Models for Continuous Variables

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.

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.

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

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

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.

Total area = 1 Total area= # rectangles  size of one rectangle Each rectangle is 2  = units of area. 1= # rectangles  # rectangles = 1 / = etc.

Interval# Rectangles 90 – – – – – Total37.5

Interval# Rectangles 90 – – – – – Total37.5 Total of 37.5 rectangles Each rectangle is area. The area under the curve is  37.5  = 0.30, or  37.5 / 125 = % of women live to at least 90 years of age. 90 years is the 70 th percentile.

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

Interval# Rectangles 60 – – – – – Total9.8 Total of 9.8 rectangles Each rectangle is area. The area under the curve is  9.8  = About 7.84% of women die between the ages of 60 and 70.

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

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

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

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

%-ile k Value x84???86

%-ile k Value x The median is about (85) years of age.

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  =

Left skewed Mode = 90 Median = 85 Mean = 83

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

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

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.

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

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.

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

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

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) = The median is a bit above 7.

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

The mode is The median is m = The mean is  = 6+ 2/3 = 

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

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

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) Rounds to 6 (between 5.5 and 6.5) 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)

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) Rounds to 6 (between 5.5 and 6.5) 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) Sum to 0.90

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

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) Rounds to 6 (between 5.5 and 6.5) 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) Sum to 1.0

Rounded Value xP(x)Mean Computation  =  =  =  =  =  =  =  =  =  =  = SUM =