Kenwood Academy



AP Statistics

Chapter 2 - The Normal Distribution

| Density Curves and the Normal |Objective: |

|Distribution |1) Know that the area under a density curve represent proportions of all observations and that the total area under a density |

| |curve is 1. |

| |2) Approximately locate the median (equal-area point) and mean (balance point) on a density curve. |

| |3) Know that the mean and median both lie at the center of a symmetric density curve and that the mean moves farther toward the |

| |long tail of a skewed curve. |

| |4) Estimate μ and ( from a normal curve |

|What You should Know From |5) Use the 68-95-99.7 rule and symmetry to state what percent of the observations from a normal distribution fall between two |

|Chapter 1: |points |

| |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) |

|Percentiles | |

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

|Examples: |describe it by a smooth curve. |

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| |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 |

|Density Curves |relative to their classmates): |

| |a) Jenny, who earned an 89. b) Norman, who earned a 72. |

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

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|Example 2.1 page 79 |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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|Median and Mean of a Density Curve | |

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| |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. See figures 2.5 and |

|Example: |2.6 page 81-82. |

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| |Density Curve Computed from actual observations |

| |Mean (μ) Mean ([pic]) |

| |S.D. (() S.D. (s) |

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

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

|Practice Problem |2. About what proportion of observations lie |

|page 83 #2.2 |between 7 and 8? |

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

| |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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|Practice Problem page 84 #2.3 | |

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| |2.2 Uniform Density Curve |

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|Normal Curve |c. |

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| |a. |

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|Normal Distributions |b. |

|N(μ,() | |

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| |e. |

|The 68-95-99.7 Rule | |

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

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

| |The mean is located in the center of the symmetric 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. |

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|Application of the | |

|68-95-99.7 Rule |A Normal distribution is described by a Normal density curve. |

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

| |The standard deviation, σ, is the deviation 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,(). |

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| |All normal curves they 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 (1() of the mean, |

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

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

| |within 3( of the mean. |

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|Practice Problem page 89 #2.6 and | |

|2.7 | |

|( ( | |

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

| |What is the mean μ? |

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| |What is the s.d. (? |

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

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

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

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|Standard Normal Calculations | |

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|The Standard Normal Distribution | |

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|Z-Score | |

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|The standard Normal Table | |

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|Normal Distribution Calculations |Objective: |

| |6) Find and interpret the z score of an observation |

| |7) Use the Z table and calculator to calculate the proportion of values above, below, or between two stated numbers |

| |8) Calculate the point having a stated proportion of all values above or below it. |

| |9) Construct and interpret a Normal Probability Plot |

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| |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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| |The standard normal distribution is the normal distribution N(0,1) with mean 0 and standard deviation 1. |

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| |[pic] |

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|How to Solve Problems Involving | |

|Normal Distributions |A z-score tells us how many standard deviations the original observation falls away from the mean, and in which direction. |

| |Observations larger than the mean are positive when standardized, and observations smaller than the mean are negative. |

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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. |

|Normal calculations | |

|Example: | |

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| |For example, lets say that we know a girl named Georgia who is 60 inches tall and a girl named Deanna that is 68 inches tall. |

| |What are Georgia’s and Deanna’s standardized heights? Women’s heights are approximately normal with N(64.5, 2.5). |

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| |a. For example, what proportion of all young women are less than 68 inches tall (in other words what is the percentile for |

| |Deanna’s height)? |

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| |b. In what percentile does Georgia fall? |

|More complicated calculations | |

|Example: | |

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| |State: Express the problem in terms of the observed variable x. |

| |Plan: Draw a picture of the distribution and shade the area of interest under the curve. |

| |Do: Perform calculations. |

| |Standardize x to restate the problem in terms of a standard Normal variable x. |

| |Use Table A and the fact that the total area under the curve is 1 to find the required area under the standard Normal curve. |

| |Conclude: Write your conclusion in context of the problem |

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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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|Using Table A in Reverse | |

|Example: | |

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

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|Normal Probability Plots | |

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|Normal Distributions on the | |

|Calculator (See Technology Box page| |

|115-117) | |

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|Assignment 2.2 page 109 | |

|#2.28-2.33,2.35 | |

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| |High levels of cholesterol in the blood increase the risk of heart disease. For 14 year old boys, the distribution of blood |

| |cholesterol is approximately Normal with mean µ = 170 milligrams of cholesterol per deciliter of blood (mg/dl) and standard |

| |deviation σ = 30 mg/dl. What is the first quartile off the distribution of blood cholesterol? |

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| |Assessing Normality using the Calculator |

| |To decide if a set of data is normal we can construct a normal probability plot. |

| |See Technology Toolbox page 105-106 |

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| |Finding Areas with ShadeNorm |

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| |Finding areas with normalcdf |

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| |Finding z values with invNorm |

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|Summary |

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