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Normal Approximation to the Binomial Business Statistics Plan for Today • The Binomial Distribution • Approximating the Binomial Distribution by the Normal Distribution • Examples 1 The Binomial Distribution • The binomial distribution is a discrete probability distribution (i.e. the random variable x has only a discrete or a finite set of possible values). • The binomial distribution requires n • repeated • independent • identical trials. Each trial is either success or failure. Probability of success = p, probability of failure = q = 1 – p . The random variable x = the number of successes. Example: tossing a coin • A fair coin is tossed or flipped n times in an identical independent manner. • The random variable x will count the number of “tails”. • Success = “tails”. • Failure = “heads”. • p=½ • q=1–½=½ • x can take values from 0 to n • What is the mean value of x ? 2 Example: rolling a die • A fair die is rolled n times in an identical, independent manner. • Declare, for example, that the random variable x counts the number of times a “6” is rolled. • Success = “6” • Failure = “1”, “2”, “3”, “4”, or “5”. • p = 1/6 and q = 5/6 = 1 – 1/6 • Again, the variable x can take values from 0 to n. • What is the mean value of x ? Example: blood type B • It is known that 9% of Canadians have blood type B. • Suppose that n Canadians are chosen at random. • x – the number of Canadians among those with blood type B. • Success = type B • Failure = type O, A, or AB • p = 0.09 and q = 1 – 0.09 = 0.91 • The random variable x can take values from 0 to n. What is its mean? 3 Example: smokers • 21% of adults in Quebec are smokers • 40 adult Quebecois are chosen at random • x – the number of smokers among these • Success = smoker (?!) • Failure = non-smoker • p = 0.21 and q = 1 – 0.21 = 0.79 • The random variable x takes values between 0 and 40. • Mean 𝜇 = 8.4. Why? The Binomial Distribution • The random variable x (the number of successes) has the following measures: • The mean 𝝁=𝒏∙𝒑 • The variance 𝝈𝟐 = 𝒏 ∙ 𝒑 ∙ 𝒒 • The standard deviation 𝝈= 𝝈𝟐 = 𝒏𝒑𝒒 4 Approximating the Binomial Distribution by the Normal Distribution • A consequence of the CLT: a Binomial Distribution 𝐵(𝑛, 𝑝) with the number of trials n and the probability of success p can be approximated by the normal distribution 𝑁(𝜇, 𝜎) if the following two conditions hold: 𝑛 ∙ 𝑝 > 5 and 𝑛 ∙ 𝑞 > 5 • Always check these conditions before applying approximation. Continuity Correction Factor • Instead of x we will use x – 0.5 or x + 0.5 depending on the contextual question. • “More than 30”: x = 30.5 • “At least 30”: x = 29.5 • “Less than 46”: x = 45.5 • “At most 46”: x = 46.5 • “Between 34 and 37 (inclusively)”: x1 = 33.5 and x2 = 37.5 • “Exactly 27”: x1 = 26.5 and x2 = 27.5 5 Example: tossing a coin • A fair coin is tossed 16 times. What is the probability that the number of “heads” is at least 11? • p = 0.5, q = 0.5, 𝜇 = 𝑛𝑝 = 8, 𝜎 = 𝑛𝑝𝑞 = 2 • Checking conditions: 𝑛𝑝 > 5, 𝑛𝑞 > 5 ? Yes. • Correction factor: x = 11 – 0.5 = 10.5 • The z-score: 𝑧 = 10.5−8 2 = 1.25 • The required probability: 𝑃 𝑧 > 1.25 = 0.5 − 𝑇 1.25 = 0.5 − 0.3944 = 0.1056 or 10.56% Example: rolling a die • A fair die is rolled 33 times. Find the probability that the number “4” was rolled less than 8 times. • p = 1/6, q = 5/6, 𝜇 = 𝑛𝑝 = 5.5, 𝜎 = 𝑛𝑝𝑞 = 2.14 • Checking conditions: 𝑛𝑝 > 5, 𝑛𝑞 > 5 ? Yes. • Correction factor: x = 8 – 0.5 = 7.5 • The z-score: 𝑧 = 7.5−5.5 2.14 = 0.93 • The required probability: 𝑃 𝑧 > 0.93 = 0.5 + 𝑇 0.93 = 0.5 + 0.3238 = 0.8238 or 82.38% 6 Example: blood type B • 90 Canadians are chosen at random. Find the probability that between 8 and 14 (inclusively) have blood type B. • p = 0.09, q = 0.91, 𝜇 = 𝑛𝑝 = 8.1, 𝜎 = 𝑛𝑝𝑞 = 2.71 • Checking conditions: 𝑛𝑝 > 5, 𝑛𝑞 > 5 ? Yes. • Correction factors: x1 = 7.5 and x2 = 14.5 • The z-scores: 𝑧1 = −0.22 and 𝑧2 = 2.36 • The required probability: 𝑃 −0.22 < 𝑧 < 2.36 = 𝑇(0.22) + 𝑇 2.36 = 0.0871 + 0.4909 = 0.5780 or 57.80% Example: smokers • 30 Quebecois are chosen at random. Find the probability that exactly 5 of those are smokers. • p = 0.21, q = 0.79, 𝜇 = 𝑛𝑝 = 6.3, 𝜎 = 𝑛𝑝𝑞 = 2.23 • Checking conditions: 𝑛𝑝 > 5, 𝑛𝑞 > 5 ? Yes. • Correction factors: x1 = 4.5 and x2 = 5.5 • The z-scores: 𝑧1 = −0.81 and 𝑧2 = −0.36 • The required probability: 𝑃 −0.81 < 𝑧 < −0.36 = 𝑇 0.81 − 𝑇 0.36 = 0.2910 − 0.1406 = 0.1504 or 15.04% 7