ap stats notes 2 - data organization

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2 4 6 8 10 12 2 4 6 8 10 12 68 95 99.7 Data Organization (Chapter 2) Date: DAY 1 Standards/Objectives: 1) know the key properties and types of density curves 2) know the standard normal density curve and the empirical rule 3) know the common methods to check for "normality" Agenda/Activities: 1A) PWRPT - Density Curves 1B) Calculator Dice Rolling: randInt(1, 6, 200) + randInt(1, 6, 200) // STATPLOT 2A) EMPIRICAL RULE and Percentiles and Areas 3A) Checks for Normality 1A) Key Properties: 1) shows 2) area 1B) roll one die 200 times, make a histogram roll two dice 200 times, make a histogram Describe the distribution. BONUS: Is there a difference between randInt(1, 6, 200) + randInt(1, 6, 200) vs. randInt(2, 12, 200)? 2A) STANDARD NORMAL DENSITY CURVE and the Empirical Rule: Area of Segments:

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2 4 6 8 10 122 4 6 8 10 12

68 95 99.7

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Data Organization (Chapter 2) Date: DAY 1Standards/Objectives:1) know the key properties and types of density curves2) know the standard normal density curve and the empirical rule3) know the common methods to check for "normality"

Agenda/Activities:1A) PWRPT - Density Curves1B) Calculator Dice Rolling: randInt(1, 6, 200) + randInt(1, 6, 200) // STATPLOT2A) EMPIRICAL RULE and Percentiles and Areas3A) Checks for Normality

1A) Key Properties: 1) shows2) area

1B) roll one die 200 times, make a histogram roll two dice 200 times, make a histogram

Describe the distribution.

BONUS: Is there a difference between randInt(1, 6, 200) + randInt(1, 6, 200) vs. randInt(2, 12, 200)?

2A) STANDARD NORMAL DENSITY CURVE and the Empirical Rule:

Area of Segments:

FUN FACT: The difference between a normal distribution and a standard normal distribution is that a standard normal distribution standardizes all normal distributions (a.k.a., makes all normal distributions same):

Process: Effects: Properties:Find z-score/standard score turns 1) z-scores =

z =

x−μσ turns 2) =

this makes all normal curves same

68%

95%

99.7%

percentile:

standard deviation count:z-score:

Calculate the percentile,standard deviation, andz-score at each mark:

BONUS: find each individual section’s area

3A) Typical Checks for Normality: 1) normal probability plot, 2) boxplot, 3) empirical rule, 4) sufficient sample size*Rank each boxplot from "least normal" to "most normal":

BONUS: Make a connection between the Empirical Rule and the Outlier Formula.

Data Organization (Chapter 2) Date: DAY 2Standards/Objectives:1) understand the concept of the z-score (and how to use the z-score table)2) students convert from raw scores to percentiles

Agenda/Activities:1A) calculating z-scores, the number of standard deviations above (+) or below (-) the mean (0)1B) converting z-scores into area/percentiles2A) practice converting raw scores into percentiles

1A) z-score =

x−μσ .

If = 13 and = 2, find the z-score of: If = 35 and = 1.3, find the z-score of:x = 15 x = 9 x = 38.9 x = 35

x = 10 x = 12.3 x = 30 x = 36.2

1B) Use the z-score table to find the area of: Use the calculator to find the area of:z < 0.8 z > -1.32 z < -2 z > 0

z > 1.78 z < 2.06 z < 1.3 z < 0.25

z > -0.35 z < -1.4 z > -1.48 z > -3.14

0.3 < z < 1.8 0.32 < z < 1.3

-2 < z < 0.1 0 < z < 2.13

0 < z < 0.57 0.7 < z < 1.9

-1.85 < z < -0.4 -2.41 < z < -1.6

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z-scores:

Percentile:

use formula

use table0

50th

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Percentile:

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= Raw scores:

z-scores:

Percentile:

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= Raw scores:

z-scores:

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= Raw scores:

z-scores:

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= Raw scores:

z-scores:

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= Raw scores:

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Percentile:

2A) draw picture, labeling and create row for raw score, z-score, percentile raw z-score (use formula); z-score percentile (use table/calculator)

Find the following:

