ࡱ>  1bjbjcTcT 42>>,,=!fY^"""!!!!!!!#%B!Q"L"""!f!fff"F!f"!ffV@  &hv; |!0!G x%% % @""f"""""!!f"""!""""%""""""""" : 6-6 Normal as Approximation to Binomial This section describes another interesting application of the normal distribution. In section 5-3, you learned how to find binomial probabilities. For instance, if a surgical procedure has an 85% chance of success and a doctor performs the procedure on 10 patients, it is easy to find the probability of exactly two successful surgeries. But what if the doctor performs the surgical procedure on 150 patients and you want to find the probability of fewer than 100 successful surgeries? To do this using the techniques described in section 5-3, you would have to use the binomial formula 100 times and find the sum of the resulting probabilities. This approach is not practical, of course. A better approach is to use a normal distribution to approximate the binomial distribution. Normal Approximation to a Binomial Distribution If np QUOTE  , then the binomial random variable x is approximately normally distributed, with mean (= np and standard deviation  =  QUOTE   . Ex (1) Decide whether you can use the normal distribution to approximate x, the number of people who reply yes. If you can, find the mean and standard deviation. If you cannot, explain why. Fifty-one percent of adults in the United States whose New Years resolution was to exercise more achieved their resolution. You randomly select 65 adults in the United States whose resolution was to exercise more and ask each if he or she achieved that resolution. Fifteen percent of adults in the United States do not make New Years resolutions. You randomly select 15 adults in the United States and ask each if he or she made a New Years resolution. Because the binomial probability distribution deals with discrete data and the normal distribution deals with continuous data, we apply the continuity correction which should be used in normal approximation to binomial distributions. To calculate exact binomial probabilities, you can use the binomial formula for each value of x and add the results. Geometrically, this corresponds to adding the areas of bars in the probability histogram. Remember that each bar has a width of one unit and x is the midpoint of the interval. When you use a continuous normal distribution to approximate binomial probability, you need to move 0.5 unit to the left and right of the midpoint to include all possible x-values in the interval. When you do this, you are making a Correction for Continuity.    Ex(2) Use a correction for continuity to convert the binomial intervals to a normal distribution interval. The probability of getting between 270 and 310 successes, inclusive. The probability of at least 158 successes. The probability of getting less than 63 successes. Guidelines: Using the Normal distribution to approximate Binomial probabilities Verify that the binomial distribution applies. (Specify n, p, and q) Determine if you can use the normal distribution to approximate x, the binomial variable. (Is np QUOTE  ?) Find the mean and standard deviation for the distribution. ((= np and  =  QUOTE   ) Apply the appropriate continuity correction. Shade the corresponding area under the normal curve. (add or subtract 0.5 from endpoints) Find the corresponding z-score(s). ( QUOTE   ) Find the probability. (use the Standard Normal Table) Ex(3) Fifty-one percent of adults in the United States whose New Years resolution was to exercise more achieved their resolution. You randomly select 65 adults in the United States whose resolution was to exercise more and ask each if he or she achieved that resolution. What is the probability that fewer than 40 of them respond yes? (Remember to apply the continuity correction for the value of x. In the binomial distribution, the possible midpoint values for fewer than 40 are ,37, 38, 39. Thus to use the normal distribution, add 0.5 to the right-hand boundary 39 to get x = 39.5.) Ex(4) A survey reports that 86% of Internet users use Windows Internet Explorer as their browser. You randomly select 200 internet users and ask each whether she or he uses Internet Explorer as his or her browser. What is the probability that exactly 176 will say yes? Interpreting Results The ultimate objective is not finding some probability value. The real objective is to make practical sense of the result. Using Probabilities to Determine when Results are Unusual Unusually high: x successes among n trial is an unusually high number of successes if P( x or more) is very small (such as 0.05 or less) Unusually low: x successes among n trials is an unusually low number of successes if P( x or fewer) is very small (such as 0.05 or less) Ex (5)( form textbook, page 308, #31) A Boeing 767-300 aircraft has 213 seats. When someone buys a ticket for a flight, there is a 0.0995 '(6 f i k l s t u v w x      " ( , 4 ŸŞŔ|ŸobŔXNXhX[CJPJaJh CJPJaJj)hhEHUj|hhEHUhrC}6CJPJaJ jmhrC}CJPJaJhrC}CJPJaJj> hhEHUjhhEHUhhCJPJaJ!jhhCJPJUaJhrC}6CJaJhrC}5CJaJhrC}CJaJh_QCJ aJ hrC}CJ aJ ({6 f ( 6 gdORgd  & Fgd gdrC}&$d%d&d'dNOPQgd )$$d%d&d'dNOPQa$gd tuwx~鮦znnnff]hy6CJaJh:lCJaJhyhy6CJaJhy5CJaJhyhyCJaJhX[CJaJhyCJaJh CJaJhc CJaJ"jhc CJUaJmHnHu&jhc CJPJUaJmHnHuhc 5CJPJaJhORCJPJaJh CJPJaJh h CJPJaJ"A:3(@B^gdyh^hgdygd:l & Fgdygdygdy & FgdygdrC}456>  "(xi^Q^ij@ohhEHUhhCJaJjhhCJUaJj`hhEHUjRhhEHU jmhyCJPJaJhyCJaJhyhyCJaJhyCJPJaJhy6CJPJaJjDhhEHUj7hhEHUhhCJPJaJ!jhhCJPJUaJ(igijnqIJX^ؽ~vmeYPYeYehT5CJaJhThT5CJaJhTCJaJh:l5CJaJhc CJaJh:lhc CJaJh:lCJPJaJh:lCJPJaJh:lh:lCJPJaJhX[CJaJh:lh:lCJaJh:lCJaJhyhyCJaJhyCJaJhyhc CJaJjhhCJUaJj}hhEHUijkl|}~J^1gdh^hgd:l & Fgdygdy^`d0<011hThTCJaJUhTCJaJhX[CJaJhCJaJprobability that the person will not show up for the flight. A ticket agent accepts 236 reservations for a flight that uses a Boeing 767-300. 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