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The Normal distribution The Practice of Statistics in the Life Sciences Third Edition © 2014 W. H. Freeman and Company Objectives (PSLS Chapter 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 quantile plots 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 Inflection point Inflection point x Normal curves are used to model many biological variables. They can describe a population distribution or a probability distribution. x A family of density curves Here means are the same (m = 15) whereas 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) whereas standard deviations are the same (s = 3). 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 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. The 68–95–99.7 rule for any N(μ,σ) All normal curves N(µ,σ) share the same properties: About 68% of all observations 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. Number of times σ from the center µ To obtain any other area under a Normal curve, use either technology or Table B. World Health Organization definitions of osteoporosis based on standardized bone density levels Normal Bone density is within 1 standard deviation (z > –1) of the young adult mean or above. Low bone mass Bone density is 1 to 2.5 standard deviations below the young adult mean (z between –2.5 and –1). Osteoporosis Bone density is 2.5 standard deviations or more below the young adult mean (z ≤ –2.5). What percent of young adults have osteoporosis or osteopenia? Population of young adults N(0,1) These are bone densities of 1 or less, representing the area to the left of the middle 68% between 1 and +1. So 16%. z Standardized bone density (no units) Young adults N(0,1) Women 70-79 N(-2,1) Women aged 70 to 79 are NOT young adults. The mean bone density in this age is about −2 on the standard scale for young adults. -6 -4 -2 0 2 4 z Standardized bone density (no units) What is the probability that a randomly chosen woman in her 70s has osteoporosis or osteopenia (< −1)? A) 97.5% B) 95% C) 84% D) 68% E) 50% The standard Normal distribution We can standardize data by computing a z-score: z (x m ) s If x has the N(m,s) distribution, then z has the N(0,1) distribution. N(64.5, 2.5) N(0,1) => x z Standardized height (no units) Standardizing: 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. Women’s heights follow the N(64.5”,2.5”) distribution. What percent of women are shorter than 67 inches tall (that’s 5’6”)? N(µ, s) = N(64.5, 2.5) Area= ??? Area = ??? mean µ = 64.5" standard deviation s = 2.5" height x = 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 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 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. N(µ, s) = N(64.5, 2.5) 84.13% of women are shorter than 67”. Area ≈ 0.84 Area ≈ 0.16 Therefore, 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 two ways of finding the area under N(0,1) curve to the right of a z-value. Area = 0.9901 Area = 0.0099 z = -2.33 area right of z = area left of –z area right of z = 1 – area left of z Using Table B to find a middle area To calculate the area between two 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 z1 – area left of z2 The area under N(0,1) for a single value of z is zero. 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 Inverse Normal calculations You may also seek the range of values that correspond to a given proportion/ area under the curve. For that, use technology or 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. The hatching weights of commercial chickens can be modeled accurately using a Normal distribution with mean μ = 45 g and standard deviation σ = 4 g. What is the third quartile of the distribution of hatching weights? 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% We find z ≈ 0.67 z (x m) s x m ( z *s ) x 45 (0.67 * 4) x 47.68 Q3 ≈ 47.7 g 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 range of values corresponds to the 10% highest cholesterol levels? A) > 175 B) > 247 C) > 269 D) > 288 Normal z Area left Area right 1.28 90% 10% 37 29 111 68 107 148 146 185 185 224 263 222 302 341 259 Cholesterol level (mg/dl) 296 333 Normal quantile plots One way to assess if a data set has an approximately Normal distribution is to plot the data on a Normal quantile plot. The data points are ranked and the percentile ranks are converted to zscores. The z-scores are then used for the horizontal axis and the actual data values are used for the vertical axis. Use technology to obtain Normal quantile plots. If the data have approximately a Normal distribution, the Normal quantile plot will have roughly a straight-line pattern. Roughly normal Right skewed (~ straight-line pattern) (most of the data points are short survival times, while a few are longer survival times)