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## Stats

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**Stats**Section 5.6 Notes**Review of Binomial Probability Distribution**• Introduced in section 4.3: • 1. The procedure must have a fixed number of trials • 2. The trials must be independent • 3. Each trial must have all outcomes classified into two categories. • 4. The probabilities must remain constant for each trial. • Our example was of a multiple choice test.**Normal Distribution as Approximation to Binomial**Distribution • If , then the binomial random variable has a probability distribution that can be approximated by a normal distribution with the mean and standard deviation given as:**Continuity Correction**• When we use the normal distribution (which is a continuous probability distribution) as an approximation to the binomial distribution (which is discrete), a continuity correction is made to a discrete whole number x in the binomial distribution by representing the single value x by the interval from • x - 0.5 to x + 0.5**Example**• Recently, American Airlines had 72.3% of its flights arriving on time. In a check of 40 randomly selected American Airlines flights, 19 arrived on time. Estimate the probability of getting 19 or fewer on-time flights among 40, assuming that the 72.3% rate is correct. Is it unusual to get 19 or fewer on-time flights among 40 randomly selected American Airlines flights?**Let x = the number that arrive on time**• n = 40 and p = .723 therefore q = .277 • np = 28.92 which is • nq = 11.08 which is • So we can use the normal distribution as an approximation to binomial. Therefore:**Therfore:**• Yes it would be an unusual even to get 19 or fewer on-time flights because .0004 < .05**Question!**• How would things change if the problem asked about 19 or more flights?