M3M8D6 Have out: Bellwork: assignment, graphing calculator, pencil, red pen, highlighter Have out: Bellwork: If the mean test score was a 75 with a standard deviation of 10.5, calculate the z–scores for the test scores in parts (a) – (c). Also, determine if they are outliers. + 1 a) 23 b) 100 c) 80 + 1 d) Victoria has a z–score of –0.95. What is her raw score. Round! + 1 + 1 + 1 + 1 + 1 outlier! outlier! + 1 + 1 + 1

Percentile Percentile percentage ____________: a ranking that indicates the ___________ of students who scored _______ that score. below Example: Consider the following Geometry test scores sorted in ascending order: {17, 42, 53, 56, 57, 58, 60, 60, 69, 71, 72, 75, 75, 75, 76, 79, 80, 81, 83, 86, 86, 89, 89, 90, 92, 93, 93, 100} a) What is Ryan’s percentile if he scored an 80? b) What is Beatriz’s percentile if she scored a 71? 16 9 57% 32% 28 28 c) What is Harry’s percentile if he scored a 53? d) What is Emily’s percentile if she scored a 92? 2 24 7% 86% 28 28

The Normal Distribution If you toss a coin 8 times, then theoretically, what is the number of possible outcomes for the number of heads? We can just look at the 8th row of Pascal’s triangle: x 1 2 3 4 5 6 7 8 P(X)

x 1 2 3 4 5 6 7 8 P(X) Make a relative-frequency histogram illustrating the number of heads and the frequency of each type. P(x) Connect the points. Notice that it forms a ____ - _______ ______. bell shaped curve Frequency x 1 2 3 4 5 6 7 8 # of heads

A distribution such as the one above is called a _________________. P(x) A distribution such as the one above is called a _________________. normal distribution Frequency x 1 2 3 4 5 6 7 8 # of heads This example is a ________ probability distribution because there is only a countable number of possible values. discrete On the other hand, a __________ probability distribution can be any value in an interval of real numbers. Continuous probability distributions are represented by ______, not histograms. continuous curves The_____ and ______________ determine the shape of the curve. mean standard deviation

On this normal curve below, place: , , , and . 68% 95% 99.7% For a normal distribution, 1 68 ____ % of the data falls within ___ σ of ____ % of the data falls within ___ σ of 95 2 99.7 3 ____ % of the data falls within ___ σ of This is called the __________ ______. Empirical Rule (by experience or results in a population) Normal curve animation

47.5% - 34% = 13.5% 50% 50% - 34% = 16% 50% - 47.5% = 2.5% 34% 68% 95% 99.7% Normal curve animation

a) What percent of men have heights over 69 inches? Example # 1: The heights of adult American men are normally distributed with a mean of 69 inches and a standard deviation of 2.5 inches. Sketch a normal curve, and label , , , and . a) What percent of men have heights over 69 inches? b) What percent of men are taller than 71.5 inches? 50% 50% – 34% = 16% percent σ = 2.5 34% // 61.5 64 66.5 69 71.5 74 76.5 Height –3 –2 –1 1 2 3 z-score

c) What percent of men are shorter than 64 inches? 50% – 13.5% – 34% = 2.5% d) What percent of men are between 64 and 71.5 inches? 13.5% + 68% = 81.5% percent σ = 2.5 34% 68% 13.5% // 61.5 64 66.5 69 71.5 74 76.5 Height –3 –2 –1 1 2 3 z-score

a) What is the percentile for a score of 110? Example # 2: The Stanford – Binet IQ Test is normally distributed with a mean of 100 and standard deviation of 10. Sketch a normal curve, and label , , , and . a) What is the percentile for a score of 110? 50% + 34% = 84% b) What percent of people score over 120? 50% – 34% – 13.5% = 2.5% Percent 34% + 13.5% σ = 10 50% 34% // Score 70 80 90 100 110 120 130 –3 –2 –1 1 2 3 z-score

c) What is the percent for a score of 80? 50% + 47.5% = 97.5% d) What percent of people score between 80 and 110? 68% + 13.5% = 81.5% Percent 50% 47.5% σ = 10 68% 13.5% // Score 70 80 90 100 110 120 130 –3 –2 –1 1 2 3 z-score

e) Forest Gump’s score was 76. What was his z–score? f) Is Forest Gump an outlier? Yes, his score is an outlier because the z-score is more than 2σ below the mean. Percent σ = 10 // Score 70 80 90 100 110 120 130 –3 –2 –1 1 2 3 z-score

Example # 3: The heights of American women, aged 18 – 24, are normally distributed, with a mean of 64.5 inches and a standard deviation of 2.5 inches. Sketch a normal curve, and label , , , and . In a random sample of 2000 young women, how many would you expect to be: a) Over 64.5 inches? b) Under 62 inches? 50% 50% – 34% = 16% 2000(0.5) = 1000 women 2000(0.16) = 320 women Percent σ = 2.5 // 57 59.5 62 64.5 67 69.5 72 Height –3 –2 –1 1 2 3 z–score

c) Between 62 and 64.5 inches? 34% 2000(0.34) = 680 women d) Under 67 inches? 50% + 34% = 84% 2000(0.84) =1680 women Percent σ = 2.5 50% 34% 34% // 57 59.5 62 64.5 67 69.5 72 Height –3 –2 –1 1 2 3 z–score

e) Over 69.5 inches? 50% – 34% – 13.5% = 2.5% 2000(0.025) = 50 women f) Under 57 inches? 50% – 49.85% = 0.15% 2000(.0015) = 3 women Percent σ = 2.5 13.5% = 49.85% 34% // 57 59.5 62 64.5 67 69.5 72 Height –3 –2 –1 1 2 3 z–score

Finish the worksheet.

The_____ and ______________ determine the shape of the curve. mean standard deviation Practice # 1: On the axes provided, sketch 2 different normal curves with the same mean but different standard deviations. Y = \ Y1 = 2nd DISTR 1: normalpdf ( ENTER X, T, , n , , 1 ) y x \ Y2 = 2nd DISTR 1: normalpdf ( ENTER X, T, , n , , 1.5 ) x σ y2 is lower because the total area under the curve is still 1 (100%). y1 Both have the same amount of data. y2 Normal curve animation