1. The Normal distributions BPS chapter 3 © 2006 W.H. Freeman and Company

2. Objectives (BPS 3) The Normal distributions  Density curves  Normal distributions  The 68-95-99.7 rule  The standard Normal distribution  Using the calculator to find Normal proportions  Finding a value given a proportion

3. Density curves A density curve is a mathematical model of a distribution. It is always on or above the horizontal axis. The total area under the curve, by definition, is equal to 1, or 100%. The area under the curve for a range of values is the proportion of all observations for that range. From: Relative area enclosed by bars of histogram To: Relative area under density curve.

4. Density curves come in any imaginable shape. Some are well-known mathematically and others aren’t.

5. Normal distributions e = 2.71828… The base of the natural logarithm π = pi = 3.14159… 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

6. The normal density curve! Shape? Center? Spread? Area Under Curve? How does the magnitude of the standard deviation affect a density curve? How does the standard deviation affect the shape of the normal density curve? Assume Same Scale on Horizontal and Vertical (not drawn) Axes.

7. Drawing Density Curves Suppose the distribution of heights, X, of young women aged 18 to 24 is approximately normal with mean mu=64.5 inches and standard deviation sigma=2.5 inches (i.e. X~N(64.5,2.5)). Lets draw the density curve for X!

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

9. mean µ = 64.5 standard deviation  = 2.5 N(µ,  ) = N(64.5, 2.5) All Normal curves N  ) share the same properties (page 71) 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.  This is called the 68-95-99.7 Rule (aka Empirical Rule)  Let’s work Problem 3.6 on Page 74 Inflection point

10. mean µ = 64.5" standard deviation  = 2.5" x (height) = 67" We calculate z, the standardized value of x: Because of the 68-95-99.7 rule, we can conclude that the percent of women shorter than 67″ should be, approximately,.68 + half of (1 −.68) =.84, or 84%. Area= ??? N(µ,  ) = N(64.5, 2.5)  = 64.5″ x = 67″ z = 0z = 1 Example: Women heights Women’s heights follow the N(64.5″,2.5″) distribution. Using the empirical rule, determine what percent of women are shorter than 67 inches tall (that’s 5′7″)?

11. How many standard deviations from the mean height is the height of a woman who is 68 inches? Who is 58 inches?

12. 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. Using the previous slide, find the z-score of a woman whose height is 70 inches, whose height is 60 inches. Using the previous slide, find the height of a woman whose z-score is -1.25, whose z-score is 2.4. Now let’s go backwards. Given a z-score find the corresponding data value x.

13. Because all Normal distributions share the same properties, we can standardize our data 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). The z-score is the number of standard deviations that a data value x is from the mean. N(0,1) => N(64.5, 2.5) Standardized height (no units)

14. Area ≈ 0.84 Area ≈ 0.16 N(µ,  ) = N(64.5”, 2.5”)  = 64.5” x = 67” z = 1 Question: What percent of women are taller than 67“?? Using Our Calculators to Find the Percent of Women Shorter than 67” TI-83 Calculator Command: Distr|normalcdf Syntax: normalcdf(left, right, mu, sigma) = area under normal curve (with mean mu and standard deviation sigma) from left to right mu defaults to 0, sigma defaults to 1 Infinity is 1E99 (use the EE key), Minus Infinity is -1E99 Normalcdf(-1E99,67,64.5,2.5) Normalcdf(-1E99,1,0,1)=Normalcdf(-1E99,1) Can we compute using z-scores??

15. The National Collegiate Athletic Association (NCAA) requires Division I athletes to score at least 820 on the combined math and verbal SAT exam to compete in their first college year. The SAT scores of 2003 were approximately normal with mean 1026 and standard deviation 209. What proportion of all students would be NCAA qualifiers (SAT ≥ 820)? Note: The actual data may contain students who scored exactly 820 on the SAT. However, the proportion of scores exactly equal to 820 being 0 for a normal distribution is a consequence of the idealized smoothing of density curves. Area right of 820= Total area − Area left of 820 =1 − 0.1611 ≈ 84% Normalcdf(820,1E99,1026,209) Normalcdf(-.99,1E99) Draw Picture Compute State Answer Answer: Approximately 84% of students who took the SAT in 2003 scored at least 820.

16. The NCAA defines a “partial qualifier” eligible to practice and receive an athletic scholarship, but not to compete, as a combined SAT score of at least 720. What proportion of all students who take the SAT would be partial qualifiers? That is, what proportion have scores between 720 and 820? About 9% of all students who take the SAT have scores between 720 and 820. Draw Picture Compute State Answer

17. N(0,1) The cool thing about working with normally distributed data is that we can manipulate it and then find answers to questions that involve comparing seemingly non- comparable distributions. We do this by “standardizing” the data. All this involves is changing the scale so that the mean now equals 0 and the standard deviation equals 1. If you do this to different distributions, it makes them comparable. Let’s work Problem 3.8 on Page 75

18. When you know the proportion, but you don’t know the x-value that represents the cut-off, you have the inverse problem. Finding a value given a proportion 1.State the problem and draw a picture. 2. Use the invNorm key to find the corresponding x or z. Syntax: invNorm(area,mu,sigma)=x x is a value of the quantitative variable area is the area to the left of x under normal curve with mean mu and standard deviation sigma. 3. If finding z, unstandardize to transform z back to the original x scale by using the formula:

19. mean µ = 64.5" standard deviation  = 2.5" proportion = area under curve=0.25 Compute: We will use our calculator to get both x and z. If we have computed z we can convert back to x State Answer: The 25 th percentile for women’s heights is 62.825”, or 5’ 2.82”. Example: Women’s heights Women’s heights follow the N(64.5″,2.5″) distribution. What is the 25 th percentile for women’s heights? Draw a picture!

20. One More Problem!  Let’s Work Problem 3.14 on Page 83