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Continuous Random Variables MATH 130, Elements of Statistics I J. Robert Buchanan Department of Mathematics Fall 2015 Objectives During this lesson we will learn to: I use the uniform probability distribution, I graph a normal curve, I state the properties of the normal curve, I explain the role of area in the normal density function. Background Recall: I A continuous random variable may take on an infinite number of values from a range without gaps. I The probability that a continuous random variable lies in particular interval depends on the shape of the distribution for the variable. Uniform Distribution Suppose that the length of time a light bulb will burn (measured in hours) is any number 0 ≤ X ≤ 1000. All intervals of equal length are equally probable. The distribution of this random variable is uniform. PDF 0.0020 0.0015 0.0010 0.0005 200 400 600 800 1000 X Equal Areas PDF 0.0020 0.0015 0.0010 0.0005 200 400 600 800 1000 Remarks: I Equal areas imply equal probability X lies in those intervals. I Total area under the line is 1. X Probability Density Function Definition A probability density function (PDF) is a function used to compute probabilities of continuous random variables. The PDF must satisfy the following two properties: 1. The total area under the graph of the PDF over all possible values of the random variable must be 1. 2. The height of the graph must be greater than or equal to 0 for all possible values of the random variable. The area under the graph of the PDF over an interval represents the probability that the random variable has a value in that interval. Example Find the probability that 0.5 ≤ X ≤ 1.5. PDF 1.0 0.8 0.6 0.4 0.2 0.5 1.0 1.5 2.0 X Solution Using the area formula for a triangle we have PDF 1.0 0.8 P(0.5 ≤ X ≤ 1.5) 1 1 = (1.5)(0.75) − (0.5)(0.25) 2 2 = 0.5625 − 0.0625 0.6 0.4 0.2 0.5 1.0 1.5 2.0 X = 0.5 Normal Random Variables Definition A continuous random variable is normally distributed or has a normal probability distribution if its probability density function has a graph with the shape of a bell-shaped curve. The normal probability density function is given by the formula 1 2 2 y = √ e−(x−µ) /2σ . σ 2π PDF Graph of the Normal Distribution Μ-Σ Μ X Μ+Σ Effect of Changing µ In the figure below both normal distributions have a standard deviation of σ = 1. The curve on the left has a mean µ = 0 while the curve on the right has mean µ = 2. 0.4 PDF 0.3 0.2 0.1 0.0 -2 0 2 X 4 Effect of Changing σ In the figure below both normal distributions have a mean of µ = 0. The lower curve has a standard deviation of σ = 3 and the more pointed curve has a standard deviation of σ = 1. 0.4 PDF 0.3 0.2 0.1 0.0 -6 -4 -2 0 X 2 4 6 Properties of the Normal Distribution 1. It is symmetric about its mean µ. 2. The mean = mode = median, and thus the peak of the curve occurs at x = µ. 3. It has inflection points at µ − σ and µ + σ. 4. The area under the curve is 1. 5. The area under the curve to the left of µ is 1/2 which is also the area under the curve to the right of µ. 6. The PDF approaches 0 but never reaches 0. 7. The Empirical Rule applies. Empirical Rule 34% PDF 34% 2.35% Μ-3Σ 13.5% Μ-2Σ Μ-Σ 13.5% Μ X Μ+Σ 2.35% Μ+2Σ Μ+3Σ Weights of M&M’s Suppose 500 M&M’s are weighed and the data is presented in the histogram below. 60 40 20 44 46 48 50 For this sample x = 47.79 and s = 1.46 grams. Question: are the data normally distributed? 52 Overlay With Normal Curve 44 46 48 50 52 Standard Normal Random Variable Definition Suppose the random variable X is normally distributed with mean µ and standard deviation σ, then the random variable Z = X −µ σ is normally distributed with mean µ = 0 and standard deviation σ = 1. The random variable Z is has the standard normal distribution. Example The mean salary of new hires at a certain company is $36,000 with a standard deviation of $1,500. The salaries of new hires are normally distributed. Suppose the salary of one new hire is X = $35, 500, what is the standardized value of this employee’s salary? Example The mean salary of new hires at a certain company is $36,000 with a standard deviation of $1,500. The salaries of new hires are normally distributed. Suppose the salary of one new hire is X = $35, 500, what is the standardized value of this employee’s salary? Z = 35, 500 − 36, 000 = −0.333 1, 500 Relation Between Normal and Standard Normal 0.5 0.00025 0.4 0.00020 0.3 0.00015 0.2 0.00010 0.1 0.00005 30 000 32 000 34 000 36 000 38 000 40 000 42 000 X The shaded areas are equal in size. -4 -2 0 2 4 Z