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Section 7.6: The Normal Distribution. MTH 245: Mathematics for Management, Life, and Social Sciences F. Patricia Medina Department of Mathematics. Oregon State University November 13, 2014 Section 7.6 Section 7.6: The Normal Distribution. The Normal Distribution. Figure : Abraham DeMoivre Section 7.6: The Normal Distribution. The normal curve Abraham DeMoivre proved that areas under the curve 1 2 fz (z) = √ e−z /2 , can be used to estimate 2π X − n 12 , where X is a random binomial random P a ≤ q 1 1 n 2 2 ≤b variable with a large n and p = 12 . The “curve” is called normal curve. We will take a look into continuous probability by studying experiments with normally distributed outcomes. For instance, we might consider choosing a newborn and observing his or her weight, chose a college student on campus and observe his or her height, etc. Associated to each experiment is the normal curve, a bell-shaped curve. Section 7.6: The Normal Distribution. 0.2 0.15 0.1 0.05 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Section 7.6: The Normal Distribution. Probability for a normal distribution The probability that the value of the random variable, X, lies between two values, a and b, is the fraction of the area under the normal curve that lies between x = a and x = b, we denote it by P(a ≤ X ≤ b). The total area under the normal curve is always 1. A normal curve is completely described by its mean µ and standard deviation σ . G iven µ and σ we write down the equation of the associated normal curve as x−µ 2 1 1 y = √ e−( 2 )( σ ) . σ 2π The standard normal curve has mean µ = 0 and standard deviation (s.d.) σ = 1. Section 7.6: The Normal Distribution. 0.8 0.6 0.4 0.2 0 −2 −1 0 1 2 3 Section 7.6: The Normal Distribution. Let Z be a random variable having the standard normal distribution. Let z be any number, A(z) denotes the area under the standard normal curve to the left of z If X is a random variable having a normal distribution with mean µ and standard deviation σ , then a−µ b−µ b−µ a−µ P(a ≤ X ≤ b) = P ≤Z≤ =A −A σ σ σ σ Section 7.6: The Normal Distribution. and x−µ x−µ P(X ≤ x) = P Z ≤ =A , σ σ where Z has the standard normal distribution and A(z) is the area under that distribution to the left of z. z-Scores allow us to use a table to find the amount of area that is to the left of a given value,x , in a normal distribution. value of x − mean x−µ = . standard deviation σ The z-score should be interpreted as the number of standard deviations above/below the mean. Z − score = Section 7.6: The Normal Distribution. Appendix A Section 7.6: The Normal Distribution. Example 1 For the case where x = 38.4, µ = 22.5, and σ = 6.2 a) Find the z-score. b) What percent of the area of the normal curve lies below x = 38.4? Section 7.6: The Normal Distribution. Example 2 Find the area under the normal curve with µ = 7, σ = 2 from x = 6 to x = 10. This represents P(6 ≤ X ≤ 10) for a random variable X having the given normal distribution. Section 7.6: The Normal Distribution. Definition 3 If a score S is in the pth percentile of a normal distribution, then p% of all scores fall bellow S, and (100 − p)% of all scores fall above S. Example 4 What is the 90th percentile of the standard normal distribution? Section 7.6: The Normal Distribution. SAT scores Example 5 Assume that SAT verbal scores for a first-year class at a university are normally distributed with mean 520 and standard deviation 75. (a) The top 10% of the students are placed into the honors program for English. What is the lowest score for admittance into the honors program? (b) What is the range of the middle 90% of the SAT verbal scores at the university? (c) Find the 98th percentile of the SAT verbal scores. Section 7.6: The Normal Distribution. Example 6 The lifetime of a certain brand of tires ins normally distributed with mean µ = 30, 000 miles and standard deviation σ = 5000. The company has decided to issue a warranty for the tires but does not want to replace more than 2% of the tires that it sells. At what mileage should the warranty expires?