1 / 34

Chapter 6

Chapter 6. The Normal Distribution. Normal Distributions. Bell Curve Area under entire curve = 1 or 100% Mean = Median This means the curve is symmetric. Normal Distributions. Two parameters Mean μ (pronounced “ meeoo ”) Locates center of curve Splits curve in half

ursala
Download Presentation

Chapter 6

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. Chapter 6 The Normal Distribution

  2. Normal Distributions • Bell Curve • Area under entire curve = 1 or 100% • Mean = Median • This means the curve is symmetric

  3. Normal Distributions • Two parameters • Mean μ (pronounced “meeoo”) • Locates center of curve • Splits curve in half • Shifts curve along x-axis • Standard deviation σ (pronounced “sigma”) • Controls spread of curve • Smaller σ makes graph tall and skinny • Larger σ makes graph flat and wide • Ruler of distribution • Write as N(μ,σ)

  4. Standard Normal (Z) World • Perfectly symmetric • Centered at zero • Half numbers below the mean and half above. • Total area under the curve is 1. • Can fill in as percentages across the curve.

  5. Standard Normal Distribution • Puts all normal distributions on same scale • z has center (mean) at 0 • z has spread (standard deviation) of 1

  6. Standard Normal Distribution • z = # of standard deviations away from mean μ • Negative z, number is below the mean • Positive z, number is above the mean • Written as N(0,1)

  7. Portal from X-world to Z-world • z has no units (just a number) • Puts variables on same scale • Center (mean) at 0 • Spread (standard deviation) of 1 • Does not change shape of distribution

  8. Standardizing Variables • z = # of standard deviations away from mean • Negative z – number is below mean • Positive z – number is above mean

  9. Standardizing • Y ~ N(70,3). Standardize y = 68. • y = 68 is 0.67 standard deviations below the mean

  10. Your Table is Your Friend • Get out your book and find your Z-table. • Look for a legend at the top of the table. • Which way does it fill from? • Find the Z values. • Find the “middle” of the table. • These are the areas or probabilities as you move across the table. • Notice they are 50% in the middle and 100% at the end.

  11. Areas under curve • Another way to find probabilities when values are not exactly 1, 2, or 3  away from µ is by using the Normal Values Table • Gives amount of curve below a particular value of z • z values range from –3.99 to 3.99 • Row – ones and tenths place for z • Column – hundredths place for z

  12. Finding Values • What percent of a standard Normal curve is found in the region Z < -1.50? • P(Z < –1.50) • Find row –1.5 • Find column .00 • Value = 0.0668

  13. Finding Values • P(Z < 1.98) • Find row 1.9 • Find column .08 • Value = 0.9761

  14. Finding values • What percent of a std. Normal curve is found in the region Z >-1.65? • P(Z > -1.65) • Find row –1.6 • Find column .05 • Value from table = 0.0495 • P(Z > -1.65) = 0.9505

  15. Finding values • P(Z > 0.73) • Find row 0.7 • Find column .03 • Value from table = 0.7673 • P(Z > 0.73) = 0.2327

  16. Finding values • What percent of a std. Normal curve is found in the region 0.5 < Z < 1.4? • P(0.5 < Z < 1.4) • Table value 1.4 = 0.9192 • Table value 0.5 = 0.6915 • P(0.5 < Z < 1.4) = 0.9192 – 0.6915 = 0.2277

  17. Finding values • P(–2.3 < Z < –0.05) • Table value –0.05 = 0.4801 • Table value –2.3 = 0.0107 • P(–2.3 < Z < –0.05) = 0.4801 – 0.0107 = 0.4694

  18. Finding values • Above what z-value do the top 15% of all z-value lie, i.e. what value of z cuts offs the highest 15%? • P(Z > ?) = 0.15 • P(Z < ?) = 0.85 • z = 1.04

  19. Finding values • Between what two z-values do the middle 80% of the obs lie, i.e. what values cut off the middle 80%? • Find P(Z < ?) = 0.10 • Find P(Z < ?) = 0.90 • Must look inside the table • P(Z<-1.28) = 0.10 • P(Z<1.28) = 0.90

  20. Solving Problems • The height of men is known to be normally distributed with mean 70 and standard deviation 3. • Y ~ N(70,3)

  21. Solving Problems • What percent of men are shorter than 66 inches? • P(Y < 66) = P(Z< ) = P(Z<-1.33) = 0.0918

  22. Solving Problems • What percent of men are taller than 74 inches? • P(Y > 74) = 1-P(Y<74) = 1 – P(Z< ) = 1 – P(Z<1.33) = 1 – 0.9082 = 0.0918

  23. Solving Problems • What percent of men are between 68 and 71 inches tall? • P(68 < Y < 71)= P(Y<71) – P(Y<68)=P(Z< )-P(Z< )=P(Z<0.33) - P(Z<-0.67)= 0.6293 – 0.2514= 0.3779

  24. Solving Problems • Scores on SAT verbal are known to be normally distributed with mean 500 and standard deviation 100. • X ~ N(500,100)

  25. Solving Problems • Your score was 650 on the SAT verbal test. What percentage of people scored better? • P(X > 650) = 1 – P(X<650) = 1 – P(Z< ) = 1 – P(Z<1.5) = 1 – 0.9332 = 0.0668

  26. Solving Problems • To solve a problem where you are looking for y-values, you need to rearrange the standardizing formula:

  27. Solving Problems • What would you have to score to be in the top 5% of people taking the SAT verbal? • P(X > ?) = 0.05? • P(X < ?) = 0.95?

  28. Solving Problems • P(Z < ?) = 0.95? • z = 1.645 • x is 1.645 standard deviations above mean • x is 1.645(100) = 164.5 points above mean • x = 500 + 164.5 = 664.5 • SAT verbal score: at least 670

  29. Solving Problems • Between what two scores would the middle 50% of people taking the SAT verbal be? • P(x1= –? < X < x2=?) = 0.50? • P(-0.67 < Z < 0.67) = 0.50 • x1 = (-0.67)(100)+500 = 433 • x2 = (0.67)(100)+500 = 567

  30. Solving Problems • Cereal boxes are labeled 16 oz. The boxes are filled by a machine. The amount the machine fills is normally distributed with mean 16.3 oz and standard deviation 0.2 oz.

  31. Solving Problems • What is the probability a box of cereal is underfilled? • Underfilling means having less than 16 oz. • P(Y < 16) = P(Z< ) = P(Z< -1.5) = 0.0668

  32. Moving to Random Sample • Even when we assume a variable is normally distributed if we take a random sample we need to adjust our formula slightly.

  33. Determining Normality • If you need to determine normality I want you to use your calculators to make a box plot and to look for symmetry.

More Related