The mean test score for the class was an 81, with a standard deviation of 6.Find the percentile rank of someone who:

scored 92 scored 70

The mean height of La Quinta girls is 64 inches, with a standard deviation of 2.5 inches.Find the percentile rank of a La Quinta girl who measures:

72 inches tall 62 inches tall

The mean La Quinta male weighs 155 lbs, with a standard deviation of 9 lbs.Find the percentile rank of a La Quinta male who weighs:

130 lbs 185 lbs

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z-scores:

Percentile:

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= Raw scores:

z-scores:

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= Raw scores:

z-scores:

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= Raw scores:

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= Raw scores:

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= Raw scores:

z-scores:

Percentile:

BONUS: The mean SAT verbal score was 496 in 2013, with a standard deviation of 80.

Jake's verbal score was at the 65th percentile. What must Jake score so that he is in the 75th

What was his verbal score? percentile?

BONUS: The mean SAT verbal score was 496 in 2013, with a standard deviation of 80. From her practice tests, Amber knows that she will score between 550-600 on that portion of the test.

A tutoring company claims that it can increase her A tutoring company claims that it can increase herSAT verbal score by 100 points. If this is true, SAT verbal rank by 10 percentile. If this is true,what would be Amber's new percentile range? what would be Amber's new score range?

BONUS: The mean SAT math score was 515 in 2007. Derrick scored a 680, which landed him in the 88th percentile.

If James scored a 630 on that test, what was his If Kerri ranked at the 40th percentile, what was herpercentile? score?

Data Organization (Chapter 2) Date: DAY 3Standards/Objectives:1) understand the graphical meaning/interpretation of the mean, median, and mode in a density curve2) practice Empirical Rule and percentiles3) practice FRQs

Agenda/Activities:1A) compare/contrast mean, median, and mode for different density curves2A) Quiz 2.1 - LQ IQ Scores3A) Practice FRQ - Test Form 2A3B) AP Flashcards 63, 69, 75

1A) mean = Symmetric Skewed Left Skewed Right

median =

mode =

2A) Quiz 2.1

3A) Chapter 2 FRQ Section II: Free Response Form A1. Scores on the Wechsler Adult Intelligence Scale (a standard IQ test) are approximately normally distributed within age groups. For the 20-34 age group, the mean is 110 and the standard deviation is 25. For the 60-64 age group, the mean is 90 and the standard deviation is 25. Sarah is 29 and her mother is 62. Sarah scores 135 on the Wechsler test, while Ann, her mother, scores 120. Who has the better score relative to her age group?

2. Use Table A to find the proportions of observations from a standard normal distribution that satisfies each of the following statements. For both problems sketch a standard normal curve and shade the area under the curve that answers the question.A) z > –1.68 B) –0.84 < z < 1.26

3. The Graduate Record Examinations are widely used to help predict the performance of applicants to graduate schools. The range of possible scores on a GRE is 200 to 900. The psychology department at a university finds that the scores of its applicants on the quantitative GRE are approximately normal with mean 544 and standard deviation 103. What minimum score would a student need in order to score in the top 10% of those taking the test?

3B) AP Flashcards 63, 69, 75

Data Organization (Chapter 2) Date: DAY 4Standards/Objectives:1) practice test2) practice checks for normality3) understand the graphical effects means vs. standard deviations4) read and calculate using a density curve of uniform distribution

Agenda/Activities:1A) Practice Test - Test 2 MC Form A2A) ACTIVITY - record individual heights into calculator, check for normality 3A) Mean vs. Standard Deviation 4A) Uniform Distribution4B) Practice Test - Quiz 2.2C

1A) Practice Test - Test 2 MC Form A

2A) class heights:

Checks for Normality:Boxplot / Histogram Normal Probability Plot Empirical Rule1) symmetric about median/mean 1) linear 1) 68% of data within 12) no outliers or unusual features 2) 2/3 of data within 1 2) 95% of data within 2

3) 99.7 of data within 3

3A) Rank the order of the graphs from lowest standard deviation to highest standard deviation (note: the location of the mean is indicated by the hashmark):

4A) If LQ girls' heights are uniformly distributed between 60" and 66", find the proportion of girls between 64" and 65".