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 |
| |a. |
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| |b. |
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|Normal Curve |c. |
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| |d. |
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| |e. |
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| |a. |
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|Normal Distributions |b. |
|N(μ,() | |
| |c. |
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| |d. |
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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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