1. The Normal distributions PSLS chapter 11 © 2009 W.H. Freeman and Company

2. 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

3. Normal distributions Normal curves are used to model many biological variables. They can describe the population distribution or density curve. Normal – or Gaussian – distributions are a family of symmetrical, bell shaped density curves defined by a mean  (mu) and a standard deviation  (sigma): N(  ). xx

4. 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!

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

6. mean µ = 64.5 standard deviation  = 2.5 N(µ,  ) = N(64.5, 2.5) The 68 – 95 – 99.7 rule for any N(μ,σ) Reminder: µ (mu) is the mean of the idealized curve, while 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.  About 68% of all observations are within 1 standard deviation (  of the mean (  ).  About 95% of all observations are within 2  of the mean .  Almost all (99.7%) observations are within 3  of the mean. Inflection point

7. Because all Normal distributions share the same properties, we can standardize to transform any Normal curve N(  ) into the standard Normal curve N(0,1). The standard Normal distribution For each x we calculate a new value, z (called a z-score). N(0,1) => N(64.5, 2.5) Standardized height, standard deviation units

8. A z-score measures the number of standard deviations that a data value x is from the mean . Standardizing: calculating z-scores When x is larger than the mean, z is positive. When x is smaller than the mean, z is negative. 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.

9. mean µ = 64.5" standard deviation  = 2.5" height x = 67" 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%. Area= ??? N(µ,  ) = N(64.5, 2.5)  = 64.5” x = 67” z = 0z = 1 X = Women heights follow the N(64.5”, 2.5”) distribution. What percent of women are shorter than 67 inches tall (that’s 5’7”)? X Z

10. 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

11. P{X < 67} = P{Z < 1} = Area ≈ 0.84 P{X > 67} = P{Z > 1} = Area ≈ 0.16 N(µ,  ) = N(64.5, 2.5)  = 64.5 x = 67 z = 1  84.13% of women are shorter than 67”.  The complementary, or 15.87% of women are taller than 67" (5'6"). 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

12. 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 right of a z value. area right of z = 1 - area left of z Area = 0.9901 Area = 0.0099 z = -2.33 area right of z = area left of -z

13. 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. area between z 1 and z 2 = area left of z 2 – area left of z 1 Don’t subtract the z values!!! Normal curves are not square! Then subtract the smaller area from the larger area.  The area under N(0,1) for a single value of z is zero

14. 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

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

17. 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.

18. 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 high cholesterol (> 240 mg/dl)? What is the probability she has an elevated cholesterol (between 200 and 240)? 68107146185224263302 39 xz area left area right 2401.4192%8% 2000.3865%35%

19. 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)? 111148185222259296333 37 xz area left area right 2400.4969%31% 200-0.5928%72%