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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

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Chapter 6

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  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.

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