ࡱ> $&#%` bjbjNN .4,,9|||||||TTT844#$hA!||||||| `9ETl0#"""|X#TT|||||| Binomial Distribution: The distribution of the count X of successes in the binomial setting is the binomial distribution with parameters n and p ( EMBED Equation.3 ). The parameter n is the number of observations, and p is the probability of a success on any one observation. The possible values of X are the whole numbers from 0 to n. As an abbreviation, we say that X is B(n, p). Define a binomial setting: 1. Each observation results in a success or failure (2 outcomes). 2. There are n observations (fixed number of trials). 3. The observations are independent of each other. 4. Each observation has the same probability p of a success. Example of a binomial setting: If both parents carry genes for the O and A blood types, each child has probability 0.25 of getting two O genes and so of having blood type O. Different children inherit independently of each other. The number of O blood types among 5 children of these parents is the count X of successes in 5 independent observations with probability 0.25 of a success on each observation. So X has the binomial distribution with n=5 and p=0.25. or B(5, 0.25) Example of a NOT binomial setting: The manager wants to know how many oysters he should expect to open to find two pearls of the appropriate size. He knows the pearls are normally distributed with a mean of 8mm and a standard deviation of 2 mm. X is the number of oysters opened in order to find two pearls of the appropriate size. Pdf: Given a discrete random variable X, the probability distribution function assigns a probability to each value of X. Cdf: Given a random variable X, the cumulative distribution function of X calculates the sum of the probabilities for 0, 1, 2, up to the value X. That is, it calculates the probability of obtaining at most X successes in n trials. Binomial Probability: If X has the binomial distribution with n observations and probability p of success on each observation, the possible values of X are 0, 1, 2, , n. If k is any one of these values:  EMBED Equation.3  Recall the solitaire game where p = 1/9 and n = 10. The sample size did not meet the conditions for CLT, therefore using the normal approximation to calculate the probability gave an incorrect estimation. In this case, the exact binomial probabilities are necessary. 1. Use the formula above to fill in the table: (Write your work below) X012345678910P(X) 2. Use the calculator to find: Binomialpdf(10, 1/9, 3)= What did this calculate? 3. Use the table to find the probability of winning 1 games or less. 4. Type in the following: Binomialcdf (10, 1/9, 1)= What did this calculate? The commands are: Binomialpdf (n, p, k) for exact probabilities. Binomialcdf (n, p, k) for continuous (n or less) probabilities. 5. Find the following probabilities by two methods: using the table(show logic), and using the calculator (write the command used). a.  EMBED Equation.3 = b.  EMBED Equation.3 = c.  EMBED Equation.3 = d.  EMBED Equation.3 = Normal approximation to binomial distributions: The formula for binomial probabilities becomes awkward and difficult to calculate as n increases. As the number of trials n gets larger, the binomial distribution gets close to a normal distribution. When n is large, we can use normal probability calculations to approximate hard-to-calculate binomial probabilities.  EMBED Equation.3   EMBED Equation.3  Example: If we were to repeat the solitaire problem with n = 200, (I dont want to create that table!!!) continuing to assign p = 1/9. Let X = number of games won. I. Because 200 is considered large: 200(1/9) > 10 and 200(1 1/9) > 10 the variable X will be approx. normally distributed with:  EMBED Equation.3   EMBED Equation.3  Use the normal probabilities to find  EMBED Equation.3 = Use the calculator to find  EMBED Equation.3 = II. Does this sound familiar?????? (Think CLT for proportions)  EMBED Equation.3   EMBED Equation.3  Use the normal probabilities to find  EMBED Equation.3  **Compare the mean and standard deviation from part I and part II.     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