Kenwood Academy



Probability and Statistics

Chapter 3 – Modeling Distribution of Data

|3.1 Measuring Location in a |Objective: |

|Distribution |MEASURE position using percentiles |

| |INTERPRET cumulative relative frequency graphs |

| |MEASURE position using z-scores |

| |TRANSFORM data |

| |DEFINE and DESCRIBE density curves |

|Measuring Position: Percentiles | |

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|Examples: | |

| |The pth percentile of a distribution is the value with p percent of the observations less than it. |

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| |Here are the scores of all 25 students in Mr. Pryor’s statistics class on their first test: |

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| |79 81 80 77 73 83 74 93 78 80 75 67 73 |

| |77 83 86 90 79 85 83 89 84 82 77 72 |

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| |Problem: Use the scores on Mr. Pryor’s test to find the percentiles for the for the following students (how did they perform |

| |relative to their classmates): |

| |a) Jenny, who earned an 86. b) Norman, who earned a 72. |

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|Example | |

| |c) Katie, who earned a 93. d) the two students who earned scores of 80. |

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| |PSAT scores |

| |In October 2007, about 1.4 million college-bound high school juniors took the PSAT. The mean score on the Critical Reading test |

| |was 46.7 and the standard deviation was 11.3. Nationally, 6 percent of test-takers earned a score higher than 65 on the Critical|

| |Reading test’s 20 to 80 scale.3tps3ech2 |

| |Scott was one of 50 junior boys to take the PSAT at his school. He scored 65 on the Critical Reading test. This placed Scott at|

| |the 68th percentile within the group of boys. Looking at all 50 boys’ Critical Reading scores, the mean was 58.2 and the |

|Measuring Position: z-Scores |standard deviation was 9.4. |

| |Write a sentence or two comparing Scott’s percentile among the national group of test takers and among the 50 boys at his school.|

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| |A z-score tells us how many standard deviations from the mean an observation falls, and in what direction. |

|Example: | |

| |To compare data from distributions with different means and standard deviations, we need to find a common scale. We accomplish |

| |this by using standard deviation units (z-scores) as our scale. Changing to these units is called standardizing. Standardizing |

| |data shifts the data by subtracting the mean and rescales the values by dividing by their standard deviation. |

| |[pic] or [pic][pic] |

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| |Standardizing does not change the shape of the distribution. It changes the center (shifts it to zero) and the spread by making |

| |the standard deviation one. |

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|Example: | |

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| |PSAT scores (continued) |

| |Refer to the previous example. Calculate and compare Scott’s z-score among these same two groups of test takers. |

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|Assignment 3.1 | |

|Part 1, page 105 #3.1-3.6 | |

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|Transforming Data | |

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|Effect of Adding (or Subtracting) a| |

|Constant | |

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|Effect of Multiplying (or Dividing)| |

|by a Constant | |

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|Example: | |

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| |Transforming converts the original observations from the original units of measurements to another scale. Transformations can |

| |affect the shape, center, and spread of a distribution. |

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|Remember: Exploring Quantitative |Adding the same number a (either positive, zero, or negative) to each observation: |

|Data |adds a to measures of center and location (mean, median, quartiles, percentiles), but |

| |Does not change the shape of the distribution or measures of spread (range, IQR, standard deviation). |

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| |Multiplying (or dividing) each observation by the same number b (positive, negative, or zero): |

| |multiplies (divides) measures of center and location by b |

| |multiplies (divides) measures of spread by |b|, but |

| |does not change the shape of the distribution |

|Density Curves | |

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|Median and Mean of a Density Curve |To describe a distribution: |

| |Make a graph |

| |Look for overall patterns (shape, center, and spread) and outliers |

| |Calculate a numerical summary to describe the center (mean, median) and spread (minimum, maximum, Q1, Q3, range, IQR, standard |

| |deviation) |

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| |In addition to the above distributions sometimes the overall pattern of a large number of observations is so regular that we can |

| |describe it by a smooth curve. |

|Examples: | |

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| |A density curve describes the |

| |overall pattern of a distribution |

| |Is always on or above the |

| |horizontal axis |

| |Has exactly 1 underneath it |

| |The area under the curve and |

| |above any range of values is the |

| |proportion of all observations |

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| |The overall pattern of this histogram of the scores of all 947 seventh-grade students in Gary, Indiana, on the vocabulary part of|

| |the Iowa Test of Basic Skills (ITBS) can be described by a smooth curve drawn through the tops of the bars. |

|Examples: | |

| |Median of a density curve is the equal areas point, the point that divides the are under the curve in half |

| |Mean of a density curve is the balance point, at which the curve would balance if made of solid material. |

