11. The Normal distributions 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 x x 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). Inflection point Inflection point Commonly called the bell curve—if you were skiing down it you would be going steeper and steeper, then start to flatten out. This is the equation—don’t have to know it—basically for every value x; you can plug it in and get f(x), the value on the y axis. The resulting curve is a density curve with total area of 1 under the curve. x x Normal curves are used to model many biological variables. They can describe a population distribution or a probability distribution.

A family of density curves Here means are the same (m = 15) whereas standard deviations are different (s = 2, 4, and 6). Here means are different (m = 10, 15, and 20) whereas standard deviations are the same (s = 3).

Human heights, by gender, can be modeled quite accurately by a Normal distribution.   Guinea pigs survival times after inoculation of a pathogen are clearly not a good candidate for a Normal model!

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. Reminder: µ (mu) is the mean of the idealized curve, and σ (sigma) is the standard deviation of the idealized curve. Number of times σ from the center µ To obtain any other area under a Normal curve, use either technology or Table B.

Standardized bone density (no units) 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%. This corresponds to bone density values of 1 or less. Between 1 and +1, there is approximately 68% of the population. Therefore, 32% are outside of this range. So, ~16 (32/2) percent of the young adult population have osteoporosis or osteopenia. Standardized bone density (no units)

Standardized bone density (no units) -6 -4 -2 2 4 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. Standardized bone density (no units) This is one standard deviation above the center 2. Because ~68% of the distribution lies within plus or minus one standard deviation from the center, the area we are looking for represents the middle 68% plus half of the remaining 32%. So, ~84% (C) of women in their 70s have either osteoporosis or osteopenia. A) 97.5% B) 95% C) 84% D) 68% E) 50% What is the probability that a randomly chosen woman in her 70s has osteoporosis or osteopenia (< −1)?

The standard Normal distribution We can standardize data by computing a z-score: If x has the N(m,s) distribution, then z has the N(0,1) distribution. N(0,1) => N(64.5, 2.5) Standardized height (no units) The distribution of women’s heights has mean 64.5 and standard deviation 2.5 inches. When you standardize a normal distribution, you change it so the mean is 0 and the standard deviation is 1. Any normal distribution can be standardized, using the formula shown here.

Standardizing: z-scores A z-score measures the number of standard deviations that a data value x is from the mean m. When x is 1 standard deviation larger than the mean, then z = 1. When x is 2 standard deviations larger than the mean, then z = 2. When x is larger than the mean, z is positive. When x is smaller than the mean, z is negative.

We calculate z, the standardized value of x: 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= ??? mean µ = 64.5" standard deviation s = 2.5" height x = 67" Area = ??? m = 64.5” x = 67” z = 0 z = 1 We calculate z, the standardized value of x: 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 First standardize x to get z, the number of standard deviations away from the mean. Once you have standardized, you can look up any z-value you want using a table. The number inside the table gives you the area to the left of that particular z-score. (…)

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) m = 64.5 x = 67 z = 1 Remember that the density curve describes the whole population. Therefore, the bottom (left) and top (right) parts of the curve must add up to 100%.  84.13% of women are shorter than 67”.  Therefore, 15.87% of women are taller than 67" (5'6").

Tips on using Table B Area = 0.9901 Area = 0.0099 z = -2.33 area right of z = area left of –z 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 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.

What percent of middle-age men have high cholesterol (> 240 mg/dl)? 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)? What percent have elevated cholesterol (between 200 and 240 mg/dl)? x z area left area right 240 0.49 69% 31% 200 -0.59 28% 72% 37 P (x > 240) = 31% (upper area, on the right of x = 240) P (200 < x < 240) = P (x < 240) – P (x < 200) = 0.69 – 0.28 = 0.41, or 41% 111 148 185 222 259 296 333

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.

What is the third quartile of the distribution of hatching weights? 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 Here, we want the lower 75%, which is what Table B gives (the area to the left of z). So the z we are looking for separates the lower 25% from the upper 75%. Q3 ≈ 47.7 g

Cholesterol level (mg/dl) 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 z Area left Area right 1.28 90% 10% 37 x = µ + z*σ = 222 + (1.28)(37) ≈ 269.4. Cholesterol values above 269 mg/dl correspond to the 10% highest cholesterol levels. 111 148 185 222 259 296 333 Cholesterol level (mg/dl)

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 z-scores. 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.

(~ straight-line pattern) Roughly normal (~ straight-line pattern) Right skewed (most of the data points are short survival times, while a few are longer survival times)