11. The Normal distributions

1. 11. The Normal distributions

2. 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

3. 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). Inflection point Inflection point x x Normal curves are used to model many biological variables. They can describe a population distribution or a probability distribution .

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

5. 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!

6. 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.

7. What percent of young adults have osteoporosis or osteopenia? Population of young adults N(0,1) Standardized bone density (no units)

8. 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) What is the probability that a randomly chosen woman in her 70s has osteoporosis or osteopenia (< −1 on the standard scale)?

9. The standard Normal distribution N(64.5, 2.5) N(0,1) => Standardized height (no units) We can standardize data by computing a z-score: If x has the N(m,s) distribution, then zhas the N(0,1) distribution.

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

11. 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%.

12. Using Table B .0062 is the area under N(0,1) left of z = –2.50 0.0052 is the area under N(0,1) left of z = –2.56 .0060 is the area under N(0,1) left of z = –2.51 Table B gives the area under the standard Normal curve to the left of any z-value. (…)

13. Using Excel • See supplement

14. N(µ, s) = N(64.5, 2.5) Area ≈ 0.84 Area ≈ 0.16 m = 64.5 x = 67 z = 1 For z = 1.00, the area under the curve to the left of z is 0.8413.  84.13% of women are shorter than 67”.  Therefore, 15.87% of women are taller than 67" (5'6").

15. area right of z = 1 – area left of z 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

16. 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

17. 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)? 37 111 148 185 222 259 296 333

20. Inverse Normal calculations For a left area of 1.25 % (0.0125), the z-value is – 2.24 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-valuefrom the left column and top row.

21. 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? upper 75% ?  The 75% longest pregnancies in this population are about 256 days or longer.

22. 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.

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