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6.2 Homework Questions Section 6.3 Binomial Random Variables Binomial Setting The four conditions for a binomial setting are: Success/Failure 2. Independent Trials 3. Constant “p” (probability of success) 4. Set number of trials, n 1. Geometric 1. 2. 3. 4. The four conditions for a geometric setting are: Success/Failure Independent Trials Constant “p” (probability of success) No set number of trials, n Binomial Random Variable The count of X successes in a binomial setting is a binomial random variable. The probability distribution of X is a binomial distribution with parameters n and p. The possible values of X are the whole numbers from 0 to n. Binomial? Genetics says that children receive genes from each of their parents independently. Each child of a particular pair of parents has probability 0.25 of having type O blood. Suppose these parents have 5 children. Let X = the number of children with type O blood. Shuffle a deck of cards. Turn over the first 10 cards, one at a time. Let Y = the number of aces you observe. Shuffle a deck of cards. Turn over the top card. Put the card back in the deck, and shuffle again. Repeat this process until you get an ace. Let W = the number of cards required. Binomial Probabilities Let’s do the children’s gene problem. n=5 , p= 0.25 or B(5, 0.25) P(X=0) P(none of the children have type O)= P(X=1) P(one child has type O) P(X=2) Building the formula P(X = k) = = P(exactly k successes in “n” trials)= = (number of arrangements)∙ 𝑝𝑘 (1 − 𝑝)𝑛−𝑘 So we need a nice way of finding the “number of arrangements” Number of arrangements: Binomial Coefficient The number of ways of arranging k successes among n observations is given by the binomial coefficient: 𝑛 𝑛! = 𝑘! 𝑛−𝑘 ! 𝑘 You may know this as nCr 𝑛 CAUTION : is NOT the fraction 𝑘 𝑛 𝑘 Binomial Probability For B(n,p) , that is to say, for any Binomial: 𝑛 𝑘 P X=k = 𝑝 (1 − 𝑝)𝑛−𝑘 𝑘 This is on the formula sheet! Examples: Find the probability that exactly 3 children have type O blood. B(5,0.25) Should the parents be surprised if more than 3 of their children have type O blood? Justify your answer. Mean and Standard Deviation of a Binomial Distribution Blood Type Probability Distribution: X 0 1 2 3 4 5 P(X) 0.23730 0.39551 0.26367 0.08789 0.01465 0.00098 Above is the binomial from the blood type problem. Find the expected value and standard deviation. Mean and Standard Deviation of Binomial Random Variables Given B(n,p): 𝜇𝑥 = 𝑛𝑝 𝜎𝑥 = 𝑛𝑝(1 − 𝑝) Remember – these formulas ONLY work for binomial distributions! Both of these are on the formula sheet as well. Examples Continued: Together, let’s do numbers Page 403: 69-72 Using the first one as an example B(20, 0.85) find: 𝑃(𝑋 = 17) 𝑃 𝑋 ≤ 12 𝑃 𝑋 = 0 + 𝑃 𝑋 = 1 + ⋯ + 𝑃(𝑋 = 12) Homework (same as before) Pg. 403 (73-75, 77, 79, 80, 82, 84-87, 89-92, 94-105) Warm Up Normal Approximation for Binomial Distributions As a rule of thumb, we will use the Normal approximation when n is so large that: 𝑛𝑝 ≥ 10 𝑛(1 − 𝑝) ≥ 10 That is, the expected number of successes and failures are both at least 10. 𝑛≤ 1 𝑁 10 (independence) Example: Suppose that exactly 60% of all adult US residents would say “agree” if asked if they think shopping is frustrating. A survey asked nationwide sampled 2500 adults. Let X = the number of people who agree. ◦ Show that X is approximately a binomial random variable. Check the conditions for using a Normal approximation in this setting. Example Continued: Use a Normal distribution to estimate the probability that 1520 or more of the sample agree. Find the mean Find the standard deviation Use the Normal curve Homework #3 (again) Pg. 403 (73-75, 77, 79, 80, 82, 84-87, 89-92, 94-105)