1. Models for Continuous Variables

2. Challenge What to do about histograms describing distributions for continuous data? Especially for large collections. Tabulating each unique value is cumbersome. Bin choices are arbitrary.

3. Models for Continuous Populations Distributions of continuous data are modeled with smooth curves (“density functions”). Nonnegative. Total area under the curve is exactly 1. The area under the curve above an interval is equal to the probability of a result in that interval.

4. Example – Female Longevity Left skewed. Median > Mean. Standard deviation  10 – 12

5. Example – Female Longevity If we want the probability a woman lives to at least 90 years of age…

6. Example – Female Longevity If we want the probability a woman lives to at least 90 years of age… …we find the area under the curve over the interval extending from 90 to the right.

7. Total area = 1 Total area= # rectangles  size of one rectangle Each rectangle is 2  0.004 = 0.008 units of area. 1= # rectangles  0.008 # rectangles = 1 / 0.008 = 125 32 34 etc.

9. Interval# Rectangles 90 – 9210.9 92 – 9410.2 94 – 968.5 96 – 985.7 98 – 1002.2 1 +0.0+ Total37.5

10. Interval# Rectangles 90 – 9210.9 92 – 9410.2 94 – 968.5 96 – 985.7 98 – 1002.2 1 +0.0+ Total37.5 Total of 37.5 rectangles Each rectangle is 0.008 area. The area under the curve is  37.5  0.008 = 0.30, or  37.5 / 125 = 0.30. 30.0% of women live to at least 90 years of age. 90 years is the 70 th percentile.

11. Example – Female Longevity What is the probability a woman dies between the ages of 60 and 70?

12. Interval# Rectangles 60 – 621.2 62 – 641.6 64 – 661.9 66 – 682.3 68 – 702.8 Total9.8 Total of 9.8 rectangles Each rectangle is 0.008 area. The area under the curve is  9.8  0.008 = 0.0784. About 7.84% of women die between the ages of 60 and 70.

13. Example – Female Longevity Determine the median. It’s below 90.And above 80.

14. Example – Female Longevity 0.50 probability above M / 0.5 below M 0.5 area (under curve) below M; 0.5 above M

15. Median Female Longevity 67.5 rectangles under curve below 86 67.5  0.008 = 0.54.  86 is 54 th %-ile 86 is too high

16. Median Female Longevity 58 rectangles under curve below 84 58  0.008 = 0.464.  84 is 46.4 th %-ile 84 is too low

17. %-ile k46.450.054.0 Value x84???86

18. %-ile k46.450.054.0 Value x8484.7586 The median is about 84.75 (85) years of age.

19. Example – Female Longevity To approximate the mean… …use midpoints and probabilities. Midpoint = 71 (rounded to nearest odd year) Area = 3.4 rectangles Probability = 3.4  0.008 = 0.0272

20. Left skewed Mode = 90 Median = 85 Mean = 83

21. Example – Female Longevity Mean = balance point  = 83.0 (Easy for symmetric distributions.)

22. Example – Female Longevity The mean and standard deviation will generally be given, or follow from formulas.  = 83.0  = 11.0

23. The curve models a population distribution for a continuous variable. The model must capture the important information in the population of data. If we think of the experiment that randomly selects a single item from the population and records a result, we call this curve a probability distribution.

24. Example Wait time (minutes) until seating at a restaurant is modeled by y = x / 50 over the range from x = 0 to 10.

25. Why is this a legitimate model for a continuous variable? It’s nonnegative. The total area is ½  b  h = 0.5(10)(0.2) = 1.

26. Determine the probability of a wait less than 6 min. If x = 6: y = 6/50 = 0.12. The shaded area is 0.5(6)(0.12) = 0.36. (6 is the 36 th percentile.)

27. Determine the probability of waiting longer than 8 min. If x = 8, y = 8/50 = 0.16. The pink area is 0.5(8)(0.16) = 0.64. (So 8 is the 64 th percentile.) The yellow area is the probability of a result greater than 8: 1 – 0.64 = 0.36.

