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The Normal distributions PSLS chapter 11 © 2009 W.H. Freeman and Company Objectives (PSLS 11) The Normal distributions Normal distributions The 68-95-99.7 rule The standard Normal distribution Using the standard Normal table (Table B) Inverse Normal calculations Normal distributions Normal – or Gaussian – distributions are a family of symmetrical, bell shaped 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 Normal curves are used to model many biological variables. They can describe the population distribution or density curve. x 18 16 14 12 10 8 6 4 2 Height (inches) Guinea pigs survival times after inoculation of a pathogen are clearly not a good candidate for a Normal model! 72 or more 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 0 under 56 Percent Human heights, by gender, can be modeled quite accurately by a Normal distribution. A family of density curves Here means are the same (m = 15) while 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 means are different (m = 10, 15, and 20) while standard deviations are the same (s = 3) 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 The 68–95–99.7 rule for any N(μ,σ) About 68% of all observations Inflection point are within 1 standard deviation (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. mean µ = 64.5 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. The standard Normal distribution Because all Normal distributions share the same properties, we can standardize 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, standard deviation units For each x we calculate a new value, z (called a z-score). 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 When x is 1 standard deviation larger than the mean, then z = 1. for x m s , z m s m s 1 s s When x is 2 standard deviations larger than the mean, then z = 2. for x m 2s , z m 2s m 2s 2 s s When x is larger than the mean, z is positive. When x is smaller than the mean, z is negative. X = Women heights follow the N(64.5”, 2.5”) N(µ, s) = N(64.5, 2.5) distribution. What percent of women are shorter than 67 inches tall (that’s 5’7”)? Area= ??? Area = ??? mean µ = 64.5" standard deviation s = 2.5" height x = 67" m = 64.5” x = 67” z=0 z=1 X Z 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 Given the 68-95-99.7 rule, the percent of women shorter than 67” should be, approximately, .68 + half of (1 - .68) = .84 or 84%. The probability of randomly selecting a woman shorter than 67” is also ~84%. Using Table B Table B gives the area under the standard Normal curve to the left of, i.e., less than, any z value. .0062 is the area under N(0,1) left of z = -2.50 .0060 is the area under N(0,1) left of z = -2.51 0.0052 is the area under N(0,1) left of z = -2.56 (…) For z = 1.00, the area under the curve to the left of z is 0.8413, i.e., P{Z < 1.00} = 0.8413 N(µ, s) = N(64.5, 2.5) 84.13% of women are shorter than 67”. P{X < 67} = P{Z < 1} = Area ≈ 0.84 P{X > 67} = P{Z > 1} = Area ≈ 0.16 The complementary, or 15.87% of women are taller than 67" (5'6"). m = 64.5 x = 67 z=1 Tips on using Table B Because of the curve’s symmetry, there are 2 ways of finding the area under N(0,1) curve to the Area = 0.9901 right of a z value. Area = 0.0099 z = -2.33 area right of z = area left of -z area right of z = 1 - area left of z More tips on using Table B To calculate the area between 2 z- values, first get the area under N(0,1) to the left for each z-value from Table B. Then subtract the smaller area from the larger area. Don’t subtract the z values!!! Normal curves are not square! area between z1 and z2 = area left of z2 – area left of z1 The area under N(0,1) for a single value of z is zero Inverse Normal calculations You may also seek the range of values that correspond to a given proportion/ area under the curve. For that, use Table B backward: first find the desired area/ proportion in the body of the table, then read the corresponding z-value from the left column and top row. For a left area of 1.25 % (0.0125), the z-value is -2.24 Vitamins and better food: The lengths of pregnancies when malnourished mothers are given vitamins and better food is approximately N(266, 15). How long are the 75% longest pregnancies in this population? We know μ, σ, and the area under the curve; we want x. Table B gives the area left of z look for the lower 25%. upper 75% We find z ≈ -0.67 z (x m) s 206 x m ( z *s ) 221 236 ? 251 266 281 296 Gestation time (days) x 266 (0.67 *15) The 75% longest pregnancies in this x 255.95 256 population are about 256 days or longer. 311 Checking your cholesterol High levels of total serum cholesterol increase the risk of cardiovascular disease. Cholesterol levels above 240 mg/dl demand medical attention because they place the subject at high risk of CV disease. In the hope of extending treatment benefits to patients with early disease, various professional societies have recommended a lower threshold value for diagnosis. Levels above 200 mg/dl are considered elevated cholesterol and may place the person at some risk of cardiovascular disease. The cholesterol levels for women aged 20 to 34 follow an approximately Normal distribution with mean 185 mg/dl and standard deviation 39 mg/dl. What is the probability that a young woman has highNormal cholesterol (> 240 mg/dl)? What is the probability she has an elevated cholesterol (between 200 and 240)? x z area left area right 240 1.41 92% 8% 200 0.38 65% 35% 39 68 29 68 107 146 107 146 185 224 185 263 302 224 341 263 302 The blood cholesterol levels of men aged 55 to 64 are approximately Normal with mean 222 mg/dl and standard deviation 37 mg/dl. What percent of middle-age men have high cholesterol (> 240 mg/dl)? Normal What percent have elevated cholesterol (between 200 and 240 mg/dl)? z area left area right 0.49 69% 31% 200 -0.59 28% 72% x 240 37 111 68 29 148 146 107 185 185 224 222 263 302 259 341 296 333