2 check the conditions for using a normal

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2) Check the conditions for using a Normal approximation. B: Success = agree, Failure = don t agree I: Because the population of U.S. adults is greater than 25,000, it is reasonable to assume the sampling without replacement condition is met. N: n = 2500 trials of the chance process. S: The probability of selecting an adult who agrees is p = 0.60. Since np = 2500(0.60) = 1500 and n (1 – p ) = 2500(0.40) = 1000 are both at least 10, we may use the Normal approximation. 3) Calculate P ( X 1520) using a Normal approximation. P ( X t 1520) P ( Z t 0.82) 1 ³ 0.7939 0.2061 21 Sampling Distribution of a Sample Proportion As n increases, the sampling distribution becomes approximately Normal . Sampling Distribution of a Sample Proportion The mean of the sampling distribution is p . Choose an SRS of size n from a population of size N with proportion p of successes. Let ˆ p be the sample proportion of successes. Then: The standard deviation of the sampling distribution is V ˆ p p (1 ³ p ) n For large n , ˆ p has approximately the N ( p , p (1 ³ p )/ n distribution. ˆ p count of successes in sample size of sample X n sample. in the successes" " of number the and ˆ proportion sample e between th connection important an is There X p 22 Sampling Distribution of a Sample Proportion ˆ p count of successes in sample size of sample X n sample. in the successes" " of number the and ˆ proportion sample e between th connection important an is There X p 23 Binomial Formula We can find a formula for the probability that a binomial random variable takes any value by adding probabilities for the different ways of getting exactly that many successes in n observations. The number of ways of arranging k successes among n observations is given by the binomial coefficient for k = 0, 1, 2, …, n. Note: n ! = n ( n – 1)( n – 2) •…• (3)(2)(1) and 0! = 1. n k § © ¨ · ¹ ¸ n ! k !( n ³ k )! 24 Binomial Probability The binomial coefficient counts the number of different ways in which k successes can be arranged among n trials. The binomial probability P ( X = k ) is this count multiplied by the probability of any one specific arrangement of the k successes. If X has the binomial distribution with n trials and probability p of success on each trial, the possible values of X are 0, 1, 2, …, n . If k is any one of these values, Binomial Probability P ( X k ) n k § © ¨ · ¹ ¸ p k (1 ³ p ) n ³ k
10/31/2012 Example Each child of a particular pair of parents has probability 0.25 of having blood type O. Suppose the parents have five children. (a) Find the probability that exactly three of the children have type O blood. b) Should the parents be surprised if more than three of their children
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( have type O blood?

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