1 / 16

Topic 3

Topic 3. The Normal Distribution. From Histogram to Density Curve. We used histogram in Topic 2 to describe the overall pattern (shape, center, and spread) of the distribution of a quantitative variable.

diane
Download Presentation

Topic 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Topic 3 The Normal Distribution

  2. From Histogram to Density Curve We used histogram in Topic 2 to describe the overall pattern (shape, center, and spread) of the distribution of a quantitative variable. A better way of describing the overall pattern is to use the smooth curve drawn through the tops of the bars in a histogram. In order to make the area under any part of the curve have an easy interpretation, we can adjust the vertical scale of the histogram so that the total area of the bars in a histogram is exactly 1. The smooth curve corresponding to such a scaled histogram is called a density curve.

  3. Estimating the Proportion (or Percent) of All Values That are within an Interval Using a Density Curve The proportion is equal to the area under the density curve and above the interval. This result is one of the most frequently used results in statistics.

  4. Normal Curves and Normal Distributions We will focus our attention on a special class of density curves. These density curves are symmetric, single-peaked, and bell-shaped. They are called normal curves and they describe normal distributions. The exact density curve for a particular normal distribution is described by its mean (denoted µ) and its standard deviation (denoted σ). We will denote a normal distribution with mean µ and standard deviation σby N(µ, σ).

  5. All Normal Distributions Obey The Famous 68-95-99.7Rule (Roughly) 68% of all data is within one standard deviation of the mean, .(I.e. - 68% of the datalies between - and + ) 68% ofData 16% ofData 16% ofData  -    + 

  6. 68-95-99 Rule (Cont’d) (Roughly) 68% of all data is within one standard deviation of the mean, . 95% of data is within two standard deviations of the mean.(I.e. - between 95% ofData 2.5% ofData 2.5% ofData   - 2  + 2

  7. 68-95-99 Rule (Cont’d) 68% of all data is within one standard deviation of the mean, . 95% of data is within two standard deviations of the mean. 99.7% of data is within threestandard deviations of the mean. 99.7% ofData 0.15% ofData 0.15% ofData   - 3  + 3

  8. The Standard Normal Distribution The standard normal distribution is the normal distribution with mean µ = 0 and standard deviation σ = 1. If x is an observation from a normal distribution with mean μ and standard deviation σ, then z = (x – μ)/σ called the standardized value or z-score of x, is an observation from the standard normal distribution. The z-score contains more information than its x value. The z-score of an observation x tells us how many standard deviations x falls away from the mean, and in which direction. Here we see that the standard deviation is used as a ruler.

  9. Examples of z-scores The distribution of heights of women aged 20 to 29 is approximately normally distributed with mean 64 inches and standard deviation 2.7 inches. Find the z-score of a woman (a) 72 inches tall. (b) 62 inches tall. (c) 64 inches tall. The z-score of a woman is – 1.4. Find her height. Find the z-scores of the two heights that mark the middle 68% of young women.

  10. Finding Proportions of Observations in a Normal Distribution That Fall in an Interval We already know that, the area under a density curve of a distribution and is above an interval on the horizontal axis, corresponds to the proportion or percent of observations in the distribution. The interval can be one of the three forms: Z ≤ b, a ≤ Z ≤ b, and Z ≥ b. Pictures that illustrate the above situations:

  11. Using the Standard Normal Table to Find Proportions • http://www.math.unb.ca/~knight/utility/NormTble.htm • Suppose that Z is a variable that has the standard normal distribution. From the standard normal distribution table, we can find proportions such as

  12. Answer

  13. Find Proportions for Any Normal Distribution Suppose that X is normally distributed with mean µ and standard deviation σ. We wish to find Pr(X < b), Pr(X > a), or Pr(a < X < b). To make use of the standard normal distribution, we first convert X to Z, the z-score of X, by the formula Z = (X - µ)/σ. Then we have Pr(X < b) = Pr(Z < b*) Pr(X > a) = Pr(Z > a*) Pr(a < X < b) = Pr(a* < Z < b*), where Z has the standard normal distribution, and a* and b* are the corresponding z-scores of a and b; that is, a* = (a - µ)/σand b* = (b - µ)/σ.

  14. Using the z-score Formula When Calculating Normal Proportions If X ~ N(2, 1.3), then (1) Pr(1.34 < X < 2.78) = Pr(- 0.5077 < Z < 0.6 ) = 0.42. (2) Pr(X < 1.52) = Pr(Z < - 0.3692) = 0.356. (3) Pr(X > 2.5) = Pr(Z > 0.3846) = 0.35.

  15. Finding Cutoff Values for Given Proportions Let X denote the score on the SAT verbal test. X ~ N(504, 111). • Draw a normal curve on which this mean and standard deviation are correctly located. • How high must a student score in order to place in the top 25% of all students taking the SAT? • A student has a score that places in the middle 50% of all students taking the SAT. Find the interval that contains the student’s score.

More Related