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4.1 Binomial Distribution Day 1. There are many experiments in which the results of each trial can be reduced to 2 outcomes.

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Presentation on theme: "4.1 Binomial Distribution Day 1. There are many experiments in which the results of each trial can be reduced to 2 outcomes."— Presentation transcript:

1 4.1 Binomial Distribution Day 1

2 There are many experiments in which the results of each trial can be reduced to 2 outcomes.

3 Binomial Experiment: There are n independent trials. Each trial has only 2 possible outcomes:  Success or failure The probability of success is the same for each trial. The probability of success is p. The probability of failure is 1-p.

4 Finding a Binomial Probability For a binomial experiment consisting of n trials, the probability of exactly k successes is: P(k successes) = n C k p k (1-p) n-k where the probability of success on each trial is p.

5 Example According to a survey taken by USA Today, about 37% of adults believe that Unidentified Flying Objects (UFOs) really exist. Suppose you randomly survey 6 adults. What is the probability that exactly 2 of them believe that UFOs exist?

6 Solution Let p = 0.37 Survey 6 adults; n = 6 k = 2 P(k = 2) = 6 C 2 (0.37) 2 (1- 0.37) 6 – 2 = 0.323 The probability that exactly 2 of the people surveyed believe that UFOs really exist is about 32%.

7 You Try At a college, 53% of students receive financial aid. In a random group of 9 students, what is the probability that exactly 5 of them receive financial aid? p=.53 (the probability of success for each trial) n=9 (number of different trials or experiments) k=5 (the probability of getting 5 successes) P(k=5) = 9 C 5.53 5 (1-.53) 9-5 ≈ 26%

8 EXAMPLE Draw a histogram of the binomial distribution for the survey from the first Example of UFOs. Then find the probability that at most 2 of the people surveyed believe that UFOs really exist. Hint: Use P(k successes) = n C k p k (1-p) n-k

9 Solution P(k = 0) = 6 C 0 (0.37) 0 (0.63) 6 ≈ 0.063 P(k = 1) = 6 C 1 (0.37) 1 (0.63) 5 ≈ 0.220 P(k = 2) = 6 C 2 (0.37) 2 (0.63) 4 ≈ 0.323 P(k = 3) = 6 C 3 (0.37) 3 (0.63) 3 ≈ 0.253 P(k = 4) = 6 C 4 (0.37) 4 (0.63) 2 ≈ 0.112 P(k = 5) = 6 C 5 (0.37) 5 (0.63) 1 ≈ 0.026 P(k = 6) = 6 C 6 (0.37) 6 (0.63) 0 ≈ 0.003

10 Solution Con’t

11 The probability of getting at most k = 2 successes is P(k < 2) = P(2) + P(1) + P(0) = 0.323 + 0.220 + 0.063 = 0.606 The probability that at most 2 of the people surveyed believe that UFOs really exist is about 61%

12 You Try Draw a histogram of the binomial distribution for the class of students from the previous You Try. Hint: Use P(k successes) = n C k p k (1-p) n-k

13 P(k=0) = 9 C 0.53 0 (1-.53) 9-0 =.001 P(k=1) = 9 C 1.53 1 (1-.53) 9-1 =.011 P(k=2) = 9 C 2.53 2 (1-.53) 9-2 =.05 P(k=3) = 9 C 3.53 3 (1-.53) 9-3 =.13 P(k=4) = 9 C 4.53 4 (1-.53) 9-4 =.23 P(k=5) = 9 C 5.53 5 (1-.53) 9-5 =.26 P(k=6) = 9 C 6.53 6 (1-.53) 9-6 =.19 P(k=7) = 9 C 7.53 7 (1-.53) 9-7 =.09 P(k=8) = 9 C 8.53 8 (1-.53) 9-8 =.03 P(k=9) = 9 C 9.53 9 (1-.53) 9-9 =.003

14 EXAMPLE

15 Find the probability that fewer than 3 students in the class receive financial aid. = P(0) + P(1) + P(2) =.001 +.011 +.05 = 0.062 EXAMPLE


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