Notes 23 - have 10 0.2 = 2 successes The spread of a binomial distribution is not so intuitive so we will not justify our formula for standard

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Normal Approximation to the Binomial Submitted by gfj100 on Wed, 11/11/2009 - 10:45 Remember binomial random variables from last week's discussion? A binomial random variable can also be approximated by using normal random variable methods discussed above. This approximation can take place as long as: 1. The population size must be at least 10 times the sample size. 2. np = 10 and n (1 − p ) = 10. [These constraints take care of population shapes that are unbalanced because p is too close to 0 or to 1.] The mean of a binomial random variable is easy to grasp intuitively: Say the probability of success for each observation is 0.2 and we make 10 observations. Then on the average we should
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Unformatted text preview: have 10 * 0.2 = 2 successes. The spread of a binomial distribution is not so intuitive, so we will not justify our formula for standard deviation. If sample count X of successes is a binomial random variable for n fixed observations with probability of success p for each observation, then X has a mean and standard deviation as discussed in section 8.4 of: Mean = np and standard deviation = And as long as the above 2 requirements are for n and p are satisfied, we can approximate X with a normal random variable having the same mean and standard deviation and use the normal calculations discussed previously in these notes to solve for probabilities for X....
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This note was uploaded on 05/31/2011 for the course STAT 200 taught by Professor Andyregards during the Spring '11 term at World College.

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