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The Normal distributions BPS chapter 3 © 2006 W.H. Freeman and Company Objectives (BPS 3) The Normal distributions Density curves Normal distributions The 68-95-99.7 rule The standard Normal distribution Using the calculator to find Normal proportions Finding a value given a proportion Density curves A density curve is a mathematical model of a distribution. It is always on or above the horizontal axis. The total area under the curve, by definition, is equal to 1, or 100%. The area under the curve for a range of values is the proportion of all observations for that range. From: Relative area enclosed by bars of histogram To: Relative area under density curve. Density curves come in any imaginable shape. Some are well-known mathematically and others aren’t. Normal distributions Normal—or Gaussian—distributions are a family of symmetrical, bellshaped density curves defined by a mean m (mu) and a standard deviation s (sigma): N (m, s). 1 f ( x) e 2 1 xm 2 s 2 x e = 2.71828… The base of the natural logarithm π = pi = 3.14159… x The normal density curve! Shape? Center? Spread? Area Under Curve? Assume Same Scale on Horizontal and Vertical (not drawn) Axes. How does the standard deviation affect the shape of the normal density curve? How does the magnitude of the standard deviation affect a density curve? Drawing Density Curves Suppose the distribution of heights, X, of young women aged 18 to 24 is approximately normal with mean mu=64.5 inches and standard deviation sigma=2.5 inches (i.e. X~N(64.5,2.5)). Lets draw the density curve for X! A family of density curves Here the means are the same (m = 15) while the standard deviations are different (s = 2, 4, and 6). 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 Here the means are different (m = 10, 15, and 20) while the standard deviations are the same (s = 3). 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 All Normal curves N (m, s) share the same properties (page 71) About 68% of all observations are within 1 standard deviation Inflection point (s) of the mean (m). About 95% of all observations are within 2 s of the mean m. Almost all (99.7%) observations are within 3 s of the mean. This is called the 68-95-99.7 Rule (aka Empirical Rule) mean µ = 64.5 Let’s work Problem 3.6 on Page 74 standard deviation s = 2.5 N(µ, s) = N(64.5, 2.5) Reminder: µ (mu) is the mean of the idealized curve, while x is the mean of a sample. σ (sigma) is the standard deviation of the idealized curve, while s is the s.d. of a sample. Example: Women heights N(µ, s) = N(64.5, 2.5) Women’s heights follow the N(64.5″,2.5″) distribution. Using the empirical rule, Area= ??? determine what percent of women are Area = ??? shorter than 67 inches tall (that’s 5′7″)? mean µ = 64.5" standard deviation s = 2.5" x (height) = 67" m = 64.5″ x = 67″ z =0 z =1 We calculate z, the standardized value of x: z (x m) s , z (67 64.5) 2.5 1 1 stand. dev. from mean 2.5 2.5 Because of the 68-95-99.7 rule, we can conclude that the percent of women shorter than 67″ should be, approximately, .68 + half of (1 − .68) = .84, or 84%. How many standard deviations from the mean height is the height of a woman who is 68 inches? Who is 58 inches? Standardizing: calculating z-scores A z-score measures the number of standard deviations that a data value x is from the mean m. z (x m ) s Using the previous slide, find the z-score of a woman whose height is 70 inches, whose height is 60 inches. When x is larger than the mean, z is positive. When x is smaller than the mean, z is negative. Now let’s go backwards. Given a z-score find the corresponding data value x. Using the previous slide, find the height of a woman whose z-score is -1.25, whose z-score is 2.4. x m zs The standard Normal distribution Because all Normal distributions share the same properties, we can standardize our data to transform any Normal curve N (m, s) into the standard Normal curve N (0,1). N(64.5, 2.5) N(0,1) => x z Standardized height (no units) For each x we calculate a new value, z (called a z-score). The z-score is the number of standard deviations that a data value x is from the mean. Using Our Calculators to Find the Percent of Women Shorter than 67” •TI-83 Calculator Command: Distr|normalcdf •Syntax: normalcdf(left, right, mu, sigma) = area under normal curve (with mean mu and standard deviation sigma) from left to right •mu defaults to 0, sigma defaults to 1 •Infinity is 1E99 (use the EE key), Minus Infinity is -1E99 Normalcdf(-1E99,67,64.5,2.5) Can we compute using z-scores?? Normalcdf(-1E99,1,0,1)=Normalcdf(-1E99,1) N(µ, s) = N(64.5”, 2.5”) Area ≈ 0.84 Question: Area ≈ 0.16 What percent of women are taller than 67“?? m = 64.5” x = 67” z=1 The National Collegiate Athletic Association (NCAA) requires Division I athletes to score at least 820 on the combined math and verbal SAT exam to compete in their first college year. The SAT scores of 2003 were approximately normal with mean 1026 and standard deviation 209. What proportion of all students would be NCAA qualifiers (SAT ≥ 820)? x 820 m 1026 Draw Picture s 209 Compute z (x m ) s (820 1026) 209 206 z 0.99 209 z Normalcdf(820,1E99,1026,209) Normalcdf(-.99,1E99) State Answer Area right of 820 = = Total area 1 − − Area left of 820 0.1611 ≈ 84% Answer: Approximately 84% of students who took the SAT in 2003 scored at least 820. Note: The actual data may contain students who scored exactly 820 on the SAT. However, the proportion of scores exactly equal to 820 being 0 for a normal distribution is a consequence of the idealized smoothing of density curves. The NCAA defines a “partial qualifier” eligible to practice and receive an athletic scholarship, but not to compete, as a combined SAT score of at least 720. What proportion of all students who take the SAT would be partial qualifiers? That is, what proportion have scores between 720 and 820? x 720 m 1026 s 209 z (x m ) Draw Picture Compute State Answer s (720 1026) 209 306 z 1.46 209 z About 9% of all students who take the SAT have scores between 720 and 820. The cool thing about working with normally distributed data is that we can manipulate it and then find answers to questions that involve comparing seemingly noncomparable distributions. We do this by “standardizing” the data. All this involves is changing N(0,1) the scale so that the mean now equals 0 and the standard deviation equals 1. If you do this to different distributions, it makes them comparable. (x m ) Let’s work Problem 3.8 on Page 75 z s Finding a value given a proportion When you know the proportion, but you don’t know the x-value that represents the cut-off, you have the inverse problem. 1. State the problem and draw a picture. 2. Use the invNorm key to find the corresponding x or z. Syntax: invNorm(area,mu,sigma)=x • • x is a value of the quantitative variable area is the area to the left of x under normal curve with mean mu and standard deviation sigma. 3. If finding z, unstandardize to transform z back to the original x scale by using the formula: x m zs Example: Women’s heights Women’s heights follow the N(64.5″,2.5″) distribution. What is the 25th percentile for women’s heights? mean µ = 64.5" standard deviation s = 2.5" proportion = area under curve=0.25 Draw a picture! Compute: We will use our calculator to get both x and z. If we have computed z we can convert back to x State Answer: The 25th percentile for women’s heights is 62.825”, or 5’ 2.82”. One More Problem! Let’s Work Problem 3.14 on Page 83