data assignment unit 6.pdf - 1 During a recent survey of...

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1. During a recent survey of ethnic backgrounds of 1000 people in a large city, 513 were Canadian, 148 were French, 72 were African and 56 were Asian and the remainder were from other groups. Calculate the probability that a person, selected at random from the population has: a. a Canadian background? P(Canadian) = Number of Favorable Outcomes/Total Number of Possible Outcomes P(Canadian) = 513/1000 P(Canadian) = 0.513 Therefore, the probability that the selected person has a Canadian Background is 0.513 b. an African background? P(African) = Number of Favorable Outcomes/Total Number of Possible Outcomes P(African) = 72/1000 P(African) = 0.072 Therefore, the probability that the selected person has an African background is 0.072 c. an “other” background? 1000 (513 + 148 + 72 + 56) = 211 P(“other”) = 211/1000 P(“other”) = 0.211 Therefore, the probability that the selected person has an “other” background is 0.211 2. A spinner is divided into three equally sized regions as shown. The spinner is spun twice. For each probability you determine, express your answer as a fraction, decimal and percent. a. What is the probability of spinning A on the first spin? P(A) = 1/3 P(A) = 0.33 P(A) = 33% Therefore the probability of spinning A on the first spin is 1/3 in a fraction, 0.33 in a decimal, or 33% in a percent.
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b. Draw a tree diagram to represent the sample space for both spins. c. What is the probability of spinning A followed by B? P (A, B) = 2/9 P (A, B) = 0.22 P (A, B) = 0.22 x 100 P (A, B) = 22.2% Therefore, the probability of spinning A followed by B is 2/9 in a fraction, 0.22 in a decimal or 22.2% in a percent. d. What is the probability of getting the same letter on both spins?
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P (A, A) = 1/9 P (B, B) = 4/9 P (A, A + B, B) = 1/9 + 4/9P (A, A + B, B) = 5/9 P (A, A + B, B) = 0.55 P (A, A + B, B) = 0.55 x 100 P (A, A + B, B) = 55.5% Therefore, the probability of getting the same letter on both spins is 5/9 in a fraction, 0.55 in a decimal, or 55.5% in a percent. 3. Binomial i. The Binomial Distribution s used when we have a fixed number of trials, in which they have two outcomes (win or loose). To use this formula, along with the number of trials and the two outcomes, we also need to know the probability of success and how many successes we want. ii. The formula for this the Binomial Distribution is: The first part of the equation n C x is used because we need to first determine the number of successes ( x ) that we can get from the total number of trials ( n ) . Order does not matter which is why we would use the combination equation for this. The second part of the equation is p x . P represents the probability of success. The reason we have to calculate the probability of success to the power of the number of wanted successes is because we want to find the probability of getting the number of successes we want. (1 p ) ( n x ) is the last part of the equation. (1 p ) is done because the difference equals the probability of not succeeding. This number is to the power of ( n x ) because since (1 p ) ( n x ) represents the probability of not succeeding, we need to minus the total number of trials by the number of wanted successes to find out how many of the total trials are not successes.
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  • Fall '19
  • Probability theory, Binomial distribution, Hypergeometric Distribution

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