28. Determine the median wait. 6 is the 36 th percentile; 8 is the 64 th. The median is the 50 th. 7 would be a good guess. However, the probability of a result less than 7 is 0.5(7)(0.14) = 0.49. The median is a bit above 7.

29. Determine the median m. Whatever m is: 0.5 m (m/50) = 0.5. So m 2 = 50. The median is m = 7.071.

30. The mode is 10.00 The median is m = 7.071. The mean is  = 6+ 2/3 = 6.667. 

31. Determine the probability of a result between 5.5 and 6.5. Area below 6.5: 0.4225 Area below 5.5: 0.3025 Area between 5.5 and 6.5: 0.1200 If the result were rounded to the nearest whole number, the probability is 0.12 that it rounds to 6.

32. Determine the probability of a result between 4.5 and 5.5. Area below 5.5: 0.3025 Area below 4.5: 0.2025 Area between 4.5 and 5.5: 0.1000 If the result were rounded to the nearest whole number, the probability is 0.10 that it rounds to 5.

33. Rounds to 0 (between 0.0 and 0.5) Rounds to 1 (between 0.5 and 1.5) Rounds to 2 (between 1.5 and 2.5) Rounds to 3 (between 2.5 and 3.5) Rounds to 4 (between 3.5 and 4.5) Rounds to 5 (between 4.5 and 5.5) 0.1000 Rounds to 6 (between 5.5 and 6.5) 0.1200 Rounds to 7 (between 6.5 and 7.5) Rounds to 8 (between 7.5 and 8.5) Rounds to 9 (between 8.5 and 9.5) Rounds to 10 (between 9.5 and 10.0)

34. Rounds to 0 (between 0.0 and 0.5) Rounds to 1 (between 0.5 and 1.5) 0.0200 Rounds to 2 (between 1.5 and 2.5) 0.0400 Rounds to 3 (between 2.5 and 3.5) 0.0600 Rounds to 4 (between 3.5 and 4.5) 0.0800 Rounds to 5 (between 4.5 and 5.5) 0.1000 Rounds to 6 (between 5.5 and 6.5) 0.1200 Rounds to 7 (between 6.5 and 7.5) 0.1400 Rounds to 8 (between 7.5 and 8.5) 0.1600 Rounds to 9 (between 8.5 and 9.5) 0.1800 Rounds to 10 (between 9.5 and 10.0) Sum to 0.90

35. Determine the probability of a result between 0 and 0.5. (Would round to 0.) Area below 0.5: 0.0025 Determine the probability of a result between 9.5 and 10.0. (Would round to 10.) Area below 10.0:1.0000 Area below 9.5: 0.9025 Area between 9.5 and 10.0:0.0975

36. Rounds to 0 (between 0.0 and 0.5) 0.0025 Rounds to 1 (between 0.5 and 1.5) 0.0200 Rounds to 2 (between 1.5 and 2.5) 0.0400 Rounds to 3 (between 2.5 and 3.5) 0.0600 Rounds to 4 (between 3.5 and 4.5) 0.0800 Rounds to 5 (between 4.5 and 5.5) 0.1000 Rounds to 6 (between 5.5 and 6.5) 0.1200 Rounds to 7 (between 6.5 and 7.5) 0.1400 Rounds to 8 (between 7.5 and 8.5) 0.1600 Rounds to 9 (between 8.5 and 9.5) 0.1800 Rounds to 10 (between 9.5 and 10.0) 0.0975 Sum to 1.0

37. Rounded Value xP(x)Mean Computation 00.00250  0.0025 = 0.0000 1 0.0200 1  0.0200 = 0.0200 2 0.0400 2  0.0400 = 0.0800 30.0600 3  0.0600 = 0.1800 4 0.0800 4  0.0800 = 0.3200 50.1000 5  0.1000 = 0.5000 60.1200 6  0.1200 = 0.7200 70.1400 7  0.1400 = 0.9800 80.1600 8  0.1600 = 1.2800 90.1800 9  0.1800 = 1.6200 10 0.0975 10  0.0975 = 0.9750 SUM = 6.6750