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| |Use the figure shown to answer the following questions. |

| |1. Explain why this is a legitimate density curve. |

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|Assignment 3.1 | |

|Part 2, page 110 #3.7-3.12 | |

| |2. About what proportion of observations lie |

| |between 7 and 8? |

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|Assignement 3.1 |3. Mark the approximate location of the median. |

|Part 3, page 113 #3.13-3.20 |4. Mark the approximate location of the mean. |

| |Explain why the mean and median have the |

| |relationship that they do in this case. |

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|3.2 Normal Distributions | |

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|Normal Distributions | |

|N(μ, σ) | |

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|The 68-95-99.7 Rule | |

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|Application of the | |

|68-95-99.7 Rule | |

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| |Objectives: |

| |DESCRIBE and APPLY the 68-95-99.7 Rule |

| |DESCRIBE the standard Normal Distribution |

| |PERFORM Normal distribution calculations |

| |ASSESS Normality |

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|Example | |

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| |All Normal curves have the same overall shape: symmetric, single-peaked, bell shaped. |

| |A Normal distribution is described by a Normal density curve. |

| |A Normal distribution can be fully described by two parameters, its mean μ and standard deviation σ |

| |The mean, µ, of a Normal distribution is at the center of the symmetric Normal curve and is the same as the median. |

| |The standard deviation σ controls the spread of a Normal curve. Curves with larger standard deviations are more spread out. |

| |The standard deviation, σ, is the distance from the center to the change-of-curvature points on either side. |

| |A short-cut notation for the normal distribution in N(μ,σ). |

|Assignment 3.2 | |

|Part 1, page 121 #3.21-3.26 | |

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|The Standard Normal Distribution |All normal curves obey the 68-95-99.7% (Empirical) Rule. |

| |This rule tells us that in a normal distribution approximately |

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| |68% of the data values fall within one standard |

| |deviation (1σ) of the mean, |

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| |95% of the values fall within 2σ of the mean, and |

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| |99.7% (almost all) of the values fall |

|Z-Score Table |within 3σ of the mean. |

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|Example |Distribution of the heights of young women aged 18 to 24 |

| |What is the mean μ? |

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| |What is the (? |

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| |What is the height range for 95% of young women? |

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| |What is the percentile for 64.5 in.? |

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|Example |What is the percentile for 59.5 in.? |

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| |What is the percentile for 67 in.? |

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|Assignment 3.2 |What is the percentile for 72 in.? |

|Part 2, page 127 #3.27-3.32 | |

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|4-Step Process | |

| |SAT performance |

| |Students’ scores on the SAT Critical Reading test follow a Normal distribution with mean 500 and standard deviation 100. What |

| |percent of students earn scores above 700? |

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|Normal calculations | |

|Example: |The standard Normal distribution is the Normal distribution with mean 0 and standard deviation 1. |

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| |If a variable x has any Normal distribution N(µ,σ) with mean µ and standard deviation σ, then the standardized variable |

| |[pic][pic] has the standard Normal distribution, N(0,1). |

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| |Because all Normal distributions are the same when we standardize, we can find areas under any Normal curve from a single table. |

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| |Table A is a table of areas under the standard Normal curve. The table entry for each value z is the area under the curve to the|

| |left of z. |

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| |Table A practice |

| |Use Table A to find the proportion of observations from a standard Normal distribution that falls in each of the following |

| |regions. In each case, sketch a standard Normal curve and shade the area representing the region. |

| |(a) [pic] |

| |(b) [pic] |

| |(c) [pic] |

| |(d) [pic] |

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| |Finding z-scores from proportions |

| |Use Table A to find the value z of a standard Normal variable that satisfies each of the following conditions. In each case, |

| |sketch a standard Normal curve with your value of z marked on the axis. |

| |(a) The point z with 70% of the observations falling below it. |

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| |(b) The point z with 85% of the observations falling above it. |

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| |(c) Find the number z such that the proportion of observations less than z is 0.8. |

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| |(d) Find the number z such that 90% of all observations are greater than z. |

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| |How to Solve Problems Involving Normal Distributions |

| |Step 1: |

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| |Step 3: |

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| |Step 4: |

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|Using Table A in Reverse | |

|Example: | |

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| |Women’s heights are approximately normal with N(64.5, 2.5). |

| |What proportion of all young women are less than 68 inches tall? |

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|Assignment 3.2 Part 3 page 132 | |

|#3.33-3.38 | |

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| |On the driving range, Tiger Woods practices his swing with a particular club by hitting many, many balls. When tiger hits his |

| |driver, the distance the balls travels follows a Normal distribution with mean 304 yards and standard deviation 8 yards. What |

| |percent of Tiger’s drives travel at least 290 yards? |

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| |What percent of Tiger’s drives travel between 305 and 325? |

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| |What distance would a ball have to travel to be at the 80th percentile f Tiger’s drive lengths? |

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|Summary |

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