P(1) = 5 C 1 (0.10) 1 (0.90) 5-1 = 0.3281 P(X = 2) = P(2) = 5 C 2 (0.10) 2 (0.90) 5-2 = 0.072 P(X = 3) = P(3) = 5 C 3 (0.10) 3 (0.90) 5-3 = 0.008 P(X = 4) = P(4) = 5 C 4 (0.10) 4 (0.90) 5-4 = 0.00045 P(X = 5) = P(5) = 5 C 5 (0.10) 5 (0.90) 5-5 = 0.00001 7
8. 8. The Binomial Experiment - Example • Let X be the binomial random variable indicating the number of late students X P(X)P(X = 0) = P(0) = 5 C 0 (0.10) 0 (0.90) 5-0 = 0.5905 0 0.5905P(X = 1) = P(1) = 5 C 1 (0.10) 1 (0.90) 5-1 = 0.3281 1 0.3281P(X = 2) = P(2) = 5 C 2 (0.10) 2 (0.90) 5-2 = 0.072 2 0.0729P(X = 3) = P(3) = 5 C 3 (0.10) 3 (0.90) 5-3 = 0.008 3 0.0081P(X = 4) = P(4) = 5 C 4 (0.10) 4 (0.90) 5-4 = 0.00045 4 0.00045P(X = 5) = P(5) = 5 C 5 (0.10) 5 (0.90) 5-5 = 0.00001 5 0.00001 ∑P(X) ≈ 1
9. 9. The Binomial Experiment - Example • Calculate the probability that 2 or less students will be late X P(X) P(X ≤ 2) 0 0.5905 = P(X = 0) + P(X = 1) + P(X = 2) 1 0.3281 2 0.0729 = 0.5905 + 0.3281 + 0.0729 3 0.0081 = 0.9915 4 0.00045 5 0.00001 ∑P(X) ≈ 1
10. 10. The Binomial Experiment - Example • Calculate the probability that less than 2 students will be late X P(X) P(X < 2) 0 0.5905 = P(X = 0) + P(X = 1) 1 0.3281 2 0.0729 = 0.5905 + 0.3281 3 0.0081 = 0.9186 4 0.00045 5 0.00001 ∑P(X) ≈ 1
11. 11. The Binomial Experiment - Example • Calculate the probability that 4 or more than 4 students will be late X P(X) P(X ≥ 4) 0 0.5905 = P(X = 4) + P(X = 5) 1 0.3281 2 0.0729 = 0.00045 + 0.00001 3 0.0081 = 0.00046 4 0.00045 5 0.00001 ∑P(X) ≈ 1
12. 12. The Binomial Experiment - Example • Calculate the probability that more than 4 students will be late X P(X) P(X > 4) 0 0.5905 = P(X = 5) 1 0.3281 2 0.0729 = 0.00001 3 0.0081 4 0.00045 5 0.00001 ∑P(X) ≈ 1
13. 13. The Binomial Experiment - Example • Calculate the probability that 3 or more students will be late X P(X) P(X ≥ 3) 0 0.5905 = P(X = 3) + P(X = 4) + P(X = 5) 1 0.3281 = 0.00856 OR 2 0.0729 3 0.0081 = 1 – P(X ≤ 2) 4 0.00045 = 1 – 0.9915 5 0.00001 = 0.0085 ∑P(X) ≈ 1
14. 14. The Binomial Experiment - Example• Calculate the probability that more than 3 students will be late X P(X) P(X > 3) 0 0.5905 = P(X = 4) + P(X = 5) 1 0.3281 = 0.00046 OR 2 0.0729 3 0.0081 = 1 – P(X ≤ 3) 4 0.00045 = 1 – 0.9996 5 0.00001 = 0.0004 ∑P(X) ≈ 1
15. 15. The Binomial Experiment • Mean and standard deviation of binomial random variable  E( X )  np    2  Var ( X )  npq REMEMBER Repeated n times(trials) under identical conditions Each trial can result in only one out of two outcomes Success – probability success p Failure – probability failure q = 1 – p Measure number of successes, x, in n trails 15
16. 16. The Binomial Experiment - Example – What is the expected number of students that come late?   E( X )  np  5(0.10)  0.5 – What is the standard deviation for the number of students who come late?     Var ( X )  npq  5(0.10)(0.90)  0.67 2 16